Irodov Problem 1.388

Suppose that a certain instant of time t the velocity of the rocket is v and its rest mass m. Suppose that the rocket ejects gas of rest mass k units/sec. Consider another instant of time t + dt, where dt is an infinitesimally small interval of time. The rocket has ejected kdt amount of rest mass. Suppose that the rocket's velocity has increased by an amount dv during this interval. Since there is no external force on the rocket, the momentum of the (ejected gas + rocket) system at t must be equal to that at t+dt.

The momentum of the rocket at time t is given by . The momentum of the rocket at t+dt is given as . The gas ejected by the rocket has a of mass and moves at a non-relativistic velocity u 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 v-u in the direction of the rocket (this is shown in the diagram). Hence, the momentum of the ejected gas will be .

Applying conservation of momentum at t and t+dt we have,






We can now use Taylor series expansion and retain only the first order terms in Eqn (1) to obtain,
















Now we know that the mass of the rocket reduces at a rate k, in other words -kdt = dm. Hence, (2) can be rewritten as,

Irodov Problem 1.387

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
pieces could move along three different directions at velocities v1,v2 and v3
as shown in the figure 1. In that case we have that,














From Eqn (1) 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 v1 to lie along the negative x-axis. This is depicted in figure 2.





















Now we can simplify (1) as,










What choice of velocities and directions will maximize the energy if particle 1? To determine this answer we can use Lagrangian Multipliers to minimize the objective function,












The conditions for local extrema are then obtain by setting the partial derivatives of J with respect to each of the unknowns to zero. Hence at extrema we have,
























Eqn 5 essentially says that in order for the energy of piece 1 to be maximum, both the the pieces 2 and 3 must fly off in the same direction and propel piece 3 in the opposite direction.

Now differentiating with respect to v2 and v3 we have,





















Eqn 5 and 7 together essentially say that the energy of piece 1 will be maximized when pieces 2 and 3 move at the same velocity and the same direction - in other words they move together. This situation is depcited in Figure 3. Let v2=v3=v.















Now equations (1) and (2) can be rewritten as,














From (8) we obtain,















From (9) we obtain,











Eqn (12) given the maximum energy that piece 1 can possible have after the disintegration.

Irodov Problem 1.386

Let the kinetic energy of the single particle be T'. In Problem 1.384 we have already determined the combined kinetic energy of the particles as,




Hence, we need to find T' such that,

Irodov Problem 1.385

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.

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!

Throughout the collision two important properties will always hold, i) the momentum will be conserved and ii) the total energy will be conserved. Let the final rest mass of the fused particle be . Let the initial velocity of the left particle be u and let v be the velocity of the combined particle after collision.

Since we know that the kinetic energy of the particle before collision is T we have,









Conservation of Momentum
Since there are no external forces on the system, its total momentum must be conserved and so we have,










Conservation of Energy
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.




From Eqn (3) and (4) we obtain,





Now we can substitute (5) in (4) to obtain,

Irodov Problem 1.384

As seen from the stationary frame of reference (in which N2 is stationary), neutron N1 is on a collision course towards neutron N2 and has a kinetic energy of .The center of inertia of the (N1+N2) system (CG) lies in between N1 and N2 and moves with velocity vCG. As seen from the reference frame of the CG, both neutrons N1 and N2 move towards the CG at equal velocities. This is shown in the Figure above.

The total energy of N1 in the stationary reference frame is given by,



We can now compute the velocity v of neutron N1 as,









Hence, the momentum of neutron N1 and hence the entire system (N1 + N2) and hence, the momentum of the CG as seen from the staionary reference frame is given by,








The total energy of the entire system (N1 + N2) and hence that of the CG, as seen from the stationary reference frame is given by the sum of kinetic energy of the neutron N1 and the sum of the rest mass energies of each of the neutrons N1 and N2. Thus, the total energy in the system in the stationary reference frame is given by,



The invariant quantity can then be computed as,







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 (CG). The momentum of the CG is 0 in the reference frame of the CG. Hence the energy of the CG, ECG can be computed as,







The kinetic energy of the system as viewed from a reference frame CG can be obtained by subtracting the rest mass energies of the two protons from total energy as,







As seen from the reference frame of the CG, both neutrons will have the same energy and hence then energy of each neutron in the CG reference frame will be,




We also know from problem 1.383 that for each neutron,








Here, is the momentum of the neutron N1 in the reference frame of the CG.

b) The total rest mass of the CG is 2mo. Hence, at velocity VCG, its energy is given by . We can thus, compute the velocity of the CG as,

Irodov Problem 1.382

The momentum of the photon is given by,



From the solution found in Problem 1.381, we can compute the energy of the photon as seen in reference frame K' as,

Irodov Problem 1.383

We can use the results from Problem 1.381 to prove this invariance result.










What Eqn (1) is indicates is an invariance since the quantity does not depend which inertial frame you are observing from!

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 P=0. The energy is equal to the rest mass energy and so we have,

Irodov Problem 1.381

From the relativistic transformations of velocities, energy and momentum we have,

Irodov Problem 1.380

a) For this problem let us denote v as the magnitude of the instantaneous velocity of the particle and 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 .

From the fundamental laws of dynamics we can write,


















Similarly, we have,







In order for (1) and (2) be parallel there are only two possible ways

1. , 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. F is parallel to velocity vector.

2. , 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 F is perpendicular to the velocity vector.

b) The solution to this part is quite evident from Eqns (1), and (2) considering each of the two cases separately.

Case 1: F is parallel to velocity vector













Case 2 : F is perpendicular to velocity vector