tag:blogger.com,1999:blog-59971478958805851792024-03-17T18:54:14.161+05:30Solutions to I E Irodov - Physical Fundamentals of MechanicsNewton single handedly brought about a new era in our understanding of the physical world, through his famous equation F = ma and the law of gravity. These laws successfully predict the behavior of all phenomena from the microscopic world to the astronomical world! Ofcourse, Einstien changed things once again but that was only after about 4 centuries. Solving Irodov dramatically helped me understand the power, elegance, simplicity and beauty of Physics. Hopefully you will enjoy the journey too.Unknownnoreply@blogger.comBlogger387125tag:blogger.com,1999:blog-5997147895880585179.post-54169033125079178262010-01-14T08:01:00.015+05:302010-01-14T16:13:33.486+05:30Irodov Problem 1.388<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhP9c0wibQwG1WUrEVUCOsaeufDhOlpxbBSLOJyaEcFJOnf8_zPj-JGdwQYgt_0Vi2K3aX4XezGt5PKoqKDUU45fbo-ItoiPoD7eN_Vrt04cHXHqTQaJnc1w0kvSmr82yNwKoc0hyPYRqDT/s1600-h/1.388_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 288px; height: 387px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhP9c0wibQwG1WUrEVUCOsaeufDhOlpxbBSLOJyaEcFJOnf8_zPj-JGdwQYgt_0Vi2K3aX4XezGt5PKoqKDUU45fbo-ItoiPoD7eN_Vrt04cHXHqTQaJnc1w0kvSmr82yNwKoc0hyPYRqDT/s400/1.388_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5426544176060482114" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Suppose that a certain instant of time <span style="font-weight: bold;">t</span> the velocity of the rocket is <span style="font-weight: bold;">v</span> and its rest mass <span style="font-weight: bold;">m</span>. Suppose that the rocket ejects gas of rest mass <span style="font-weight: bold;">k</span> units/sec. Consider another instant of time <span style="font-weight: bold;">t</span> <span style="font-weight: bold;">+ dt</span>, where <span style="font-weight: bold;">dt</span> is an infinitesimally small interval of time. The rocket has ejected <span style="font-weight: bold;">kdt</span> amount of rest mass. Suppose that the rocket's velocity has increased by an amount <span style="font-weight: bold;">dv</span> during this interval. Since there is no external force on the rocket, the momentum of the (ejected gas + rocket) system at <span style="font-weight: bold;">t</span> must be equal to that at <span style="font-weight: bold;">t+dt</span>.<br /><br />The momentum of the rocket at time <span style="font-weight: bold;">t</span> is given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjt3kR99WTonjOM0G_mZ15Ozi1PO2BHK8xgUAHMWCmXJnmebrYzklQAx260wIq4bKzIL2W4_ZdmIbHko7DPw8b_mAS8BdxvAQYcjdGJNu5ABYOBITUY-vUAkBtFR_ZJO4IvSN61Tzzx0AUe/s1600-h/1.388_1.jpg"><img style="cursor: pointer; width: 77px; height: 44px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjt3kR99WTonjOM0G_mZ15Ozi1PO2BHK8xgUAHMWCmXJnmebrYzklQAx260wIq4bKzIL2W4_ZdmIbHko7DPw8b_mAS8BdxvAQYcjdGJNu5ABYOBITUY-vUAkBtFR_ZJO4IvSN61Tzzx0AUe/s400/1.388_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5426527242633672002" border="0" /></a>. The momentum of the rocket at <span style="font-weight: bold;">t+dt</span> is given as <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgc16wAFJ1yOzYcFa404JNMtvJZIf87S0czkyIUELjuXEmIInchL8ORdaMpVpXahi5S25-p09hBLgjXCeVXM0a6H2eDUN68j6qh4GNiLwsbebwhOYy4EqLZi-dIkTmYhkzXQR-BUa_W1UV-/s1600-h/1.388_1a.jpg"><img style="cursor: pointer; width: 107px; height: 62px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgc16wAFJ1yOzYcFa404JNMtvJZIf87S0czkyIUELjuXEmIInchL8ORdaMpVpXahi5S25-p09hBLgjXCeVXM0a6H2eDUN68j6qh4GNiLwsbebwhOYy4EqLZi-dIkTmYhkzXQR-BUa_W1UV-/s400/1.388_1a.jpg" alt="" id="BLOGGER_PHOTO_ID_5426527849775180946" border="0" /></a>. The gas ejected by the rocket has a of mass <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrfRXUP0QrCYbeDpuLaa7RFgEwx3dT44snKjkkqaITIEKwpr4zBtBJGC_V2M9tEeIFJooFCWecPBoCcTPMdpJn1-oaQieGTbxny0wMoCT_L4WqDMTh3TZi0Prd75ZJj1BMl-XQ-sTv-Za1/s1600-h/1.388_1b.jpg"><img style="cursor: pointer; width: 79px; height: 49px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrfRXUP0QrCYbeDpuLaa7RFgEwx3dT44snKjkkqaITIEKwpr4zBtBJGC_V2M9tEeIFJooFCWecPBoCcTPMdpJn1-oaQieGTbxny0wMoCT_L4WqDMTh3TZi0Prd75ZJj1BMl-XQ-sTv-Za1/s400/1.388_1b.jpg" alt="" id="BLOGGER_PHOTO_ID_5426534849579585890" border="0" /></a> and moves at a non-relativistic velocity <span style="font-weight: bold;">u</span> relative to the rocket in the opposite direction to the rocket. In other words, its velocity as seen by an observer on Earth will be <span style="font-weight: bold;">v-u</span> in the direction of the rocket (this is shown in the diagram). Hence, the momentum of the ejected gas will be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWqtrsx0WPrSfPQWd_Kzp_DLPcJ6LvEWShpGW6YnGoBeXHeMJPVzSS3zsfJC-573q7b7qTs4m-2TWnTl2mMhoNKJa0Pq8JILlRbVocNH_zcllFnvGMrPIw22lCkVhcz9VZWMCh6VtCJOJt/s1600-h/1.388_1c.jpg"><img style="cursor: pointer; width: 79px; height: 49px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWqtrsx0WPrSfPQWd_Kzp_DLPcJ6LvEWShpGW6YnGoBeXHeMJPVzSS3zsfJC-573q7b7qTs4m-2TWnTl2mMhoNKJa0Pq8JILlRbVocNH_zcllFnvGMrPIw22lCkVhcz9VZWMCh6VtCJOJt/s400/1.388_1c.jpg" alt="" id="BLOGGER_PHOTO_ID_5426535423176507490" border="0" /></a>.<br /><br />Applying conservation of momentum at <span style="font-weight: bold;">t</span> and <span style="font-weight: bold;">t+dt</span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhE4yrn8OijvrpPwdUNlO5Wgy7cz-K-Rc18SHQMkUFdDt96RTJtIWsfyB5ay9d5HTp5wtLV8sUG1VQSJzZ9njWuNJsVAid7KdCbIc5XQ8O0LeOjDSlqaS3IosYWJkFBrNqCD3IB20pYhCf1/s1600-h/1.388_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 360px; height: 74px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhE4yrn8OijvrpPwdUNlO5Wgy7cz-K-Rc18SHQMkUFdDt96RTJtIWsfyB5ay9d5HTp5wtLV8sUG1VQSJzZ9njWuNJsVAid7KdCbIc5XQ8O0LeOjDSlqaS3IosYWJkFBrNqCD3IB20pYhCf1/s400/1.388_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5426536417756188690" border="0" /></a><br /><br /><br /><br /><br />We can now use <a href="http://en.wikipedia.org/wiki/Taylor_series">Taylor series</a> expansion and retain only the first order terms in Eqn <span style="font-weight: bold;">(1)</span> to obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOhW37M7M38_gzz7CNJ0xDWmOLhe0-lE8zpqD2s_HwP-aY5qKJUo_B3rJpzM2Hyo9dlKmNSRfNqjpjc-pTaydDH4mWiug4jwsPXNQw4RoTxLwWmQm3T8EBdDWTZpT_6XW6mgNlEbbhkw9H/s1600-h/1.388_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 257px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOhW37M7M38_gzz7CNJ0xDWmOLhe0-lE8zpqD2s_HwP-aY5qKJUo_B3rJpzM2Hyo9dlKmNSRfNqjpjc-pTaydDH4mWiug4jwsPXNQw4RoTxLwWmQm3T8EBdDWTZpT_6XW6mgNlEbbhkw9H/s400/1.388_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5426537826597389922" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now we know that the mass of the rocket reduces at a rate <span style="font-weight: bold;">k</span>, in other words <span style="font-weight: bold;">-kdt = dm</span>. Hence, <span style="font-weight: bold;">(2)</span> can be rewritten as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgiEF5AgUkBih4kLVU7zeEz04vDp3lh79mF635yGbu9_rOY1ZLrrOhU3b8_z2hs8vcpT_VtKnFLzrD3g6OjeriFw8YrjJ8y8MjLYthtF24HmBIFFyvnL0xEUWN3Vmo95dn3Ea4fHF6KRpFU/s1600-h/1.388_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 240px; height: 374px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgiEF5AgUkBih4kLVU7zeEz04vDp3lh79mF635yGbu9_rOY1ZLrrOhU3b8_z2hs8vcpT_VtKnFLzrD3g6OjeriFw8YrjJ8y8MjLYthtF24HmBIFFyvnL0xEUWN3Vmo95dn3Ea4fHF6KRpFU/s400/1.388_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5426543494735307202" border="0" /></a>Unknownnoreply@blogger.com5tag:blogger.com,1999:blog-5997147895880585179.post-27777265882911829662010-01-13T13:48:00.009+05:302010-01-13T13:58:28.618+05:30Irodov Problem 1.387<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBPlBdIT0MHubfsBf2XoytC1rf7dzj7O7m1BDmWf11fsZSU7ikEuQdHYDFOhDyYbuqVepj_K1H44-EH-ewc7OgTK3R6ACVSGgwNl5IKBYM8jcc23rsO4CiERpbvhtfVay598uTshxa3Ehn/s1600-h/1.387_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 257px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBPlBdIT0MHubfsBf2XoytC1rf7dzj7O7m1BDmWf11fsZSU7ikEuQdHYDFOhDyYbuqVepj_K1H44-EH-ewc7OgTK3R6ACVSGgwNl5IKBYM8jcc23rsO4CiERpbvhtfVay598uTshxa3Ehn/s400/1.387_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5426136002348993522" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Since the mass spontaneously breaks up into three pieces without any external forces both linear mommentum and energy must be conserved. In general, the three<br />pieces could move along three different directions at velocities <span style="font-weight: bold;">v<span style="font-size:78%;">1</span></span>,<span style="font-weight: bold;">v<span style="font-size:78%;">2</span></span> and <span style="font-weight: bold;">v<span style="font-size:78%;">3</span></span><br />as shown in the figure 1. In that case we have that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKSV7a1nznEDa71DJO0dDbPbwhUvFuy-4RLTHyFJZx0ygXOwqdoyrJNSiBDdpxOnn-0txbVrAi7KREHoQ21X4TeNMMz9BD34BzqqIaf96zuQqXltg1g6FciWag2jjuU0RKLG3r1AKzEoHZ/s1600-h/1.387_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 357px; height: 214px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKSV7a1nznEDa71DJO0dDbPbwhUvFuy-4RLTHyFJZx0ygXOwqdoyrJNSiBDdpxOnn-0txbVrAi7KREHoQ21X4TeNMMz9BD34BzqqIaf96zuQqXltg1g6FciWag2jjuU0RKLG3r1AKzEoHZ/s400/1.387_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5426136318032473458" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />From Eqn <span style="font-weight: bold;">(1)</span> since the sum of a linear combination of the three velocity vectors is zero (they are linearly dependent) they must be co-planar. This means that without a loss of generality we can restrict the velocity vectors to lie on a plane and choose the direction of velocity <span style="font-weight: bold;">v<span style="font-size:78%;">1</span></span> to lie along the negative <span style="font-weight: bold;">x-axis</span>. This is depicted in figure 2.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcGcDaeal3vCP9QNVnTCcdSXd8U8UN-lVdqDgnpGg65IDgw0jAocyxL-b9T9njaT9uWqc-peveJzSSyI2AaJVWsOaPXS1_C69MnB77y6x2qig6JyRK9TNDbITr72p33rQDiktbsxpShR98/s1600-h/1.387_fig3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 347px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcGcDaeal3vCP9QNVnTCcdSXd8U8UN-lVdqDgnpGg65IDgw0jAocyxL-b9T9njaT9uWqc-peveJzSSyI2AaJVWsOaPXS1_C69MnB77y6x2qig6JyRK9TNDbITr72p33rQDiktbsxpShR98/s400/1.387_fig3.jpg" alt="" id="BLOGGER_PHOTO_ID_5426136599634824386" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now we can simplify <span style="font-weight: bold;">(1)</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2w4F6SeirqjjVFOc-SSkiHyyhRuo7HFU8TlVEwyQLwfOkaxIB0zsBndTfVpJwfCKUDwRLVxeyYUuD-Kxg5NIqM9Iy_tR8qwOeSjRznlMgXGU_fpCYv-0Eo7TH9OF7onsWBEdrCTk-QyhT/s1600-h/1.387_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 334px; height: 144px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2w4F6SeirqjjVFOc-SSkiHyyhRuo7HFU8TlVEwyQLwfOkaxIB0zsBndTfVpJwfCKUDwRLVxeyYUuD-Kxg5NIqM9Iy_tR8qwOeSjRznlMgXGU_fpCYv-0Eo7TH9OF7onsWBEdrCTk-QyhT/s400/1.387_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5426136749897821778" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br />What choice of velocities and directions will maximize the energy if particle 1? To determine this answer we can use <a href="http://en.wikipedia.org/wiki/Lagrange_multipliers">Lagrangian Multipliers</a> to minimize the objective function,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9cQE-XNn0RYBd1rvfQl6VJtf-WFVHQL2rIiIkNFL19AG5BbUEDFYgLFHfP2RnPNpuHJ-dTV8TULmrtcKVYkJ8MXdqakQtZgS_xCghxcnS6jNofZfso8oYDildfRRGvlRE2czNcWP7m8au/s1600-h/1.387_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 201px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9cQE-XNn0RYBd1rvfQl6VJtf-WFVHQL2rIiIkNFL19AG5BbUEDFYgLFHfP2RnPNpuHJ-dTV8TULmrtcKVYkJ8MXdqakQtZgS_xCghxcnS6jNofZfso8oYDildfRRGvlRE2czNcWP7m8au/s400/1.387_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5426137084618859554" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />The conditions for local extrema are then obtain by setting the partial derivatives of <span style="font-weight: bold;">J</span> with respect to each of the unknowns to zero. Hence at extrema we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-CjSWYKV2CJazgrxe1pODx-QWsgEWG_thUEdJ2anKJVjbDnttT9KrPAWhUpTZINAv0ZfxkyPoOC22Y6XhkSEo77qbgdNUK61OZ8gbrN0xQCVhSKXFH7Glwy7ZZsRHCeXKC8OjDol8PaGA/s1600-h/1.387_5.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 263px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-CjSWYKV2CJazgrxe1pODx-QWsgEWG_thUEdJ2anKJVjbDnttT9KrPAWhUpTZINAv0ZfxkyPoOC22Y6XhkSEo77qbgdNUK61OZ8gbrN0xQCVhSKXFH7Glwy7ZZsRHCeXKC8OjDol8PaGA/s400/1.387_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5426137251146601346" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Eqn <span style="font-weight: bold;">5</span> essentially says that in order for the energy of piece <span style="font-weight: bold;">1</span> to be maximum, both the the pieces <span style="font-weight: bold;">2</span> and <span style="font-weight: bold;">3</span> must fly off in the same direction and propel piece <span style="font-weight: bold;">3</span> in the opposite direction.<br /><br />Now differentiating with respect to <span style="font-weight: bold;">v<span style="font-size:78%;">2</span></span> and <span style="font-weight: bold;">v<span style="font-size:78%;">3</span></span> we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhu3zG8U491df5Hf5X4gL3gNbxNFk2cZjymYXCwwAdEwKM79PJtZY99ksdcu-k3Ekmhihzkd04cY1CnKyT4M1ReHMerWIuYj8Wsx6B_8W8m98Y7W75u6Yevuu_seveEWjl6LL60nBVjn781/s1600-h/1.387_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 345px; height: 335px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhu3zG8U491df5Hf5X4gL3gNbxNFk2cZjymYXCwwAdEwKM79PJtZY99ksdcu-k3Ekmhihzkd04cY1CnKyT4M1ReHMerWIuYj8Wsx6B_8W8m98Y7W75u6Yevuu_seveEWjl6LL60nBVjn781/s400/1.387_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5426137580828160482" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Eqn <span style="font-weight: bold;">5</span> and <span style="font-weight: bold;">7</span> together essentially say that the energy of piece <span style="font-weight: bold;">1</span> will be maximized when pieces <span style="font-weight: bold;">2</span> and <span style="font-weight: bold;">3</span> move at the same velocity and the same direction - in other words they move together. This situation is depcited in Figure <span style="font-weight: bold;">3</span>. Let<span style="font-weight: bold;"> v<span style="font-size:78%;">2</span>=v<span style="font-size:78%;">3</span>=v</span>.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqstDxzRNIfP8ZglqxgzCTrY6jMxRSpCaTFiBrsYRmq2CuaSL81U3JxKFwyGaI7VuxQyEUpX0mFAYIEj-6r-6Kn0RcJt4moMi72ytLdVXsz8WjSUdfGlTJoqPCR8sZat65XOs_-nNV-tbi/s1600-h/1.387_fig2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 225px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqstDxzRNIfP8ZglqxgzCTrY6jMxRSpCaTFiBrsYRmq2CuaSL81U3JxKFwyGaI7VuxQyEUpX0mFAYIEj-6r-6Kn0RcJt4moMi72ytLdVXsz8WjSUdfGlTJoqPCR8sZat65XOs_-nNV-tbi/s400/1.387_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5426137869748593874" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Now equations <span style="font-weight: bold;">(1)</span> and <span style="font-weight: bold;">(2)</span> can be rewritten as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgeVk5_n4Del5d82y5ryW_98sZM_EGOD56xIth_gemoL-zDDcJb1PWi4i18aBG_21RlVn_6avcWHYQH79NExdXUadi7-OZLbwwDTa-3MObYLiJi8pZ1aC5Gsk3FfeSYOKDxFwGmbSKN1yQd/s1600-h/1.387_7.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 252px; height: 211px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgeVk5_n4Del5d82y5ryW_98sZM_EGOD56xIth_gemoL-zDDcJb1PWi4i18aBG_21RlVn_6avcWHYQH79NExdXUadi7-OZLbwwDTa-3MObYLiJi8pZ1aC5Gsk3FfeSYOKDxFwGmbSKN1yQd/s400/1.387_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5426138009012912546" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />From <span style="font-weight: bold;">(8)</span> we obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjtgmtfXHwT4MJXsvP0sFHPzntRBzgArawBxvGkkcYbtGj0RaP651Ba0M_Nz6isof71Jhk1gt_xKfDCDlXxIzvzZbab9uBM5dJQJBMI4no5-sNIA9SwHIOQW2ZuS5Hqj26LJOF_xoIWFwb3/s1600-h/1.387_8.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 342px; height: 235px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjtgmtfXHwT4MJXsvP0sFHPzntRBzgArawBxvGkkcYbtGj0RaP651Ba0M_Nz6isof71Jhk1gt_xKfDCDlXxIzvzZbab9uBM5dJQJBMI4no5-sNIA9SwHIOQW2ZuS5Hqj26LJOF_xoIWFwb3/s400/1.387_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5426138177547825170" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />From <span style="font-weight: bold;">(9)</span> we obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwQmzSqzIYmeBj_4dFjdB7skPN68hWBHjSeneyPNL3saTnYAF8Rm2bYx5TYlgWxS7243XQK9bQi_AzrKCzYcezN1gAun9OVGlQI-NtWnj-8Ab6h9l2-bsPQZvNCN36fN4TGEMOAbzre2k0/s1600-h/1.387_9.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 345px; height: 170px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwQmzSqzIYmeBj_4dFjdB7skPN68hWBHjSeneyPNL3saTnYAF8Rm2bYx5TYlgWxS7243XQK9bQi_AzrKCzYcezN1gAun9OVGlQI-NtWnj-8Ab6h9l2-bsPQZvNCN36fN4TGEMOAbzre2k0/s400/1.387_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5426138308234111794" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Eqn <span style="font-weight: bold;">(12)</span> given the maximum energy that piece <span style="font-weight: bold;">1</span> can possible have after the disintegration.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-19026298475471361982010-01-12T07:30:00.001+05:302010-01-12T07:32:47.403+05:30Irodov Problem 1.386Let the kinetic energy of the single particle be <span style="font-weight: bold;">T'</span>. In Problem <a href="http://irodovsolutionsmechanics.blogspot.com/2010_01_11_archive.html"><span style="font-weight: bold;">1.384</span></a> we have already determined the combined kinetic energy of the particles as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKpKzMoyFamIDnAr87ccxK-Krk0aRHoVlLe_KpematJV9hyGhVpRdmiGoJpZAprUJ11d8n-Wh2amomWSQhWblBo-7pJd56xQt8ge-KzMXy4vVRBN-35GNDZRqBSl8S6ENSzARDLQhazCZM/s1600-h/1.386_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 240px; height: 42px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKpKzMoyFamIDnAr87ccxK-Krk0aRHoVlLe_KpematJV9hyGhVpRdmiGoJpZAprUJ11d8n-Wh2amomWSQhWblBo-7pJd56xQt8ge-KzMXy4vVRBN-35GNDZRqBSl8S6ENSzARDLQhazCZM/s400/1.386_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5425667690341278466" border="0" /></a><br /><br /><br />Hence, we need to find <span style="font-weight: bold;">T'</span> such that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjejbS8pG3XGyfIDPj2tTH59C1ZcJCAZKC62pqvoD49bWsnW1Kp1LmsXbLUdfnpMY7MOmXqStxGy-2kRZiZVDn1D6z5Y7uObq83T3CJxXDInrv6vmEGWwFQnTJNki7P1QgpQRWP0srrRR28/s1600-h/1.386_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 268px; height: 99px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjejbS8pG3XGyfIDPj2tTH59C1ZcJCAZKC62pqvoD49bWsnW1Kp1LmsXbLUdfnpMY7MOmXqStxGy-2kRZiZVDn1D6z5Y7uObq83T3CJxXDInrv6vmEGWwFQnTJNki7P1QgpQRWP0srrRR28/s400/1.386_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5425667780967760850" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-44532586627651330842010-01-11T17:11:00.009+05:302010-01-11T17:53:26.858+05:30Irodov Problem 1.385<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjF-R7CRVM2KnzQp1OfEL7JICQiluh6tSVs7d9vA4OyNHjEn5YNixPeB3Q0eQ4C7nf1prk2wrXEIs5ueSiqJuV3-hJ5ua1ovaQnxcKORQtQ2aKQHr3q2SKbNtCP9pb1DkK2v0dsjMGx0afS/s1600-h/1.385_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 160px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjF-R7CRVM2KnzQp1OfEL7JICQiluh6tSVs7d9vA4OyNHjEn5YNixPeB3Q0eQ4C7nf1prk2wrXEIs5ueSiqJuV3-hJ5ua1ovaQnxcKORQtQ2aKQHr3q2SKbNtCP9pb1DkK2v0dsjMGx0afS/s400/1.385_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5425447289604108402" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br />This is a very interesting basic problem that demonstrates how different relativistic collisions are compared to Newtonian collisions - the essential difference being that mass can change in relativity.<br /><br />After the two masses collide and fuse into a new mass, its rest mass is not simply the sum of rest masses of the two constituent particles but something entirely different. In fact it depends on the energy of the collision!<br /><br />Throughout the collision two important properties will always hold, <span style="font-weight: bold;">i)</span> the momentum will be conserved and <span style="font-weight: bold;">ii)</span> the total energy will be conserved. Let the final rest mass of the fused particle be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzP_oVYQUv1qSPnE12CmZ2K2Ht1Cu2HCkI6W0MiEctLO2vy0NUIv48e2LGXO5qHwZW0h0O28McsJBX-X5PPIh7-X4mw1Aj7GNmQ97Wq9KlYD4u3AmVZ5s0WbicJseokD85SjsBQEO_6BeS/s1600-h/1.385_1.jpg"><img style="cursor: pointer; width: 23px; height: 19px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzP_oVYQUv1qSPnE12CmZ2K2Ht1Cu2HCkI6W0MiEctLO2vy0NUIv48e2LGXO5qHwZW0h0O28McsJBX-X5PPIh7-X4mw1Aj7GNmQ97Wq9KlYD4u3AmVZ5s0WbicJseokD85SjsBQEO_6BeS/s400/1.385_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5425448416027547506" border="0" /></a>. Let the initial velocity of the left particle be <span style="font-weight: bold;">u</span> and let <span style="font-weight: bold;">v</span> be the velocity of the combined particle after collision.<br /><br />Since we know that the kinetic energy of the particle before collision is <span style="font-weight: bold;">T</span> we have,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiy3Z8SEf0CR6XBLOmtlyiJ9OtHqbnvYfR93UUJa-5rBViiDDZOoud6tmbz-rKv2oTNqX7DgJ0bQQWHaztrWJwE3mxO3thlFHvvNBF26klmr3wq7jcAfyulmoRNzlkRXBBtkgVsdJflGhx4/s1600-h/1.385_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 186px; height: 125px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiy3Z8SEf0CR6XBLOmtlyiJ9OtHqbnvYfR93UUJa-5rBViiDDZOoud6tmbz-rKv2oTNqX7DgJ0bQQWHaztrWJwE3mxO3thlFHvvNBF26klmr3wq7jcAfyulmoRNzlkRXBBtkgVsdJflGhx4/s400/1.385_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5425452668375647362" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;">Conservation of Momentum</span><br />Since there are no external forces on the system, its total momentum must be conserved and so we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN_qZe3FwuyHewzjKfCPisEIaDCAj7OcuWg73-K-IN8xrXxhDGYY9ovZe3-vNzAEzkYXjjs87vHCwO92kGkkA6oas_5Q9RN-_IV1jMbqVhPpH37HAFSzucijPPLvk379x76TyGVqFrTVqL/s1600-h/1.385_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 253px; height: 146px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN_qZe3FwuyHewzjKfCPisEIaDCAj7OcuWg73-K-IN8xrXxhDGYY9ovZe3-vNzAEzkYXjjs87vHCwO92kGkkA6oas_5Q9RN-_IV1jMbqVhPpH37HAFSzucijPPLvk379x76TyGVqFrTVqL/s400/1.385_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5425453601992827266" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;"><br />Conservation of Energy</span><br />The total energy of the system before collision must be equal to that after collision. The total energy is the sum of the rest mass energies of each of the particles and the kinetic energy of the colliding particle.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1alpFcNkw3ckx6-xtSGpOFFWLA_aZWQPoW4reCqLdJ3qRc90nwXiHvQ-cIBsrPgUYftZ6y3Pq9T3DJvJJLJYBKfRXx3Bj0oubfAVLzQuH69ngI2EASP8sFAqJJ6NN0cY1QMbyibGUEvIW/s1600-h/1.385_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 209px; height: 55px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1alpFcNkw3ckx6-xtSGpOFFWLA_aZWQPoW4reCqLdJ3qRc90nwXiHvQ-cIBsrPgUYftZ6y3Pq9T3DJvJJLJYBKfRXx3Bj0oubfAVLzQuH69ngI2EASP8sFAqJJ6NN0cY1QMbyibGUEvIW/s400/1.385_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5425454520160156930" border="0" /></a><br /><br /><br /><br />From Eqn <span style="font-weight: bold;">(3)</span> and <span style="font-weight: bold;">(4)</span> we obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2lmfok1e9QlRcxD3-NxosrovrVduHwg1B2B8ODEkgOOwKaZslgFA9W2-SAQUEj8YMS2DsiumoofGU69gyCAd6iPPWyg2xewshcGeL8wzSZr4K5juWd5c_SSmCIqn4q5RhSjMXUU3J8i5H/s1600-h/1.385_5.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 183px; height: 53px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2lmfok1e9QlRcxD3-NxosrovrVduHwg1B2B8ODEkgOOwKaZslgFA9W2-SAQUEj8YMS2DsiumoofGU69gyCAd6iPPWyg2xewshcGeL8wzSZr4K5juWd5c_SSmCIqn4q5RhSjMXUU3J8i5H/s400/1.385_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5425455779008984098" border="0" /></a><br /><br /><br /><br />Now we can substitute <span style="font-weight: bold;">(5)</span> in <span style="font-weight: bold;">(4)</span> to obtain,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEice0jcAHDF4oyzYjpzSOq2nBNKve1HDcP9tjwdZDbTFOxi3pWrA2u_0Fv8NyI_4WRMb9t1Gb3fFNMBHI-o7L8PZhXN3sJ38a3s3pImUQXnM3rRP-g8qO1CtOpT5oHjbBZZlgT1kcE9HppC/s1600-h/1.385_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 186px; height: 49px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEice0jcAHDF4oyzYjpzSOq2nBNKve1HDcP9tjwdZDbTFOxi3pWrA2u_0Fv8NyI_4WRMb9t1Gb3fFNMBHI-o7L8PZhXN3sJ38a3s3pImUQXnM3rRP-g8qO1CtOpT5oHjbBZZlgT1kcE9HppC/s400/1.385_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5425456710171368706" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-35243219650867811792010-01-11T07:11:00.012+05:302010-01-11T07:39:41.890+05:30Irodov Problem 1.384<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzEbIdAOUjc32OkcGRc41H97nQ1d-_KEUxZNMRHZe5rZ3g2Ls9LCrz1Yxyxg7MgdjjkoOM7zg4HwUUW9kn8P_m2DTRDopZ8dy5M8tDxDL3uy1FhPhIyUb6xQROh3Z0rDcBYkcJ3qpq7OgN/s1600-h/1.384_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 312px; height: 285px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzEbIdAOUjc32OkcGRc41H97nQ1d-_KEUxZNMRHZe5rZ3g2Ls9LCrz1Yxyxg7MgdjjkoOM7zg4HwUUW9kn8P_m2DTRDopZ8dy5M8tDxDL3uy1FhPhIyUb6xQROh3Z0rDcBYkcJ3qpq7OgN/s400/1.384_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5425291711076963474" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />As seen from the stationary frame of reference (in which <span style="font-weight: bold;">N2</span> is stationary), neutron <span style="font-weight: bold;">N1</span> is on a collision course towards neutron <span style="font-weight: bold;">N2</span> and has a kinetic energy of .The center of inertia of the (<span style="font-weight: bold;">N1+N2</span>) system (<span style="font-weight: bold;">CG</span>) lies in between <span style="font-weight: bold;">N1</span> and <span style="font-weight: bold;">N2</span> and moves with velocity <span style="font-weight: bold;">v<span style="font-size:78%;">CG</span></span>. As seen from the reference frame of the <span style="font-weight: bold;">CG</span>, both neutrons <span style="font-weight: bold;">N1</span> and <span style="font-weight: bold;">N2</span> move towards the <span style="font-weight: bold;">CG</span> at equal velocities. This is shown in the Figure above.<br /><br />The total energy of <span style="font-weight: bold;">N1</span> in the stationary reference frame is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPK57jOBljXMalXCo40DDkd0LMcJif9589rPCvdVpgSSrZfQPF3p4k7Knz4s1PbXLO01gBg9E6eiEWczDCO18ag_c8OBH5zCFLUrdtZRbjVfGLKGC7RqHtp_tOTnJD3f79JB3OJHuuxdZQ/s1600-h/1.384_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 120px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPK57jOBljXMalXCo40DDkd0LMcJif9589rPCvdVpgSSrZfQPF3p4k7Knz4s1PbXLO01gBg9E6eiEWczDCO18ag_c8OBH5zCFLUrdtZRbjVfGLKGC7RqHtp_tOTnJD3f79JB3OJHuuxdZQ/s400/1.384_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5425291919745212642" border="0" /></a><br /><br />We can now compute the velocity <span style="font-weight: bold;">v</span> of neutron <span style="font-weight: bold;">N1</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioGlYEqoknjPVcjuliqMaNKfhEDKrLHGsQ_MGxf-rW7FMLpVRwMH_LQ1hxNdeMFZvRat35p0Pvf3CJk8zmq3SSQbSXyTfPoWK7gp3bqYKnq0gHIhGBeAmZ4b6NIcARd7fQQyyOPuQFEZBy/s1600-h/1.384_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 188px; height: 126px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioGlYEqoknjPVcjuliqMaNKfhEDKrLHGsQ_MGxf-rW7FMLpVRwMH_LQ1hxNdeMFZvRat35p0Pvf3CJk8zmq3SSQbSXyTfPoWK7gp3bqYKnq0gHIhGBeAmZ4b6NIcARd7fQQyyOPuQFEZBy/s400/1.384_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5425292401262958754" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br />Hence, the momentum of neutron<span style="font-weight: bold;"> N1</span> and hence the entire system (<span style="font-weight: bold;">N1 + N2</span>) and hence, the momentum of the CG as seen from the staionary reference frame is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigfYmrgCufNe47c9qvF0izzctR-yo5wTU1em84MHtNVTJ_CWJCCVhG8SizhyphenhyphenL6J9dMeC78uRP8aY3N-aCfY_4_aXmC_3e9PPTqhvpahMWmeHLlskdb7XKAROAccKDYdUKcyzU4QnHpuXLl/s1600-h/1.384_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 179px; height: 101px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigfYmrgCufNe47c9qvF0izzctR-yo5wTU1em84MHtNVTJ_CWJCCVhG8SizhyphenhyphenL6J9dMeC78uRP8aY3N-aCfY_4_aXmC_3e9PPTqhvpahMWmeHLlskdb7XKAROAccKDYdUKcyzU4QnHpuXLl/s400/1.384_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5425292583752213970" border="0" /></a><br /><br /><br /><br /><br /><br /><br />The total energy of the entire system (<span style="font-weight: bold;">N1 + N2</span>) and hence that of the <span style="font-weight: bold;">CG</span>, as seen from the stationary reference frame is given by the sum of kinetic energy of the neutron <span style="font-weight: bold;">N1</span> and the sum of the rest mass energies of each of the neutrons <span style="font-weight: bold;">N1</span> and <span style="font-weight: bold;">N2</span>. Thus, the total energy in the system in the stationary reference frame is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBJMUhvHkTgSguH1l4pvirGQ6ANToJ9xDE5H9100wugn81c7qNQmZ2E2HBe9e2wYfhHuG58KC3QhN8u-GmaI2ukL-kY6Q39wbqX04QvplKLtgsj8PFEHEKTqGf_qbYKgwEpqZ57OP2vXX3/s1600-h/1.384_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 118px; height: 25px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBJMUhvHkTgSguH1l4pvirGQ6ANToJ9xDE5H9100wugn81c7qNQmZ2E2HBe9e2wYfhHuG58KC3QhN8u-GmaI2ukL-kY6Q39wbqX04QvplKLtgsj8PFEHEKTqGf_qbYKgwEpqZ57OP2vXX3/s400/1.384_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5425293253106215042" border="0" /></a><br /><br />The invariant quantity <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj3Rj6nmAFbWPCVfDeMWOwXYJbksjlo_vdYZ8xEWijNwGyDyP3Kdq6H5KbsoO9fvdxDa9aC2XZ7FWt_-Zy6a7tGsuCSuOQqhfmGJ2kXc4LJWQjt9SiI-1X3rmUVocRFuyvPQ5FVUF_OOm6/s1600-h/1.384_5.jpg"><img style="cursor: pointer; width: 58px; height: 20px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj3Rj6nmAFbWPCVfDeMWOwXYJbksjlo_vdYZ8xEWijNwGyDyP3Kdq6H5KbsoO9fvdxDa9aC2XZ7FWt_-Zy6a7tGsuCSuOQqhfmGJ2kXc4LJWQjt9SiI-1X3rmUVocRFuyvPQ5FVUF_OOm6/s400/1.384_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5425293397987574482" border="0" /></a> can then be computed as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRwQDGf127it8vW1uziBiBU_yaSr_OnkLxeXA0vW10E0AMic_hAKqQImP6wasLHpADnPp1B1tCfCGJCUzjhy78Y16EJ9eyn9z0X4yFsnN526GZPV7HsRAWl5_3sFNgUByWj8Ulaq_a7JN8/s1600-h/1.384_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 188px; height: 91px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRwQDGf127it8vW1uziBiBU_yaSr_OnkLxeXA0vW10E0AMic_hAKqQImP6wasLHpADnPp1B1tCfCGJCUzjhy78Y16EJ9eyn9z0X4yFsnN526GZPV7HsRAWl5_3sFNgUByWj8Ulaq_a7JN8/s400/1.384_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5425293513271955314" border="0" /></a><br /><br /><br /><br /><br /><br />This quantity is invariant with respect to the choice of any inertial frame and hence must be the same even in the frame of the center of inertia (<span style="font-weight: bold;">CG</span>). The momentum of the <span style="font-weight: bold;">CG</span> is <span style="font-weight: bold;">0</span> in the reference frame of the <span style="font-weight: bold;">CG</span>. Hence the energy of the <span style="font-weight: bold;">CG</span>, <span style="font-weight: bold;">E<span style="font-size:78%;">CG</span></span> can be computed as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2BoXNhZKa2vubRXBFlRxe2o02t6iYsINz9Dy_tqwb0Be4I5FbmVevFdhzg_o3vpNrES3KVwaFYReBdx3nBa7oUlnYtjIT74RabDafSDgMfsetmzo5EkJsVqv4qN8zXnu7atR8J9oxPAlE/s1600-h/1.384_7.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 217px; height: 95px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2BoXNhZKa2vubRXBFlRxe2o02t6iYsINz9Dy_tqwb0Be4I5FbmVevFdhzg_o3vpNrES3KVwaFYReBdx3nBa7oUlnYtjIT74RabDafSDgMfsetmzo5EkJsVqv4qN8zXnu7atR8J9oxPAlE/s400/1.384_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5425294005567837010" border="0" /></a><br /><br /><br /><br /><br /><br />The kinetic energy of the system as viewed from a reference frame <span style="font-weight: bold;">CG</span> can be obtained by subtracting the rest mass energies of the two protons from total energy as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjD2Z-OoDyh2VbJ-CfizWkqjAwpCqV8jg1EMQ2GXM-we1TH3BZKKfZRst3YgXA_-6NNSVV5RxFrdptSo0nWfsctCT01rP60yCNHZMmj5QKoSrm91UOD4SjxoPFYZVowIOWDWkYSnWhhVgBy/s1600-h/1.384_8.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 276px; height: 86px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjD2Z-OoDyh2VbJ-CfizWkqjAwpCqV8jg1EMQ2GXM-we1TH3BZKKfZRst3YgXA_-6NNSVV5RxFrdptSo0nWfsctCT01rP60yCNHZMmj5QKoSrm91UOD4SjxoPFYZVowIOWDWkYSnWhhVgBy/s400/1.384_8.jpg" alt="" id="BLOGGER_PHOTO_ID_5425294188526027698" border="0" /></a><br /><br /><br /><br /><br /><br />As seen from the reference frame of the <span style="font-weight: bold;">CG</span>, both neutrons will have the same energy and hence then energy of each neutron in the <span style="font-weight: bold;">CG</span> reference frame will be,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOACh8kRK7muBNxWfhUOo2MTeyj_nkzFIz_z66YkvAZt11KNLQF9i5uMNN_9fIi5PF9A9VoRM8tHg9xBqXuXcPvu1Uppu3q9KLLiRH0D7zKCJl5uyKij-r6QOpNa33KCWpzWzTCNxwO3QV/s1600-h/1.384_9.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 143px; height: 32px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOACh8kRK7muBNxWfhUOo2MTeyj_nkzFIz_z66YkvAZt11KNLQF9i5uMNN_9fIi5PF9A9VoRM8tHg9xBqXuXcPvu1Uppu3q9KLLiRH0D7zKCJl5uyKij-r6QOpNa33KCWpzWzTCNxwO3QV/s400/1.384_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5425297628622852850" border="0" /></a><br /><br /><br />We also know from problem <span style="font-weight: bold;">1.383</span> that for each neutron,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiC5V1ZmHMuy2D11t0dUNgLZJuIEJL0lumwMAkQvIEXm5Vbpj8oG2ZjRa4NPRYsEfLB2uPyUdYlFk8w4HEZuvc8CJJbV-gqc1zbj5YVTsjDo2voumDhYZ812zkMJfCI7rjQfGrmD8T9Phi6/s1600-h/1.384_10.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 309px; height: 115px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiC5V1ZmHMuy2D11t0dUNgLZJuIEJL0lumwMAkQvIEXm5Vbpj8oG2ZjRa4NPRYsEfLB2uPyUdYlFk8w4HEZuvc8CJJbV-gqc1zbj5YVTsjDo2voumDhYZ812zkMJfCI7rjQfGrmD8T9Phi6/s400/1.384_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5425297846865851410" border="0" /></a><br /><br /><br /><br /><br /><br /><br />Here, <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3NxAgAoRUPqAAH7zB4IWFlQgj9v_HUsMG1NAnQn6hTQOdoSPxmXbNGkWs_-FyUDsv8dlElod0HtuZG0v2xIPaCHkz3dtDtUX18aRKl-6bKdxwDxwZrKGQ84U-1UL2gM9lS4xl-P8H7oda/s1600-h/1.384_11.jpg"><img style="cursor: pointer; width: 30px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3NxAgAoRUPqAAH7zB4IWFlQgj9v_HUsMG1NAnQn6hTQOdoSPxmXbNGkWs_-FyUDsv8dlElod0HtuZG0v2xIPaCHkz3dtDtUX18aRKl-6bKdxwDxwZrKGQ84U-1UL2gM9lS4xl-P8H7oda/s400/1.384_11.jpg" alt="" id="BLOGGER_PHOTO_ID_5425298021613412226" border="0" /></a> is the momentum of the neutron <span style="font-weight: bold;">N1</span> in the reference frame of the <span style="font-weight: bold;">CG</span>.<br /><br /><span style="font-weight: bold;">b)</span> The total rest mass of the <span style="font-weight: bold;">CG</span> is <span style="font-weight: bold;">2m<span style="font-size:78%;">o</span></span>. Hence, at velocity <span style="font-weight: bold;">V<span style="font-size:78%;">CG</span></span>, its energy is given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgu8ehQZaNGeUmcKF7IlLTpYJGlgsIWHVUl1AMnRFQhoW0DyVdEejDsnri_qpqBxKS62QkDWeO6igHUC_cuCmAIpmZnzaWww4vvl2dyI64RRcgRqQ7mzLYhqLb4AuHNr8DEtcHZjjVr8Mn9/s1600-h/1.384_12.jpg"><img style="cursor: pointer; width: 83px; height: 52px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgu8ehQZaNGeUmcKF7IlLTpYJGlgsIWHVUl1AMnRFQhoW0DyVdEejDsnri_qpqBxKS62QkDWeO6igHUC_cuCmAIpmZnzaWww4vvl2dyI64RRcgRqQ7mzLYhqLb4AuHNr8DEtcHZjjVr8Mn9/s400/1.384_12.jpg" alt="" id="BLOGGER_PHOTO_ID_5425298329662392306" border="0" /></a>. We can thus, compute the velocity of the <span style="font-weight: bold;">CG</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkEP_VOr9fnqcBHM5nVnmzo8UxhEocGzXPfQ8d8n0V6qUtZF2-TMYQ0JjyARCYHbr4Vm7UbTOFX2R8SuUA1Kube9Ye_b1PtBemEiZ_w88YBNVmachdyYiEMYviMd0TiwBcenAblUzp2uem/s1600-h/1.384_13.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 299px; height: 126px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkEP_VOr9fnqcBHM5nVnmzo8UxhEocGzXPfQ8d8n0V6qUtZF2-TMYQ0JjyARCYHbr4Vm7UbTOFX2R8SuUA1Kube9Ye_b1PtBemEiZ_w88YBNVmachdyYiEMYviMd0TiwBcenAblUzp2uem/s400/1.384_13.jpg" alt="" id="BLOGGER_PHOTO_ID_5425298562245624450" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-91256474487331554162010-01-08T20:55:00.004+05:302010-01-08T21:04:43.841+05:30Irodov Problem 1.382The momentum of the photon is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXocVoWuB03HlKD9QlREzwS2EI7ss0sgjpcDEC4WJ4NXjaPDczhucDI5FQklT6cZcarGcrGQFg1KvPEKrAv8vc6tThm7tMtlAIitSmQr1XvCLT_hZxnQFVuchk7DvlejjCUfVmO-fiEvf9/s1600-h/1.382_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 103px; height: 25px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXocVoWuB03HlKD9QlREzwS2EI7ss0sgjpcDEC4WJ4NXjaPDczhucDI5FQklT6cZcarGcrGQFg1KvPEKrAv8vc6tThm7tMtlAIitSmQr1XvCLT_hZxnQFVuchk7DvlejjCUfVmO-fiEvf9/s400/1.382_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5424391281845153538" border="0" /></a><br /><br />From the solution found in Problem <span style="font-weight: bold;">1.381</span>, we can compute the energy of the photon as seen in reference frame K' as,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_IlaZOzMT2BpVU3lUsp2q0GpnNyhdqmPL8OvukoeNKvD9FabZRvTqsbCpkInACGdfs9uasMovbFq1JhaL9Xu7X2uFltTGfW5te3-PvKu1RglspleLnSbSkiRHpyBkPP3vnRhXIuxIYAiG/s1600-h/1.382_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 228px; height: 60px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_IlaZOzMT2BpVU3lUsp2q0GpnNyhdqmPL8OvukoeNKvD9FabZRvTqsbCpkInACGdfs9uasMovbFq1JhaL9Xu7X2uFltTGfW5te3-PvKu1RglspleLnSbSkiRHpyBkPP3vnRhXIuxIYAiG/s400/1.382_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5424392786446783490" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-85048113562850434702010-01-08T20:36:00.004+05:302010-01-08T21:05:02.151+05:30Irodov Problem 1.383We can use the results from Problem <span style="font-weight: bold;">1.381</span> to prove this invariance result.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBbfm80VmyPHy-UZQimIFZNVRO74tfKlJZYYVbAoVFmrE8PfPfx3917jN9V_zIN1ygEhkb-vzYtlIsgm9b2W8wvg1V85Uw1Djc6JBLnr9v1-xYmm4BbFmDj1maCFO9lGRK573Fs3H_hs2P/s1600-h/1.382_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 229px; height: 166px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBbfm80VmyPHy-UZQimIFZNVRO74tfKlJZYYVbAoVFmrE8PfPfx3917jN9V_zIN1ygEhkb-vzYtlIsgm9b2W8wvg1V85Uw1Djc6JBLnr9v1-xYmm4BbFmDj1maCFO9lGRK573Fs3H_hs2P/s400/1.382_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5424388752601107746" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />What Eqn <span style="font-weight: bold;">(1)</span> is indicates is an invariance since the quantity does not depend which inertial frame you are observing from!<br /><br />The value of the constant can be determined simply by considering a frame that is moving with the particle itself. In such a frame, the particle is absolutely stationary, in other words <span style="font-weight: bold;">P=0</span>. The energy is equal to the rest mass energy and so we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnRf5KQHqltADn5rogqGonueBnjG5UfTXhS5Og5GwWjCTHP2jDJV3xf9PGD9zBJHFuwji8cdovrGkARcAV33jfqxRc4uKRws618b9c_jFftw1yY8NgMeO-rXGxeHdegsItmBW2oF4hemt6/s1600-h/1.382_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 270px; height: 27px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnRf5KQHqltADn5rogqGonueBnjG5UfTXhS5Og5GwWjCTHP2jDJV3xf9PGD9zBJHFuwji8cdovrGkARcAV33jfqxRc4uKRws618b9c_jFftw1yY8NgMeO-rXGxeHdegsItmBW2oF4hemt6/s400/1.382_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5424390102307743154" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-8202174219684672252010-01-08T19:29:00.007+05:302010-01-08T20:25:32.581+05:30Irodov Problem 1.381From the relativistic transformations of velocities, energy and momentum we have,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj8xTvupG1SL5dH9JYo_0ay3qssgoGd4d0xEd9gfGV_CagvvAqdkyGg5TRWST9Nbnkh6i38cj9syaA6RXOtIJC5FLzXqq3d3iWpQeImYzZpLv4rHu6uihjPLLxNWB6IQbHlqOzmyfgB6sWf/s1600-h/1.381_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 159px; height: 256px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj8xTvupG1SL5dH9JYo_0ay3qssgoGd4d0xEd9gfGV_CagvvAqdkyGg5TRWST9Nbnkh6i38cj9syaA6RXOtIJC5FLzXqq3d3iWpQeImYzZpLv4rHu6uihjPLLxNWB6IQbHlqOzmyfgB6sWf/s400/1.381_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5424381814561835074" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgdQWJ3lFK5vRUcqkvGVxaRRVOxkv_wbALjanuVYNoakLmjaob5LrsJuLWgsip8etu5osRM7eEBF84dXypTayP3f4FntBFZXmZjUa-SmXT_bQRro4lDR_nNcSquGfTOCx9Duy88N8j0z2Fw/s1600-h/1.381_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 183px; height: 231px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgdQWJ3lFK5vRUcqkvGVxaRRVOxkv_wbALjanuVYNoakLmjaob5LrsJuLWgsip8etu5osRM7eEBF84dXypTayP3f4FntBFZXmZjUa-SmXT_bQRro4lDR_nNcSquGfTOCx9Duy88N8j0z2Fw/s400/1.381_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5424382139426559154" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGddoskiScsJnDx-vTOJaKDR2O6_TBTBF6xXku6Fu9gP8UYdAovJx15m866PCXqSBzJLGpAK7PsJSoaJ2lJffuwd3PQ-IAEGK0KSaupHTJhE9Unt4ET9pcRkk3o2DqkXo6llbbiIHGxXpd/s1600-h/1.381_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 189px; height: 154px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGddoskiScsJnDx-vTOJaKDR2O6_TBTBF6xXku6Fu9gP8UYdAovJx15m866PCXqSBzJLGpAK7PsJSoaJ2lJffuwd3PQ-IAEGK0KSaupHTJhE9Unt4ET9pcRkk3o2DqkXo6llbbiIHGxXpd/s400/1.381_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5424382524527001250" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-23133153421150047642010-01-07T19:31:00.022+05:302010-01-08T18:26:15.496+05:30Irodov Problem 1.380<span style="font-weight: bold;">a)</span> For this problem let us denote<span style="font-weight: bold;"> v</span> as the magnitude of the instantaneous velocity of the particle and <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFwHrZavE3A_VVysYzw3PB5FGR6QtAgrSBYXtC4DtxCcrHGXVBnVgiBdSwien5inMFpWt0X-M077_A08_4J-0NAW4vO1cNEGFMrhoNK-BSRSWATdLhCycAorb8tHZC_tD0yDME625faMMk/s1600-h/1.380_1.jpg"><img style="cursor: pointer; width: 15px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFwHrZavE3A_VVysYzw3PB5FGR6QtAgrSBYXtC4DtxCcrHGXVBnVgiBdSwien5inMFpWt0X-M077_A08_4J-0NAW4vO1cNEGFMrhoNK-BSRSWATdLhCycAorb8tHZC_tD0yDME625faMMk/s400/1.380_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423998789621547666" border="0" /></a>as the unit vector along the instantaneous direction of motion of the particle. In other words the instantaneous velocity vector of the particle is given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhT02eplMqAOey_QXK2iUguQvElRE9bWt4ac4AEBO12XgjgU0D35DqUAF9g8Mr1aqumHjXZD2aDK8MpPdsSWQoFwxUnSDo_jmu9TvYKWMmNlaldmFvuCVd7AEHiOjeqdUGU0ZRkXSUjs1P-/s1600-h/1.380_2.jpg"><img style="cursor: pointer; width: 25px; height: 21px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhT02eplMqAOey_QXK2iUguQvElRE9bWt4ac4AEBO12XgjgU0D35DqUAF9g8Mr1aqumHjXZD2aDK8MpPdsSWQoFwxUnSDo_jmu9TvYKWMmNlaldmFvuCVd7AEHiOjeqdUGU0ZRkXSUjs1P-/s400/1.380_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423999520198929042" border="0" /></a>.<br /><br />From the fundamental laws of dynamics we can write,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDHfUZ0uMI8YijnrQuWYdDX_RuA2gOMGhkGpgz7UlYXqVNED718E04EW0gjMzpWY55UwV3azJbRVwMrwsaEzNXYPs66D8KJzZyZsdsjFInA5FnaVm9dJ1AO-P_Rwb2pY4nFlA24USbXqd_/s1600-h/1.380_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 340px; height: 286px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDHfUZ0uMI8YijnrQuWYdDX_RuA2gOMGhkGpgz7UlYXqVNED718E04EW0gjMzpWY55UwV3azJbRVwMrwsaEzNXYPs66D8KJzZyZsdsjFInA5FnaVm9dJ1AO-P_Rwb2pY4nFlA24USbXqd_/s400/1.380_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5424015463785519122" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Similarly, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitGIvMjf6QsTHCpGTH-NGZfm6MuCOgGDbDc4xLgaEZ2po6fEV6nG92xSA-cxXgzzcQIhuqTHKEIy2tPdzG7METyqY2hfq3Lf_JSGrmbLoFkLxByJf7CSges56V1jQ39_-z3pQCzJGskiZb/s1600-h/1.380_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 150px; height: 85px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitGIvMjf6QsTHCpGTH-NGZfm6MuCOgGDbDc4xLgaEZ2po6fEV6nG92xSA-cxXgzzcQIhuqTHKEIy2tPdzG7METyqY2hfq3Lf_JSGrmbLoFkLxByJf7CSges56V1jQ39_-z3pQCzJGskiZb/s400/1.380_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5424013906991011378" border="0" /></a><br /><br /><br /><br /><br /><br />In order for <span style="font-weight: bold;">(1)</span> and <span style="font-weight: bold;">(2)</span> be parallel there are only two possible ways<br /><br /><span style="font-weight: bold;">1.</span> <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilJNralwID0Cz-gAeXqcme5VMnkSxqCTUIkrSd-NOwwoP0LbslwWJHhn5YL3p6BLIY5ed2xwu7RoaMw52o3wioYSd6Jkiis0bXtd9TNPOvbSysZeTqLUCVDSu-g62pTHMwFkIUKPhneIqF/s1600-h/1.380_9.jpg"><img style="cursor: pointer; width: 57px; height: 48px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilJNralwID0Cz-gAeXqcme5VMnkSxqCTUIkrSd-NOwwoP0LbslwWJHhn5YL3p6BLIY5ed2xwu7RoaMw52o3wioYSd6Jkiis0bXtd9TNPOvbSysZeTqLUCVDSu-g62pTHMwFkIUKPhneIqF/s400/1.380_9.jpg" alt="" id="BLOGGER_PHOTO_ID_5424034832424089746" border="0" /></a>, in which case both vectors get trivially aligned. However, this also means that the particle never changes it direction. In other words, the force has to be aligned in the instantaneous direction of motion of the particle i.e. <span style="font-weight: bold;">F</span> is parallel to velocity vector.<br /><br /><span style="font-weight: bold;">2.</span> <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgOrI2VVALkYTRhRzHh7qaWa4YuiF6KQs436Gq5UmnhrGI6YL3I35OtWUfvVBo7vnrq0drWbRXrVFkJlnkkCj2p_9Pk9yWO-Z1UPyyBRIxgt4o8W4OEwmpJWzdiAhJ0b6cYfMiGfDKmXW-A/s1600-h/1.380_10.jpg"><img style="cursor: pointer; width: 57px; height: 48px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgOrI2VVALkYTRhRzHh7qaWa4YuiF6KQs436Gq5UmnhrGI6YL3I35OtWUfvVBo7vnrq0drWbRXrVFkJlnkkCj2p_9Pk9yWO-Z1UPyyBRIxgt4o8W4OEwmpJWzdiAhJ0b6cYfMiGfDKmXW-A/s400/1.380_10.jpg" alt="" id="BLOGGER_PHOTO_ID_5424348675958953218" border="0" /></a>, in which case the magnitude of the velocity of the particle never changes. This can only happen if the force acting on the particle always acts perpendicular to the instantaneous direction of its motion i.e <span style="font-weight: bold;">F</span> is perpendicular to the velocity vector.<br /><br /><span style="font-weight: bold;">b)</span> The solution to this part is quite evident from Eqns (1), and (2) considering each of the two cases separately.<br /><br /><span style="font-weight: bold;">Case 1: F is parallel to velocity vector</span><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRTM3rtU7R5MPAbbgHShtm9dSskS7GuZB_mwwhRfjKdjO50DAk0QBpAZ4xJn0WTeDDzNcWtpmEwY41kZQv0qfLBK63S-gHS88OuBkxriHUipKsUmPuVCyNUP2Q-DaMKgwaxCZ8y8mI6KJA/s1600-h/1.380_11.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 178px; height: 189px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRTM3rtU7R5MPAbbgHShtm9dSskS7GuZB_mwwhRfjKdjO50DAk0QBpAZ4xJn0WTeDDzNcWtpmEwY41kZQv0qfLBK63S-gHS88OuBkxriHUipKsUmPuVCyNUP2Q-DaMKgwaxCZ8y8mI6KJA/s400/1.380_11.jpg" alt="" id="BLOGGER_PHOTO_ID_5424351032257643634" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;">Case 2 : F is perpendicular to velocity vector</span><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgliAtRV4IMSAoa7ojfr-umCKj77tx-wdQLWj0vv913802WuKGY441MbtfayxnQQnz4dAfDbZETva7y3GSpow_OYW6cShqlssGEriblXmNLw-qmwPh4cJwAlHmRkxB9M6IG8FvK9OsFCwJk/s1600-h/1.380_12.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 173px; height: 189px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgliAtRV4IMSAoa7ojfr-umCKj77tx-wdQLWj0vv913802WuKGY441MbtfayxnQQnz4dAfDbZETva7y3GSpow_OYW6cShqlssGEriblXmNLw-qmwPh4cJwAlHmRkxB9M6IG8FvK9OsFCwJk/s400/1.380_12.jpg" alt="" id="BLOGGER_PHOTO_ID_5424351950118060386" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-68449423258243327172010-01-07T12:44:00.003+05:302010-01-07T12:49:07.736+05:30Irodov Problem 1.379This problem can be solved from a very simple inspection of solution to problem <span style="font-weight: bold;">1.378</span> and seeing that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZMkpMK3_ThjtYwz0RlCuNK6gFrGQhHvhxKHcKy32Xb9o_71gtqWY-NdeQZKa0wfH4DjucYvk7TMPG6jJNKus7GWsjn-BsESXwLTgx35_mwA5kSM_HochOPnZCkd-E9XO5eZKjdJTNfwcD/s1600-h/1.379_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 100px; height: 58px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZMkpMK3_ThjtYwz0RlCuNK6gFrGQhHvhxKHcKy32Xb9o_71gtqWY-NdeQZKa0wfH4DjucYvk7TMPG6jJNKus7GWsjn-BsESXwLTgx35_mwA5kSM_HochOPnZCkd-E9XO5eZKjdJTNfwcD/s400/1.379_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423893824312272882" border="0" /></a><br /><br /><br /><br />The constant offset is not important since the origin could always be displaced by a constant value without effecting the dynamics.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-13881544284086922242010-01-07T12:32:00.002+05:302010-01-07T12:44:13.206+05:30Irodov Problem 1.378We can solve this using the fundamental equation of dynamics as follows:<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhK31jCWaGvzwHgCbM5f4hxKuRlFeWq4lIjR0E6B6fkbzXCRiAyBo3HwC9ltMreFPaWRiDyt5gLw8YLgFppB0jHpJaGq6ETmUzFrZY-xRJl54h7zAWf0SlIrI0Z4Z4RHchpPzRUImBg_cI9/s1600-h/1.378_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 209px; height: 300px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhK31jCWaGvzwHgCbM5f4hxKuRlFeWq4lIjR0E6B6fkbzXCRiAyBo3HwC9ltMreFPaWRiDyt5gLw8YLgFppB0jHpJaGq6ETmUzFrZY-xRJl54h7zAWf0SlIrI0Z4Z4RHchpPzRUImBg_cI9/s400/1.378_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423892700328678738" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-17538151348747058872010-01-07T11:15:00.012+05:302010-01-07T12:31:39.844+05:30Irodov Problem 1.377<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjy8BXu-hZRKJZ_wVlaSEniLMSgR8x5bTAE0z5J3DuBuZMyZGinsEj-gHnRa2gdPBlrVtgIliUMVsRxzuWRjK-2-lLaIvJyvXnsTv1XXKaANP-R1HXTQiUc2a0BM-Iud-R-sGVvy_F-bdLH/s1600-h/1.377_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 176px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjy8BXu-hZRKJZ_wVlaSEniLMSgR8x5bTAE0z5J3DuBuZMyZGinsEj-gHnRa2gdPBlrVtgIliUMVsRxzuWRjK-2-lLaIvJyvXnsTv1XXKaANP-R1HXTQiUc2a0BM-Iud-R-sGVvy_F-bdLH/s400/1.377_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423876313438623554" border="0" /></a><br /><span style="font-weight: bold;"></span><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;">In the frame of reference of the sphere</span><br /><br />As seen from the frame of reference of the sphere, the sphere itself is stationary. The gas particles are however moving at a speed <span style="font-weight: bold;">v</span> towards it. Let us now consider the collision of the sphere with a single particle. As asked in the question consider a section of the sphere that is normal to the direction of velocity. Assuming that the sphere is extremely large compared to the gas particles, in case of an elastic collision, the gas particle will simply bounce off with the same velocity <span style="font-weight: bold;">v</span> in the opposite direction. The net change in momentum of the gas particle is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgccqFAbGCZhKYKZ743J9iJSvV_-961CEutZvKufi_7XANWi-IPTXyVvZ7ai521G1uDIGC4GQvJZ4iLPCGc1UkApx3PySnj3LFudehwIUo03jpTSsQDcVHA2gVBODwdWytCYRnK5rs8bhyphenhyphenw/s1600-h/1.377_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 184px; height: 108px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgccqFAbGCZhKYKZ743J9iJSvV_-961CEutZvKufi_7XANWi-IPTXyVvZ7ai521G1uDIGC4GQvJZ4iLPCGc1UkApx3PySnj3LFudehwIUo03jpTSsQDcVHA2gVBODwdWytCYRnK5rs8bhyphenhyphenw/s400/1.377_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423879242969077218" border="0" /></a><br /><br /><br /><br /><br /><br /><br />As seen from the reference frame of the particles there are <span style="font-weight: bold;">n</span> particles per unit volume. However, the same unit volume with <span style="font-weight: bold;">n</span> particles as seen by an observer moving with the sphere is shrunk by a factor of <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_NhY6TAmK1oHha0zXcf0cqooqOjqgZCfRssOS4C0A4_a8L_uC0kYC_AxC6aPaUsckbvyKlo3-D9ZPlUy0cXDYjdtY89x8U2UQhGnakIqH7JdLSWUy_i53Y3ceCYxq56xb2wMhQ4cxgNmA/s1600-h/1.377_2.jpg"><img style="cursor: pointer; width: 58px; height: 31px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_NhY6TAmK1oHha0zXcf0cqooqOjqgZCfRssOS4C0A4_a8L_uC0kYC_AxC6aPaUsckbvyKlo3-D9ZPlUy0cXDYjdtY89x8U2UQhGnakIqH7JdLSWUy_i53Y3ceCYxq56xb2wMhQ4cxgNmA/s400/1.377_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423880416244321682" border="0" /></a>since space itself would have shrunk by this factor in the direction of motion. In other words, the density of the gas a seen from the observer on the sphere would be<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7SFtL0Gxgm-bUZl7Jna9mR9Hh4UPQN1Du1I5sxGUQM7a-1aUFV02IJlIU4ZWXNeczzFYBKmmPfXtApMxX3Bbk-amOLs6dND9n36csEu9avC5VozG_4r4d2eU1vC2lv9lSKlkkC_8kpaUR/s1600-h/1.377_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 136px; height: 49px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7SFtL0Gxgm-bUZl7Jna9mR9Hh4UPQN1Du1I5sxGUQM7a-1aUFV02IJlIU4ZWXNeczzFYBKmmPfXtApMxX3Bbk-amOLs6dND9n36csEu9avC5VozG_4r4d2eU1vC2lv9lSKlkkC_8kpaUR/s400/1.377_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5423881658408344098" border="0" /></a><br /><br /><br /><br />Since <span style="font-weight: bold;">n'</span> particles change their momentum each second, the net force exerted on the sphere as a reaction will be<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBNDsGsVe3jYpZfcl_IRRo4fzBy_yBAz8hlFWuElKRIcGnhCmCLLzEcnJLEibbIyR1HX7DxLP2S2y1GnvxfBlCm4vfUrX7bhYj3Rc4dKIbT7_sDtTVfR4j4Kr8hZ4YcjaKBDBuUJAMcW42/s1600-h/1.377_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 123px; height: 47px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBNDsGsVe3jYpZfcl_IRRo4fzBy_yBAz8hlFWuElKRIcGnhCmCLLzEcnJLEibbIyR1HX7DxLP2S2y1GnvxfBlCm4vfUrX7bhYj3Rc4dKIbT7_sDtTVfR4j4Kr8hZ4YcjaKBDBuUJAMcW42/s400/1.377_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5423882853969508754" border="0" /></a><br /><br /><br /><br /><span style="font-weight: bold;">Solution in reference frame of the particles</span><br />As seen from the view of the gas particles, they were initially stationary and the sphere was moving at a velocity <span style="font-weight: bold;">v</span>. Since as seen from the sphere's reference frame, the final velocity (velocity after collision) was <span style="font-weight: bold;">2v</span>, their final velocity as seen by a stationary observer can be obtained by using the relativistic transformation as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzvFbPJ-Rq0dg5aaldnhLqIIt5sIJ09SqYQ36Be-Gp5r3dTDTI5pCWG65a17ha5FD9oiuQEMbWqPojRJXBgjD1dEa9d9mf1CngB_aWSzkVIEeMasZFmfsFgRjUVD1eg545ZGbKdr1_0MdF/s1600-h/1.377_5.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 139px; height: 47px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzvFbPJ-Rq0dg5aaldnhLqIIt5sIJ09SqYQ36Be-Gp5r3dTDTI5pCWG65a17ha5FD9oiuQEMbWqPojRJXBgjD1dEa9d9mf1CngB_aWSzkVIEeMasZFmfsFgRjUVD1eg545ZGbKdr1_0MdF/s400/1.377_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5423886853642337122" border="0" /></a><br /><br /><br /><br />This means that the change in momentum from the point of view of a stationary observer is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjR_FEKVy4S2ccCAXY-SqB_roSltZn2jAI1g05cqAA190Nz-jksoCOC74-Zm36bZUBWR8HkWYutqtN02m89ZfHQ4roR7L8xgUYBPhjScOdYqCr26PJDfOho1M3saNkQ4xz74s_JeVcNVbfz/s1600-h/1.377_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 165px; height: 147px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjR_FEKVy4S2ccCAXY-SqB_roSltZn2jAI1g05cqAA190Nz-jksoCOC74-Zm36bZUBWR8HkWYutqtN02m89ZfHQ4roR7L8xgUYBPhjScOdYqCr26PJDfOho1M3saNkQ4xz74s_JeVcNVbfz/s400/1.377_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5423888603683135762" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br />Since the number of particles colliding with the sphere per unit time is <span style="font-weight: bold;">n</span>, the net force exerted on the sphere is given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRURt3rqifyLCaw6IR7QjAx_Bm_yow5C8gqMrtoj8dCaiAbnQLSQ2zDPf5V2o2WlTVEDc0xfZjG5tUAlFm0yzmKLxzSb6hXPAUCsFsWX5EhUHUZEcM50dUb26ndTq4ZLA_5-ydEYEExxvw/s1600-h/1.377_7.jpg"><img style="cursor: pointer; width: 44px; height: 26px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRURt3rqifyLCaw6IR7QjAx_Bm_yow5C8gqMrtoj8dCaiAbnQLSQ2zDPf5V2o2WlTVEDc0xfZjG5tUAlFm0yzmKLxzSb6hXPAUCsFsWX5EhUHUZEcM50dUb26ndTq4ZLA_5-ydEYEExxvw/s400/1.377_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5423889284393246514" border="0" /></a>, which evaluates exactly the same as calculated from the point of view of an observer on the sphere!Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-55127974519839173322010-01-06T15:05:00.006+05:302010-01-06T15:16:29.189+05:30Irodov Problem 1.376The number of particles impinging upon the target each second is given by<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGluYlhyphenhyphenMXNM-mEgF0hsWEmCd4Jv0_krcu6VjwVFoVMyQb-ioIB2YIiYY1O8tdxHstXF-rdWXz82X6UxusB5zMEkOJ3KSmew3eyUxWf3v4TbCUElr7ZCHF2N6in7W1xIK2aSkPNXo-JK3y/s1600-h/1.376_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 84px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGluYlhyphenhyphenMXNM-mEgF0hsWEmCd4Jv0_krcu6VjwVFoVMyQb-ioIB2YIiYY1O8tdxHstXF-rdWXz82X6UxusB5zMEkOJ3KSmew3eyUxWf3v4TbCUElr7ZCHF2N6in7W1xIK2aSkPNXo-JK3y/s400/1.376_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423558871112706722" border="0" /></a><br /><br />The momentum of each particle is given by the solution in Problem <span style="font-weight: bold;">1.375</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjt7zB6Kjx_FiKUsD8Kv1L9UNvKgfajbuKluXdEpaRRZCYUt2UAKZkPCpJzcVqY_3DpEjgDPRgI70Y12hfD78bm_vxZ96WIl72SYlRDgAZILwuIuaCLS16J3vn9QnLflOscZiBMjLv3bLkX/s1600-h/1.376_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 185px; height: 32px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjt7zB6Kjx_FiKUsD8Kv1L9UNvKgfajbuKluXdEpaRRZCYUt2UAKZkPCpJzcVqY_3DpEjgDPRgI70Y12hfD78bm_vxZ96WIl72SYlRDgAZILwuIuaCLS16J3vn9QnLflOscZiBMjLv3bLkX/s400/1.376_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423559296937255394" border="0" /></a><br /><br /><br />The total change in momentum of all the <span style="font-weight: bold;">n</span> particles each second is the force imparted in the target and is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZMqj3CnU1ZicZXmh3QeGBemKHmbtE2fPmBNYrqq3uz9g5L0OXHL7oEjixSfstVqjAOtOaE3KyTJu7Bi6vyuNlB9twiusz1yMx-7ShKFawQyVmf0wTqIt31KgQY1yOUWelCM8vqw3_rcOO/s1600-h/1.376_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 188px; height: 36px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZMqj3CnU1ZicZXmh3QeGBemKHmbtE2fPmBNYrqq3uz9g5L0OXHL7oEjixSfstVqjAOtOaE3KyTJu7Bi6vyuNlB9twiusz1yMx-7ShKFawQyVmf0wTqIt31KgQY1yOUWelCM8vqw3_rcOO/s400/1.376_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5423560127898945010" border="0" /></a><br /><br /><br />The total energy of the <span style="font-weight: bold;">n</span> particles getting absorbed into the target is<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj_VDz8GqYg5BgP9KNIgt3F-matKHEPIkpp-BHS4wWWAy8e8Pd1XnVCZymmbO5wDJip61WcnmcoBmsB1N_tycrUYKjC_ZMQKODZIrVazkS4_bmZHFaJlo0XI7zESEHYIRbEqGXw9pkhBO7/s1600-h/1.376_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 64px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjj_VDz8GqYg5BgP9KNIgt3F-matKHEPIkpp-BHS4wWWAy8e8Pd1XnVCZymmbO5wDJip61WcnmcoBmsB1N_tycrUYKjC_ZMQKODZIrVazkS4_bmZHFaJlo0XI7zESEHYIRbEqGXw9pkhBO7/s400/1.376_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5423560874821460498" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-35002382823031023622010-01-06T14:39:00.005+05:302010-01-06T14:57:14.750+05:30Irodov Problem 1.375From the solution to Problem <span style="font-weight: bold;">1.374</span> we know that given that the particle has a kinetic energy <span style="font-weight: bold;">T</span> and a rest mass of m<span style="font-size:78%;">0</span>,<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgks6UnSnTYHV2UHGxyk_vMDkhfQk2XrDBMd2nqVAZRAeREQbpOkKFHaqJARBJe61ODqgC-dSwcOA4thBQsrf1ABUYeXbeeYRmFwtHdu8UDhI1AHZG3zKC660xuh_0Oxal55b1yGP8b3OMt/s1600-h/1.375_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 160px; height: 147px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgks6UnSnTYHV2UHGxyk_vMDkhfQk2XrDBMd2nqVAZRAeREQbpOkKFHaqJARBJe61ODqgC-dSwcOA4thBQsrf1ABUYeXbeeYRmFwtHdu8UDhI1AHZG3zKC660xuh_0Oxal55b1yGP8b3OMt/s400/1.375_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423553268199561618" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br />Also we know that the momentum of the particle is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQJIUPIF0LwVHfajTXQMVaYx0Mz5_bROVZUAmFM8uCwWqDdCYvb1qeGs08khioZ7tX1yw8LzrlzzUqNhf9szY7jxIaB-LYEiRN-0PJ5Q9Kht-Y62RFY0cq6PRCVKdCmayONBHxAeDnKhgg/s1600-h/1.375_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 142px; height: 61px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQJIUPIF0LwVHfajTXQMVaYx0Mz5_bROVZUAmFM8uCwWqDdCYvb1qeGs08khioZ7tX1yw8LzrlzzUqNhf9szY7jxIaB-LYEiRN-0PJ5Q9Kht-Y62RFY0cq6PRCVKdCmayONBHxAeDnKhgg/s400/1.375_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423554402794026034" border="0" /></a><br /><br /><br /><br />From <span style="font-weight: bold;">(1), (2)</span> and <span style="font-weight: bold;">(3)</span> we can write,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj89S84tVu_rKBkmGe3YFDGZFzIlzrL8mf3sn0Db1O082yILSY5Inb7OatC6X_px90kJmvcVtUuOU3UmgrcLsXDf7Dy1-N7fAlr_5mMK07ywmL54LJj9NWNzeUYFB0hsBULihCA6EbZGGFf/s1600-h/1.375_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 142px; height: 32px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj89S84tVu_rKBkmGe3YFDGZFzIlzrL8mf3sn0Db1O082yILSY5Inb7OatC6X_px90kJmvcVtUuOU3UmgrcLsXDf7Dy1-N7fAlr_5mMK07ywmL54LJj9NWNzeUYFB0hsBULihCA6EbZGGFf/s400/1.375_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5423555863981641730" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-4058361831507424122010-01-06T13:37:00.008+05:302010-01-06T14:23:04.556+05:30Irodov Problem 1.374Let <span style="font-weight: bold;">T</span> be the kinetic energy of the particle. Let its rest mass be <span style="font-weight: bold;">m</span><span style="font-weight: bold;font-size:78%;" >o</span>. Let the ratio of its kinetic energy to rest mass energy be equal to,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGQ8PLu34heGclBWbSDeJFqGf7hSODh-Z_hy53lh_k-vbMwCUUbSV0NT8W_ZR-jqjTLQ-MehfzoCqImOFsrCVT4GwdR1kkQ4VynCVBuaj8NC7r8Z4vHHJSLSV-lK8qqftGeQrgGjwcIXdS/s1600-h/1.374_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 125px; height: 27px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGQ8PLu34heGclBWbSDeJFqGf7hSODh-Z_hy53lh_k-vbMwCUUbSV0NT8W_ZR-jqjTLQ-MehfzoCqImOFsrCVT4GwdR1kkQ4VynCVBuaj8NC7r8Z4vHHJSLSV-lK8qqftGeQrgGjwcIXdS/s400/1.374_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423536185083864322" border="0" /></a><br /><br /><br />If we used Newtonian mechanics to determine the velocity we would have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhz1a4TSOJu4cldfhFnXmJvg3r0cRAuJ2MfdfmKIRqYGaYtbrQvS3MQ2HVkY7ojTRavuPXvFAi6oN4PSfAsSmbaDH0CR13kaAGxL-WDBYftiK5KgEUxPWggDBEI6FYWo3pUCxBVpV55sZyR/s1600-h/1.374_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 213px; height: 59px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhz1a4TSOJu4cldfhFnXmJvg3r0cRAuJ2MfdfmKIRqYGaYtbrQvS3MQ2HVkY7ojTRavuPXvFAi6oN4PSfAsSmbaDH0CR13kaAGxL-WDBYftiK5KgEUxPWggDBEI6FYWo3pUCxBVpV55sZyR/s400/1.374_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423537604731905938" border="0" /></a><br /><br /><br /><br />If we used Relativistic mechanics on the other hand we would have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0T820g8V6-bPTOtIl-Yg4Uc8HNSWSXDAg1bgENmSQtBHgAl6e4mQAjzCNMZEZSnIw8Dn3_AQlLiTh2N-sY2MqES4cP8JhTOA8vlAJC3vCiSg9EBuQKH1AbUikI5jczuR-BI_SvFs2BGif/s1600-h/1.374_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 221px; height: 245px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0T820g8V6-bPTOtIl-Yg4Uc8HNSWSXDAg1bgENmSQtBHgAl6e4mQAjzCNMZEZSnIw8Dn3_AQlLiTh2N-sY2MqES4cP8JhTOA8vlAJC3vCiSg9EBuQKH1AbUikI5jczuR-BI_SvFs2BGif/s400/1.374_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5423540323012913586" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />The relative error in the estimation of velocity would then be,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHhpneW3h2gaO2kyupQ2rZNsxI0o5fZNiEJi_A4hVnR7c95MU4Q7TEhneBhiRETq7pn_Vn49QalygCgjzWGSZ0H1VRzPJxDeO8yfZoRCahBziZhvzIxRgYw9QKsGLV6DXwqRUce3O2f1AC/s1600-h/1.374_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 254px; height: 207px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHhpneW3h2gaO2kyupQ2rZNsxI0o5fZNiEJi_A4hVnR7c95MU4Q7TEhneBhiRETq7pn_Vn49QalygCgjzWGSZ0H1VRzPJxDeO8yfZoRCahBziZhvzIxRgYw9QKsGLV6DXwqRUce3O2f1AC/s400/1.374_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5423542290904709970" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />For small values of <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiIft8tI08ARbSC6nWeDAWy_UCYV6q3mxOmuSf5OxYimHq35ugKl9Ma_1e45MnlSm0wTzxe7dKsm5kjTBHPzdqCEb_MmcCPp9n764RNUEuRIQd-OeOJE_T50HkcpkRm8zXSTnnIzK3ukdYT/s1600-h/1.374_5.jpg"><img style="cursor: pointer; width: 13px; height: 16px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiIft8tI08ARbSC6nWeDAWy_UCYV6q3mxOmuSf5OxYimHq35ugKl9Ma_1e45MnlSm0wTzxe7dKsm5kjTBHPzdqCEb_MmcCPp9n764RNUEuRIQd-OeOJE_T50HkcpkRm8zXSTnnIzK3ukdYT/s400/1.374_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5423542791841812162" border="0" /></a>we can consider only the linear terms in equation <span style="font-weight: bold;">(4)</span> and we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNWcFPb7Qqzr9DgZWTWSulK6TVXyTgAeEWLXeLDou-t7ltMT-Col8fnfHu8DBbZL4v9SHliZHr2bjBFx8sMXajoYgyDnnX1pfMKR0yBubG-bYuk7fziDHgMSe0U5lqv9g2ejiEbP6MSEzO/s1600-h/1.374_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 222px; height: 127px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNWcFPb7Qqzr9DgZWTWSulK6TVXyTgAeEWLXeLDou-t7ltMT-Col8fnfHu8DBbZL4v9SHliZHr2bjBFx8sMXajoYgyDnnX1pfMKR0yBubG-bYuk7fziDHgMSe0U5lqv9g2ejiEbP6MSEzO/s400/1.374_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5423547017644584802" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-92045923100742156332010-01-06T12:16:00.003+05:302010-01-06T12:18:07.575+05:30Irodov Problem 1.373When the particle moves at velocity <span style="font-weight: bold;">v</span>, its kinetic energy is given by<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHjDAKKaCNr6azFjfJ2U6743eeHk-UcBSe2kSY1Rcw8L3gfgJDOMnml_YpCP6PBtOUOVU9neQL9zU8iAv4kA0lmP2tyAGGhJg5vbJtRJARDBO2AUtBT2eDOwtU2DyQ_FeyQANA94NwekjY/s1600-h/1.373_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 127px; height: 54px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHjDAKKaCNr6azFjfJ2U6743eeHk-UcBSe2kSY1Rcw8L3gfgJDOMnml_YpCP6PBtOUOVU9neQL9zU8iAv4kA0lmP2tyAGGhJg5vbJtRJARDBO2AUtBT2eDOwtU2DyQ_FeyQANA94NwekjY/s400/1.373_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423514730117058594" border="0" /></a><br /><br /><br /><br />Thus, if its kinetic energy is equal to its rest mass energy, then we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGnxOe-d1o8CUAy88zp63B6ezsoGGIsJeQtSPkAqQeD9T5pB5TbPjOADEoPqisgk6rj34fsW7rPAyb1FCKml0k61c6jRdClwEmmiR6dsNUz49Mmrdl7H3_V6l5hefaj7xY8hqDHEK1OZDQ/s1600-h/1.373_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 182px; height: 170px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGnxOe-d1o8CUAy88zp63B6ezsoGGIsJeQtSPkAqQeD9T5pB5TbPjOADEoPqisgk6rj34fsW7rPAyb1FCKml0k61c6jRdClwEmmiR6dsNUz49Mmrdl7H3_V6l5hefaj7xY8hqDHEK1OZDQ/s400/1.373_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423514940831913474" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-46711814407054475462010-01-06T12:13:00.001+05:302010-01-06T12:15:25.552+05:30Irodov Problem 1.372All the work done on the particle translates into kinetic energy of the particle.<br />The Newtonian kinetic energy is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnaNBvmZm2zA-QyRye0TzcIr4P8C6sGFOd5AXK7j2unQKJvmvPdRy2sRqp_lJyz59qMy4JfvtdwnSVQ6IbF4nQXb5RCxbu462JeRmzVI9ruwMCtrGTo484GJ-6JlnxJ8bZl8axifNRkXUE/s1600-h/1.372_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 93px; height: 22px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnaNBvmZm2zA-QyRye0TzcIr4P8C6sGFOd5AXK7j2unQKJvmvPdRy2sRqp_lJyz59qMy4JfvtdwnSVQ6IbF4nQXb5RCxbu462JeRmzVI9ruwMCtrGTo484GJ-6JlnxJ8bZl8axifNRkXUE/s400/1.372_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423514097659421346" border="0" /></a><br /><br />The relativistic kinetic energy is given by the change in total energy in the particle given by,<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgeeoenjTu5R5-vyGSmf08d8ivOzMMELagEd5Nh-PXkVBGC-hbe5DFWN7JW1tDjdpjyEOVP54c5ABq11rrADtdwnDcvPH63nVUSJQsSRHj02VN5EhY8yNGTWxjbNAT494-J5Lg8MbLtDeYu/s1600-h/1.372_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 244px; height: 137px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgeeoenjTu5R5-vyGSmf08d8ivOzMMELagEd5Nh-PXkVBGC-hbe5DFWN7JW1tDjdpjyEOVP54c5ABq11rrADtdwnDcvPH63nVUSJQsSRHj02VN5EhY8yNGTWxjbNAT494-J5Lg8MbLtDeYu/s400/1.372_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423514164336314866" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-73531285655422174972010-01-06T12:12:00.002+05:302010-01-06T12:13:28.796+05:30Irodov Problem 1.371The ratio of relativistic momentum to Newtonian momentum is given by,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAFpOes3qxQQ1q8ODrqI5f_D9p_Dqj8arD127a_5n0ZFRjsIHbPTm3AbTrGx0Eb-3Dy8_Hcij3HeIyNPRDAg7p2fxdxfwtpUP5gDaEgF-dx4cvcNkiKTozLo4YS5e_7u6KhV9dmjse3wBD/s1600-h/1.371_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 245px; height: 141px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAFpOes3qxQQ1q8ODrqI5f_D9p_Dqj8arD127a_5n0ZFRjsIHbPTm3AbTrGx0Eb-3Dy8_Hcij3HeIyNPRDAg7p2fxdxfwtpUP5gDaEgF-dx4cvcNkiKTozLo4YS5e_7u6KhV9dmjse3wBD/s400/1.371_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423513683483671778" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-2902908850843532612010-01-06T12:04:00.004+05:302010-01-06T12:11:39.159+05:30Irodov Problem 1.370As the particle moves its mass increases. When the velocity of the particle is <span style="font-weight: bold;">v</span>, its mass is given by, <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglbH1mRhFGAyRkeDf1U5TPgt-NrlvrfK4mpo3zl4RrPqeZ4U7JMkqw_30fezP3IF-Z680Y8nqTtHQO3YIzNBpUTAAl02ha6Z1dcT0aiMrGWnuvUFr59D9Om4-gokhBUZfwArJpUix8UYos/s1600-h/1.370_1.jpg"><img style="cursor: pointer; width: 63px; height: 40px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglbH1mRhFGAyRkeDf1U5TPgt-NrlvrfK4mpo3zl4RrPqeZ4U7JMkqw_30fezP3IF-Z680Y8nqTtHQO3YIzNBpUTAAl02ha6Z1dcT0aiMrGWnuvUFr59D9Om4-gokhBUZfwArJpUix8UYos/s400/1.370_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5423512120154393122" border="0" /></a>.<br /><br />Consequently its momentum is given by<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGJVID-rjusNklPQtwBsRaQLOxEI5zEOdDQusy6kRQ_xrTw041hKOHX1q5uXflYZ3n_A2Sk7pztj3X1zsqAacI6t2_kBkNbuc-Vhyphenhyphen0RdHVaFVMiRQzMowADM0SP3UhBoN0hvIlcmhIEv2A/s1600-h/1.370_2.jpg"><img style="cursor: pointer; width: 130px; height: 46px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGJVID-rjusNklPQtwBsRaQLOxEI5zEOdDQusy6kRQ_xrTw041hKOHX1q5uXflYZ3n_A2Sk7pztj3X1zsqAacI6t2_kBkNbuc-Vhyphenhyphen0RdHVaFVMiRQzMowADM0SP3UhBoN0hvIlcmhIEv2A/s400/1.370_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5423512209198083778" border="0" /></a><br /><br />Now we can solve for <span style="font-weight: bold;">v</span> from equation from<span style="font-weight: bold;"> (1)</span> and obtain,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFN7seU7v0V3r1sBGXty1r29S_VLSLESIZG_7RHj_nhICsgsuAfE01FaxdanqWp998Tg0K2PPjPad-Qf4L-Vn6YUDgTkn-qLrUkkByVxYJPnG2d3n3CfXVChnxPITapl-BSoIOpufuT7Yf/s1600-h/1.370_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 171px; height: 114px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFN7seU7v0V3r1sBGXty1r29S_VLSLESIZG_7RHj_nhICsgsuAfE01FaxdanqWp998Tg0K2PPjPad-Qf4L-Vn6YUDgTkn-qLrUkkByVxYJPnG2d3n3CfXVChnxPITapl-BSoIOpufuT7Yf/s400/1.370_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5423513221284757058" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-81086227564931779462009-12-31T18:44:00.010+05:302009-12-31T18:54:57.992+05:30Irodov Problem 1.369<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEoGwjra_ZGuiFDWLH-kecFxknml7cHG2Z6PNvezHh5Mt6kJiZCRHmEwhNTYuqrmqeV8lWt5-vtnVYRgcHi_R5I0Luec-ppa5KmCO8_lZ8jjBRU3sJrK2qGtJhuu3iqcRLQwXX2RYd-xu9/s1600-h/1.369_fig.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 194px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEoGwjra_ZGuiFDWLH-kecFxknml7cHG2Z6PNvezHh5Mt6kJiZCRHmEwhNTYuqrmqeV8lWt5-vtnVYRgcHi_R5I0Luec-ppa5KmCO8_lZ8jjBRU3sJrK2qGtJhuu3iqcRLQwXX2RYd-xu9/s400/1.369_fig.jpg" alt="" id="BLOGGER_PHOTO_ID_5421388370536084258" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br />Without loss of generality let us assume that the observer is moving at a speed with velocity <span style="font-weight: bold;">v</span> along the positive<span style="font-weight: bold;"> x-axis</span>. As seen by the observer, the body is moving along the negative <span style="font-weight: bold;">x-axis</span> with a speed <span style="font-weight: bold;">v</span>. Let be its density be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhSEvdpfs1ysn3wHkVWuJuUdSZDMLuahMo_aGVToVvE_is0SwUjtWCq8GujsbhxGaWeFpntaXzU5reat_lhqywKxC7mD0kz6Q3c6YMMMN88rxI751FVzhE14ICwwpEHMiwFMfgoOnKIsuOo/s1600-h/1.369_1.jpg"><img style="cursor: pointer; width: 19px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhSEvdpfs1ysn3wHkVWuJuUdSZDMLuahMo_aGVToVvE_is0SwUjtWCq8GujsbhxGaWeFpntaXzU5reat_lhqywKxC7mD0kz6Q3c6YMMMN88rxI751FVzhE14ICwwpEHMiwFMfgoOnKIsuOo/s400/1.369_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5421388554639657058" border="0" /></a>.<br /><br />Consider an infinitesimally small cuboidal piece of the body of dimensions <span style="font-weight: bold;">dx, dy</span> and <span style="font-weight: bold;">dz</span>. The mass <span style="font-weight: bold;">dm</span> of this infinitesimally small piece is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjviYts_cjkt3Y6LJ-iL2z0CST_m_bssWZ2ExEPrFUIQdzMaiureZvauT5GoNZaDSIyQleLyRCzOfMZfyY2b9wtGCjXQ5YHQR4muD1Gi1hMZTHRL_P6qr5vHSQGoqRwr2pmccrnCcVRPJNZ/s1600-h/1.369_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 164px; height: 26px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjviYts_cjkt3Y6LJ-iL2z0CST_m_bssWZ2ExEPrFUIQdzMaiureZvauT5GoNZaDSIyQleLyRCzOfMZfyY2b9wtGCjXQ5YHQR4muD1Gi1hMZTHRL_P6qr5vHSQGoqRwr2pmccrnCcVRPJNZ/s400/1.369_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5421388811685068978" border="0" /></a><br /><br />As seen by the observer, the mass of the inifinitesimally small cuboid would be<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgg5ShOWMRjS0ybe6ahh692IT-OqJKY_CzghR3WdMrkjedDAjMk8xB9mlAKw_Von7vwgg9EAsTHuY0Zx50gq4s-0IlGCqC1gvM2IYAcBqC7Cxb73JiMbaNzJ2Uxx93Cz31J-Md5jsnoxoiv/s1600-h/1.369_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 164px; height: 53px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgg5ShOWMRjS0ybe6ahh692IT-OqJKY_CzghR3WdMrkjedDAjMk8xB9mlAKw_Von7vwgg9EAsTHuY0Zx50gq4s-0IlGCqC1gvM2IYAcBqC7Cxb73JiMbaNzJ2Uxx93Cz31J-Md5jsnoxoiv/s400/1.369_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5421388921571226514" border="0" /></a><br /><br /><br /><br />Further, also as seen by the observer, the cuboid would appear to have shrunk along the <span style="font-weight: bold;">x </span>dimension to <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrd2ZugFvyVu4egvRUlQ8R89kPqq6oGTWaccJkyi6YmwcFHudFM1TMgRHdPM5kAmN60k-yGqe_18HK64hL7r00e0l293uEYd1u4VOwrnOezKeGfZZjHYsH0hhyXVvqwWQAULP-g6E2kCnk/s1600-h/1.369_4.jpg"><img style="cursor: pointer; width: 77px; height: 27px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrd2ZugFvyVu4egvRUlQ8R89kPqq6oGTWaccJkyi6YmwcFHudFM1TMgRHdPM5kAmN60k-yGqe_18HK64hL7r00e0l293uEYd1u4VOwrnOezKeGfZZjHYsH0hhyXVvqwWQAULP-g6E2kCnk/s400/1.369_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5421389218625333554" border="0" /></a>.<br /><br />Let the density of body as seen by the moving observer be <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEie1u2YAgqRclFc9bfXrZ-XViCrL0HN49mwSDOTKEdaMKDNTCFlVNM0UAm7SkOmD5Qw2Fmkf2s5LXQ_1OlujanbGvTXCMNREste-Y2MSMuwcdLbxILx1PXwqE-lVZu6-SkKblnfxdwDwSgD/s1600-h/1.369_5.jpg"><img style="cursor: pointer; width: 13px; height: 23px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEie1u2YAgqRclFc9bfXrZ-XViCrL0HN49mwSDOTKEdaMKDNTCFlVNM0UAm7SkOmD5Qw2Fmkf2s5LXQ_1OlujanbGvTXCMNREste-Y2MSMuwcdLbxILx1PXwqE-lVZu6-SkKblnfxdwDwSgD/s400/1.369_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5421389493457878226" border="0" /></a> then we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiknZjDPMrmWjs9NoJvmOVc6GmKnZ4YOqyQJptT4IZ1t9PTIwQexcpXgBdsYXKyjkuajNk3s1tOwgtRtzKDYEBLZ1n-HeJClWQmzZCXFWbYRN6_b5dafW15aX3Ian4vrzDj2QhXyKIr3TZn/s1600-h/1.369_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 190px; height: 268px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiknZjDPMrmWjs9NoJvmOVc6GmKnZ4YOqyQJptT4IZ1t9PTIwQexcpXgBdsYXKyjkuajNk3s1tOwgtRtzKDYEBLZ1n-HeJClWQmzZCXFWbYRN6_b5dafW15aX3Ian4vrzDj2QhXyKIr3TZn/s400/1.369_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5421389956256303506" border="0" /></a>Unknownnoreply@blogger.com4tag:blogger.com,1999:blog-5997147895880585179.post-88275859798753009592009-12-31T18:18:00.003+05:302009-12-31T18:44:22.685+05:30Irodov Problem 1.368The mass of the moving particle as seen by the observer is given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJH1jFw31HkmxVhHpe6YjDlzol2LTTjw-itliF82YLxzzwijrPt9DexvaT0n6cUH7lMRQXhfxVdvHoULTEr3Ogxa4wbIowAQRTtGfEyd7scOxwrK02fY043Bk15VV4Jgn3f3en1rReR0qp/s1600-h/1.368_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 139px; height: 82px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJH1jFw31HkmxVhHpe6YjDlzol2LTTjw-itliF82YLxzzwijrPt9DexvaT0n6cUH7lMRQXhfxVdvHoULTEr3Ogxa4wbIowAQRTtGfEyd7scOxwrK02fY043Bk15VV4Jgn3f3en1rReR0qp/s400/1.368_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5421387594548085266" border="0" /></a><br /><br /><br /><br /><br /><br />At <span style="font-weight: bold;">v/c = 1 - 0.0001 = 0.9999</span> we have<span style="font-weight: bold;"> m = 70.712m<span style="font-size:78%;">0</span> </span>.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-83665978349271817742009-12-29T06:35:00.003+05:302009-12-29T06:54:25.772+05:30Irodov Problem 1.367This problem is an extension of <span style="font-weight: bold;">1.366</span> and essentially asks, how long did the rocket travel according to an observer in the rocket. As the rocket accelerates and moves faster, time gets progressively slower. Let <span style="font-weight: bold;">t'</span> be the time in the reference frame of the rocket. We know that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuaWVGt06TB4G0TqTQ64ZbYh2VMuFTTR2GqwXP8BuIHZE8hh5onkvc45oA9GYiGyW4PSJSPGGZYwuYOMsn7tSAuip_nht69hjh1tQouEG_sKGEHmsZ-tSzLzpjS9BVPDOnqnZmqhb6ei2c/s1600-h/1.367_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 265px; height: 194px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuaWVGt06TB4G0TqTQ64ZbYh2VMuFTTR2GqwXP8BuIHZE8hh5onkvc45oA9GYiGyW4PSJSPGGZYwuYOMsn7tSAuip_nht69hjh1tQouEG_sKGEHmsZ-tSzLzpjS9BVPDOnqnZmqhb6ei2c/s400/1.367_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5420462836879738978" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-87516128211959371962009-12-29T05:29:00.005+05:302009-12-29T06:33:22.027+05:30Irodov Problem 1.366This problem can be considered as an extension of the problem <span style="font-weight: bold;">1.365(a)</span>. We have to simply replace <span style="font-weight: bold;">V=v</span>, to obtain the relationship between <span style="font-weight: bold;">w'</span> and the acceleration of the rocket <span style="font-weight: bold;">w</span> as seen from Earth. This is because at each instant, the moving reference frame <span style="font-weight: bold;">K'</span> has the velocity <span style="font-weight: bold;">v</span> same as that of the rocket as seen from Earth. Thus, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMLbxUyyjiA3GQZsxwOTD107XLVS7ljst7l7cZE62JIo8HmhRfkumtlPfh66ZHWB8oniMTrGaeDqAUGjTxvChDx0GOMqHSEaM9pwNgLQ20EtimK8TPRotRH_2lWpJA3vJKNm9ZdxEpeeR4/s1600-h/1.366_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 236px; height: 312px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMLbxUyyjiA3GQZsxwOTD107XLVS7ljst7l7cZE62JIo8HmhRfkumtlPfh66ZHWB8oniMTrGaeDqAUGjTxvChDx0GOMqHSEaM9pwNgLQ20EtimK8TPRotRH_2lWpJA3vJKNm9ZdxEpeeR4/s400/1.366_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5420456979161728514" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />The distance traveled by the body can be obtained using <span style="font-weight: bold;">(2)</span> and the velocity of the body is obtained from <span style="font-weight: bold;">(1)</span>. The fraction by which the velocity differs from light speed can be obtained by computing <span style="font-weight: bold;">1-v/c</span>.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-87258809981140693732009-12-29T04:30:00.006+05:302009-12-29T05:22:05.763+05:30Irodov Problem 1.365<span style="font-weight: bold;">a)</span> Let the direction of motion of the frame <span style="font-weight: bold;">K'</span> be assigned as the <span style="font-weight: bold;">x-axis</span>. Let <span style="font-weight: bold;">v'</span>, <span style="font-weight: bold;">x'</span> and <span style="font-weight: bold;">t'</span> be the instantaneous velocity, position and time of the moving body as seen by the observer in frame <span style="font-weight: bold;">K'.</span> Since, the direction of motion of the accelerating body coincides with the motion of the frame <span style="font-weight: bold;">K'</span> , <span style="font-weight: bold;">x -axis</span> and <span style="font-weight: bold;">x' axis</span> coincide. From the relativistic transformations for velocity and time we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzlQxnutfhjwFMJscvO_RTU3txyBp1zrC1OQBvi8lrRejRBothLkNJJhvSFvOjlEDshQhB-vJE4W7zMlj1oYz9W9AxoyfbeWsjbkImJ3e5TOeYTohMudjUsaTNTg6jWty7s4h6YLkm7f-y/s1600-h/1.365_1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 128px; height: 105px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzlQxnutfhjwFMJscvO_RTU3txyBp1zrC1OQBvi8lrRejRBothLkNJJhvSFvOjlEDshQhB-vJE4W7zMlj1oYz9W9AxoyfbeWsjbkImJ3e5TOeYTohMudjUsaTNTg6jWty7s4h6YLkm7f-y/s400/1.365_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5420436554426786354" border="0" /></a><br /><br /><br /><br /><br /><br /><br />The acceleration of the body as seen from frame <span style="font-weight: bold;">K'</span> is then given by,<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvMCKqHmhbW6SAEQYhOtRRrw0VOKltFTDs3A0_JRAwPgo9qeJipOOS1OTNIOgkKavarysVDarhMkFNxlFQX7fHYr0riSZaL-QKxzdot0Qyc8a3_T9ZNVBNJe30W3iX8DtbjJgBnlQ7Y3Zz/s1600-h/1.365_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 179px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvMCKqHmhbW6SAEQYhOtRRrw0VOKltFTDs3A0_JRAwPgo9qeJipOOS1OTNIOgkKavarysVDarhMkFNxlFQX7fHYr0riSZaL-QKxzdot0Qyc8a3_T9ZNVBNJe30W3iX8DtbjJgBnlQ7Y3Zz/s400/1.365_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5420435070424151058" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><span style="font-weight: bold;">(b) </span>In this problem the body moves along the <span style="font-weight: bold;">y-axis</span> while the frame <span style="font-weight: bold;">K'</span> moves along the <span style="font-weight: bold;">x-axis</span>. Thus, using relativistic transformations we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhx0GvySATzxXW2TNEAgD_hgzh_2AOcx89E1KdacsFiJFIcUUpFP3YVS7TDcK41xKL1oiv2LvRoqUY1F3VbYVdLEWwxh3nohd5tkOoCrjcYUah8rG_e1jYr6pJCxJo7Rzg4jtIGLt14Nl-F/s1600-h/1.365_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 151px; height: 46px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhx0GvySATzxXW2TNEAgD_hgzh_2AOcx89E1KdacsFiJFIcUUpFP3YVS7TDcK41xKL1oiv2LvRoqUY1F3VbYVdLEWwxh3nohd5tkOoCrjcYUah8rG_e1jYr6pJCxJo7Rzg4jtIGLt14Nl-F/s400/1.365_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5420437120261007570" border="0" /></a><br /><br /><br /><br />The acceleration of the body as seen in frame <span style="font-weight: bold;">K'</span> is then given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizox08EYspNpCQtkLMSoRELAEfWah5clNpmYyeVU75kb3S5NB-vb4T0rLfraDhi7N37owO43yk6Cx4LbE5pkEW8YQopNAX8p9L7tFkA4rVf9f3WSkUFnztTvl5_bDhhLB2gyqGizjHuZWv/s1600-h/1.365_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 188px; height: 400px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizox08EYspNpCQtkLMSoRELAEfWah5clNpmYyeVU75kb3S5NB-vb4T0rLfraDhi7N37owO43yk6Cx4LbE5pkEW8YQopNAX8p9L7tFkA4rVf9f3WSkUFnztTvl5_bDhhLB2gyqGizjHuZWv/s400/1.365_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5420439018357636530" border="0" /></a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-5997147895880585179.post-52103772369941854082009-12-27T06:00:00.011+05:302009-12-27T07:48:33.664+05:30Irodov Problem 1.364<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIoocYqCoz-mlwuTx877xSHNs63ivBPOo9352bmUl5aUIZbd9YTeU_BUz4in0hzDK3oloohZSw5bAfnfftN7adg_42_WnUOz11L7pSe_F_aQGNoMmK4iBfaX2oOlPfx5qjTlU7kJ7f0iiI/s1600-h/1.364_fig1.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 276px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIoocYqCoz-mlwuTx877xSHNs63ivBPOo9352bmUl5aUIZbd9YTeU_BUz4in0hzDK3oloohZSw5bAfnfftN7adg_42_WnUOz11L7pSe_F_aQGNoMmK4iBfaX2oOlPfx5qjTlU7kJ7f0iiI/s400/1.364_fig1.jpg" alt="" id="BLOGGER_PHOTO_ID_5419713929853723138" border="0" /></a><span style="font-weight: bold;">As seen by observer O' :</span><br />Without loss of generalization let us assume that observer <span style="font-weight: bold;">O'</span> is at the origin (<span style="font-weight: bold;">x'=0,y'=0,t'=0</span>). <span style="font-weight: bold;">x'</span>,<span style="font-weight: bold;">y'</span> and <span style="font-weight: bold;">t'</span> being the coordinates in frame <span style="font-weight: bold;">K'</span> as seen by <span style="font-weight: bold;">O'</span>. As seen by <span style="font-weight: bold;">O'</span>, the rod moves only along the <span style="font-weight: bold;">y'</span> axis at a speed <span style="font-weight: bold;">v'</span>. In other words <span style="font-weight: bold;">y'=v't'</span>. The rod however, has no motion along the <span style="font-weight: bold;">x'-axis</span>. Further assume that the rod is places such that the <span style="font-weight: bold;">x'</span> coordinate of its mid point is at <span style="font-weight: bold;">x'=0</span> (this only makes solving the problem easier but does not fundamentally change the result).<br /><br />At time <span style="font-weight: bold;">t'=0</span>, <span style="font-weight: bold;">O'</span> sees the two ends of the rod and finds them perfectly parallel to the <span style="font-weight: bold;">x'-axis</span>. The ends of the rod <span style="font-weight: bold;">A</span> and <span style="font-weight: bold;">B</span> as seen by <span style="font-weight: bold;">O'</span> are actually due to two different rays of light that started some times in the past <span style="font-weight: bold;">t</span><span style="font-weight: bold;">'</span><span style="font-weight: bold;font-size:78%;" >A</span> and <span style="font-weight: bold;">t</span><span style="font-weight: bold;">'</span><span style="font-weight: bold;font-size:78%;" >B</span> and both reached <span style="font-weight: bold;">O'</span> simultaneously at <span style="font-weight: bold;">t'=0</span>. Since, the rod is not moving along the <span style="font-weight: bold;">x'-axis</span>, and <span style="font-weight: bold;">O '</span> saw the rod to be parallel to the <span style="font-weight: bold;">x'-axis</span>, both these times must have been exactly the same i.e. <span style="font-weight: bold;">t'<span style="font-size:78%;">A</span> = t'<span style="font-size:78%;">B</span></span>.<br /><br />The space-time diagram of the entire situation is depicted in the figure beside. The world-line of the rod is a plane tilted at an angle to the <span style="font-weight: bold;">x'y'</span> plane. The world-lines of the ends of the rod only change their <span style="font-weight: bold;">y',t'</span> coordinates but not their <span style="font-weight: bold;">x'</span> coordinates. The light rays are emitted along the surface of a cone but we depict only those rays that reach the observer <span style="font-weight: bold;">O'</span>. These rays of light move at <span style="font-weight: bold;">45 degrees</span> to the <span style="font-weight: bold;">x'y'</span> plane but lie entirely in the world-line of the rod.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJBTTkwcxLSFaRB0Jss-hTJbPHNMRkta50j-BB7q4PQrkJR-BWiOICWNe0iR6w72MUfRfM0wAm9_0DVjxYxbNkRF3AVUlY1bIUPV6p_ADk_a21bsTilvrXdzx0njeMpaPz3hT6eORMfi7I/s1600-h/1.364_fig2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 400px; height: 276px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJBTTkwcxLSFaRB0Jss-hTJbPHNMRkta50j-BB7q4PQrkJR-BWiOICWNe0iR6w72MUfRfM0wAm9_0DVjxYxbNkRF3AVUlY1bIUPV6p_ADk_a21bsTilvrXdzx0njeMpaPz3hT6eORMfi7I/s400/1.364_fig2.jpg" alt="" id="BLOGGER_PHOTO_ID_5419721748442109682" border="0" /></a><br /><span style="font-weight: bold;">As seen by observer O :<br /></span><span>Now lets us see the same </span><span>phenomenon through the eyes of an observer <span style="font-weight: bold;">O</span> located in Frame <span style="font-weight: bold;">K</span> at the origin i.e. (<span style="font-weight: bold;">x=0,y=0,t=0</span>), where <span style="font-weight: bold;">x,y,t</span> are the space time coordinates of frame <span style="font-weight: bold;">K</span>. Let us assume that the origins of frames <span style="font-weight: bold;">K</span> and <span style="font-weight: bold;">K'</span> coincide. As seen by <span style="font-weight: bold;">O</span>, the rod moves along the <span style="font-weight: bold;">x-axis</span> with a speed <span style="font-weight: bold;">V</span>. It also moves long the <span style="font-weight: bold;">y-axis</span> with a speed </span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5axg7k2LmjisdC7RN3RfcvHPMkWbShcYuvUPVgmj-TUK7eAVwbk8o8OjPGgn4ihqC122UJGlw2vT7ol3cyUxgkfm72wmNC_aITK73DZBljv02FYDPIfWwBHNPQyDeo1p8l0OeUzWSK9RK/s1600-h/1.364_1.jpg"><img style="cursor: pointer; width: 104px; height: 28px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5axg7k2LmjisdC7RN3RfcvHPMkWbShcYuvUPVgmj-TUK7eAVwbk8o8OjPGgn4ihqC122UJGlw2vT7ol3cyUxgkfm72wmNC_aITK73DZBljv02FYDPIfWwBHNPQyDeo1p8l0OeUzWSK9RK/s400/1.364_1.jpg" alt="" id="BLOGGER_PHOTO_ID_5419719740127408818" border="0" /></a>(<span style="font-weight: bold;">v</span><span style="font-weight: bold;font-size:78%;" >y</span> is less than <span style="font-weight: bold;">v'</span> because time <span style="font-weight: bold;">t'</span> in frame is running slower as seen by <span style="font-weight: bold;">O</span>, in other words the same distance along the <span style="font-weight: bold;">y-axis</span> is covered in a shorter time when seen in frame <span style="font-weight: bold;">O</span>. The same can also be derived the same way as it was derived in Problem <a style="font-weight: bold;" href="http://irodovsolutions.blogspot.com/2009_11_21_archive.html">1.358</a>) . In other words, the coordinates of the center of the rod are given by,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgl0t06pRY1REnGfG19IPycCLK-hsgvjilABYtNfqlAA3rRC9rLAhMPAqiqkXe4TgX53Bdd6KH84k8Fg8cF7EVO9S0Z-DLTA1Sv3PmuIoGEUi2tIcpmjmeVVmDoCef2pGTQzU5mrwZzO0Qz/s1600-h/1.364_2.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 136px; height: 65px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgl0t06pRY1REnGfG19IPycCLK-hsgvjilABYtNfqlAA3rRC9rLAhMPAqiqkXe4TgX53Bdd6KH84k8Fg8cF7EVO9S0Z-DLTA1Sv3PmuIoGEUi2tIcpmjmeVVmDoCef2pGTQzU5mrwZzO0Qz/s400/1.364_2.jpg" alt="" id="BLOGGER_PHOTO_ID_5419720298167766514" border="0" /></a><br /><br /><br /><br /><br />The world-line of the rod as seen by <span style="font-weight: bold;">O</span> in frame <span style="font-weight: bold;">K</span> is depicted in the figure. The light rays traveling from the ends of the rod reach the origin at the same time but unlike O', O arrives at different conclusion, namely the light ray that left end <span style="font-weight: bold;">A</span> started at a time <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >A</span>, that is before the time <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >B, </span>when the ray from end <span style="font-weight: bold;">B</span> started its journey<span style="font-weight: bold;font-size:78%;" ></span>. This also means that the end <span style="font-weight: bold;">A</span> was a little behind along the <span style="font-weight: bold;">y-axis</span> than the end <span style="font-weight: bold;">B</span>, in other words the rod was not parallel to <span style="font-weight: bold;">x-axis</span>.<br /><br />Now let us compute the angle the rod is at to the <span style="font-weight: bold;">x-axis</span> as seen by <span style="font-weight: bold;">O</span>. First we will start by computing the difference <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >A</span> - <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >B.<br /><br /></span>Let the length of the rod as seen in frame <span style="font-weight: bold;">K</span> be <span style="font-weight: bold;">l</span>. Thus, both the rays of light have to cover a distance of <span style="font-weight: bold;">l/2</span> along the rod.<br /><br />As we trace back the light ray starting from end <span style="font-weight: bold;">A</span> in time, the ray moving at speed <span style="font-weight: bold;">c</span> has to "catch up" with the moving rod at it moves at a rate <span style="font-weight: bold;">V</span> to the left. This means that,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-sORO-k_jEBVb60KXvQuVF3PSWI8WYC8XkhS2xCu8zdSgklaHeiskdS-1bqGmI1uqU4rahp5T3ZOTQvBSjB3HlV1BC6f0aUVtI6OLjWftAiDaqyFkYAOHmwWNPg5_kjYv__xyv3qKCbTV/s1600-h/1.364_3.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 116px; height: 40px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-sORO-k_jEBVb60KXvQuVF3PSWI8WYC8XkhS2xCu8zdSgklaHeiskdS-1bqGmI1uqU4rahp5T3ZOTQvBSjB3HlV1BC6f0aUVtI6OLjWftAiDaqyFkYAOHmwWNPg5_kjYv__xyv3qKCbTV/s400/1.364_3.jpg" alt="" id="BLOGGER_PHOTO_ID_5419725397198123426" border="0" /></a><br /><br /><br />The negative sign comes because, the time <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >A </span>is in the past to t=0.<br /><br />The situation is different when we trace back to find out when the ray from rod <span style="font-weight: bold;">B</span> started its journey. In this case, the rod is aiding the ray in catching up by moving towards it at a rate <span style="font-weight: bold;">V</span>. Hence, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh49SuL538klvjpp45-u3REYp0yzk5N-A1IH61XO2_AwnVOisIj602qPBIFAUXUbVYspnFDygRx_B1bLNB0il8Bhtv66AJQZYBsGCvA04HJJf_g62G5omZ6iQ6t_99Ore7nvMxb2mbdNuDa/s1600-h/1.364_4.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 114px; height: 42px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh49SuL538klvjpp45-u3REYp0yzk5N-A1IH61XO2_AwnVOisIj602qPBIFAUXUbVYspnFDygRx_B1bLNB0il8Bhtv66AJQZYBsGCvA04HJJf_g62G5omZ6iQ6t_99Ore7nvMxb2mbdNuDa/s400/1.364_4.jpg" alt="" id="BLOGGER_PHOTO_ID_5419726360427040018" border="0" /></a><br /><br /><br />Thus, we have,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiXvVnBL3u63wjwbKs37xFMcXV03r2_FQqM7gF2h-IHeOHN5SUU-8OS8FkXQ4LynmaDTmB0jGG3GhBtA1OrSibPZUAOkGmAqn2gYaDs_Gdx1ZmlJDy5sFIcUSzjMFvzsMQ5h_NGVIv0P4qp/s1600-h/1.364_5.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 128px; height: 44px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiXvVnBL3u63wjwbKs37xFMcXV03r2_FQqM7gF2h-IHeOHN5SUU-8OS8FkXQ4LynmaDTmB0jGG3GhBtA1OrSibPZUAOkGmAqn2gYaDs_Gdx1ZmlJDy5sFIcUSzjMFvzsMQ5h_NGVIv0P4qp/s400/1.364_5.jpg" alt="" id="BLOGGER_PHOTO_ID_5419727590856996274" border="0" /></a><br /><br /><br />During this time interval <span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >B </span><span style="font-weight: bold;">- </span><span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >A </span>the end <span style="font-weight: bold;">B</span> moved a distance of <span style="font-weight: bold;">v</span><span style="font-weight: bold;font-size:78%;" >y</span><span style="font-weight: bold;">(</span><span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >B </span><span style="font-weight: bold;">- </span><span style="font-weight: bold;">t</span><span style="font-weight: bold;font-size:78%;" >A</span><span style="font-weight: bold;">) </span>farther along the <span style="font-weight: bold;">y-axis</span> than end <span style="font-weight: bold;">A</span>.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjLE26I0D3sgU9fLRt-WC3bkm_l83XprZVaOygk0kEW1NCpBKgkJ3ltAZtLmhO5Gn1lwiBvwZzd9fhH_5i9qcAIPD-p4TGcvuGX43vyHOEy7nCKQBC7kJn4YghgAVmk_DOeJjfI0WPepxVc/s1600-h/1.364_6.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 334px; height: 186px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjLE26I0D3sgU9fLRt-WC3bkm_l83XprZVaOygk0kEW1NCpBKgkJ3ltAZtLmhO5Gn1lwiBvwZzd9fhH_5i9qcAIPD-p4TGcvuGX43vyHOEy7nCKQBC7kJn4YghgAVmk_DOeJjfI0WPepxVc/s400/1.364_6.jpg" alt="" id="BLOGGER_PHOTO_ID_5419732601333023618" border="0" /></a><br /> As shown in the figure thus, we can compute the angle to the <span style="font-weight: bold;">x-axis</span> as,<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGfFxITANDCAGI2Ie9C4bQXwEBVUHJBFUOGfXqGCkBT7ghg0wet5kJH_hMt48kmlM8wOHeP1-qyxmHoC8jD67Rz92twmRS5sYHGAGGY3eesnBVdhPXDJJ1PyyvaspiW5kBLRE9aziQa_Bo/s1600-h/1.364_7.jpg"><img style="cursor: pointer; width: 171px; height: 98px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGfFxITANDCAGI2Ie9C4bQXwEBVUHJBFUOGfXqGCkBT7ghg0wet5kJH_hMt48kmlM8wOHeP1-qyxmHoC8jD67Rz92twmRS5sYHGAGGY3eesnBVdhPXDJJ1PyyvaspiW5kBLRE9aziQa_Bo/s400/1.364_7.jpg" alt="" id="BLOGGER_PHOTO_ID_5419734560728749602" border="0" /></a>Unknownnoreply@blogger.com0