# Elastic collision against a moving wall

AlexmBIOSS
Hello,
A very basic question, with a bit of confusion.
a ball hitting a wall which is not moving with velocity v in an elastic collision has its momentum conserved, opposite velocity; -v.
if the ball is to collide with a wall which has a velocity -v, equal speed towards the ball, what is the return velocity for the ball of mass m?

Can we just say that the ball is moving relative to the wall and has a velocity of 2v who's momentum will be conserved? Or is there some other relationship?

Alex M.

Mentor
If the ball bounces off the wall, its momentum' is not conserved. In an elastic collision, energy is conserved.

Hint: Transform to a frame in which the answer is obvious. Then transform back to the original frame.

robertm
If the ball bounces off the wall, its momentum' is not conserved. In an elastic collision, energy is conserved.

Hint: Transform to a frame in which the answer is obvious. Then transform back to the original frame.

Obvious as a "Wall"... hint hint...

AlexmBIOSS
"An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed. This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterward."

So the momentum for the ball is conserved. v=v and v'=-v
But what effect does the wall's movement have on the end velocity?
This is a question of curiosity, not for any school assignment. A change of frame just creates a linear relationship where the wall's velocity is additive to that of the ball.

Mentor
"An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed. This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterward."
Sure, if you include the entire system--ball + wall (and attached planet)--then momentum is conserved.
So the momentum for the ball is conserved. v=v and v'=-v
The momentum of the ball itself is not conserved. Recall that momentum is a vector. The ball switches direction. (Sure the magnitude of the ball's momentum doesn't change.)
But what effect does the wall's movement have on the end velocity?
This is a question of curiosity, not for any school assignment. A change of frame just creates a linear relationship where the wall's velocity is additive to that of the ball.
Assuming the ball is a normal ball (not something the size of the moon), then you realize that in an elastic collision with the wall that all of the kinetic energy remains with the ball--to a very close approximation. (Of course the wall and planet do move a teensy little bit--way too small to notice.)

Do as I suggest: Transform to a frame in which the wall is at rest. Solve the problem. Then transform back. Just do it!

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mlohbihler
I know I'm coming late to this party, but i have a similar problem to solve and ended up here. I picked up some starting equations from Wikipedia, which i am trusting at the moment, but they look good at first glace. These are for a one dimensional collision where m1, u1, and v1 are the mass, the velocity before collision, and the velocity after collision of first object, and m2, u2, and v2 are the analogous for the second object. m and u are given, and v is what we are trying to find out. The equations are:

m1u1 + m2u2 = m1v1 + m2v2

... and...

v1 - v2 = u2 - u1

So, when we solve for v1 and v2, we end up with this:

v1 = (m1u1 + 2m2u2 - m2u1) / (m1 + m2)
v2 = (m2u2 + 2m1u1 - m1u2) / (m1 + m2)

It's been a while since i had to do any heavy equation solving, but the result is satisfyingly mirrored, so i'll assume it's right and proceed. So, the thing about the "wall" is that it's mass is assumed to be effectively infinite, such that it is not moved by the collision. Using the concept of a limit, we can observe what happens to the equations as m1 approaches infinity. We end up with this:

As m1 --> infinity and m2 remains constant:
v1 = u1
v2 = 2u1 - u2

So, we see that the velocity of the wall is unchanged by the collision, which is what we would expect.

Ok, let's say u1=3 and u2=-2. If we transform to a frame where the wall is at rest (a'=a-3 where a is any variable), we get u1'=0 and u2'=-5. For a wall at rest we know v2=-u2, so v2'=5. Transforming back (a=a'+3) we get v2=8. This is the same result as when we use the above equation v2 = 2u1 - u2 = 2*3 + 2 = 8.

Ok, so all i did was just back up Doc Al, but i thought the process was interesting, and wanted to share.

Cheers,
m@

Bob_for_short
"An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed. This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterwards."

So the momentum for the ball is conserved. v=v and v'=-v
But what effect does the wall's movement have on the end velocity?
This is a question of curiosity, not for any school assignment. A change of frame just creates a linear relationship where the wall's velocity is additive to that of the ball.

If you are in a reference frame (RF) where the wall is moving, its potential of interaction is time-dependent (a step-wise form). Any system with a time-dependent potential is not conservative - the energy and momentum do not conserve.

In a RF where the wall is still you can apply the conservation laws. There the energy conserves and the momentum changes the sign (does not conserve due to presence of a force!).

TurtleMeister
m1u1 + m2u2 = m1v1 + m2v2

... and...

v1 - v2 = u2 - u1

So, when we solve for v1 and v2, we end up with this:

v1 = (m1u1 + 2m2u2 - m2u1) / (m1 + m2)
v2 = (m2u2 + 2m1u1 - m1u2) / (m1 + m2)

I used the same equations from Wikipedia to solve a problem involving the collision of two vehicles in this thread: https://www.physicsforums.com/showthread.php?p=2304842#post2304842