Irodov Problem 1.32

Rather than numbers I will solve this problem in greek alphabet. Let V0 be the initial velocity,
and d be the distance to target. Let initial angle at which the cannon ball was shot be . Then the range of the cannon ball must be exactly equal to d. From previous problems we know that,


















So in general there two possible angles which can be used to hit the target. Thus, we have,







Now we can determine the time taken by the cannon ball to reach the target based on fact that the total time of flight for the cannon ball is,







Thus, there are two possible values of time depending on + or - sign. Note that the expressions and the derivations of total time of flight and range of a projectile can be found in any std physics textbook that describes projectile motion. So I did not bother to explain it here.

Irodov Problem 1.31

The entire event is depicted in the figure. The ball first falls from a height h on to the inclined plane and then reflects off of the plane. The ball will hit the plane at an angle and thus will at an angle that forms an equal angle with the normal to the plane. Thus, the ball will reflect off of the inclined plane at an angle of with the horizontal as shown in the figure.

Let the point of contact of the ball with the inclined plane for the first time be the origin in our Cartesian coordinate system and let t=0. Then the position of the ball after reflection at any time is given by,






Here V0 is the velocity at first impact on the inclined plane.

Now this ball must meet the inclined plane. Any point on the inclined plane satisfies the relation


This can be seen from the figure drawn beside.

Thus where ever the ball meets the plane, the inclined plane relation must also hold true. In other words,









Now knowing the time we can calculate the x coordinate of where the ball hits the inclined plane as,




Now the distance l traveled by the ball along the inclined plane is given as,






Since the ball fell a height h before hitting the inclined plane, it must have attained a velocity of . This relation can be obtained by either equating the potential energy lost by the balls (mgh) during the fall and equating to its kinetic energy upon impact (½ mv₀²) . Now substituting this value of initial velocity in to our found value of l we have,

Irodov Problem 1.30

We know that,





The angle between v and w is given as,
















Having found the tangential component, now we can find the normal component knowing the magnitude of the acceleration is g.













In the above equation we choose the -ive sign since we already know that g is acting in the -ive direction and cannot have a projection along the normal to the velocity. Thus,





The velocity vector is directed along the tangent to the path of the particle, thus,











These curves can now be plotted

Irodov Problem 1.29

(a) time of flight can be found in any textbook explaining projectile motion and there is no need to explain it here.
(b) the range and the max height can also be found in any basic physics text book that explains projectile motion and I donot think I need to solve them here. For the angle at which maximum height is equal to range we have,







(c) The equation of motion of the particle is given by,










(d)I will provide two complete different ways to determining the radius of curvature, i) a purely mathematical approach and ii) a purely physics based approach.

The purely mathematics based approach: The raidus of curvature of a planar curve in Cartesian coordinates is given by,





Here x' and y' are first derivatives and x'' and y'' are second derivatives. To find out how (1) came about please go here. I will not explain this since we focus on physics not math.

At time t = 0,







Max height occurs at time. At this time,








The purely physics based approach:
A more physics way to find the radius of curvature is to use the force diagram of the particle.

Basically the idea is that at any point the moving particle experiences a centrifugal force (a pseudo force invented for convenience) that trying to push the particle in a direction perpendicular to its line of motion. This centrifugal force is to be balanced by gravity.

Initially the body moves at an angleto the horizontal at speed V0 and so the centrifugal force acts perpendicular to this and is equal to. Here, ofcourse the radius of curvature of the particle is unknown and needs to be determined. The component of gravity that opposes the centrifugal force, as shown in the figure, is given by, . Since we know that the particle is not acceleration along this direction,







Using the same explanation as above for the case when the particle is at its maximum height we have,







This is exactly what we obtained using the purely math approach!!! The use of pseudo force definitely makes things very easy dosen't it!!

Irodov Problem 1.28

(a) Let g be the acceleration vector of the body due to gravitation, then in vector notation, the displacement of the body with respect to time is simply given by, r(t) = v₀t + ½ gt².
(b) The mean velocity vector is simply displacement over time r(t)/t = v₀ + ½ gt (1).
As seen in the figure, the cosine of the angle between the vector g and the vector v₀ , is given by,




Note that the cosine may be a negative quantity since V0 is actually directed above the horizontal in the diagram while g is pointing downwards.
The component of v₀ along the direction of g is then given by




Note that this value of itself will be negative since the cosine is negative but it will directed towards g, in other words the negative value will indicate that it is directed away from g.

The component of velocity in the direction of g will be become zero when,






The small object will take double this time to reach back to the horizontal, thus, the total time the body will travel is .

Now, knowing the total time the body traveled, we can use (1) to calculate the value of as,