Question

A tennis ball is thrown against a wall and rebounced. M(mass of wall)*v (for ball) =2*m(mass of ball) * V(for wall), KE(wall)=1/2 * Mv²; KE(in)=1/2 *mV² and

KE(out)=1/2 *mV²+1/2*Mv² ; If KE(in)=KE(out), Where did 1/2*Mv² go? Is the KE conderved? Prove it.

Answer 1

The scenario put by you is that of a collision between the wall and the ball.

In general, we can view this as head on collision between a small and a massive object.

now your query is

where did 1/2 M v2^2 go

1/2 M v2^2 = 0 (its KE of wall after collision)

Did the wall move? No it will not move (you must have banged ur ball multiple times into the wall, but the wall must have never moved) So the kinetic energy of wall is zero.

But Yes still the total energy of the system is conserved, as the ball rebounce back with the same speed but in apposite direction, so the KE of the system is conserved.

Lets prove this mathematically,

Let mass of ball be m and wall be M, v1 be the intial velocity of the ball and the wall is ofcourse at rest.

We know that the final velocities, in case of elastic collison, target at rest are

v1' = (m - M)v1/(m + M) ; v2 = 2mv1/(m + M)

We know that M >>> m, so when we solve the eqn we get

v1' = -v1 and v2 = 0

Intial KE = 1/2 m v1^2 (that of the ball only as wall is at rest)

Final KE = 1/2 m (-v1^2) + 1/2 m v2^2 = 1/2 m v1^2 + 0 = Intial KE

Hence it is proved that the energy is conserved.