What will be the ball's velocity after a perfectly elastic collision?

[Lotto]

TL;DR Summary

I throw a ball with a velocity ##v## against a bus moving towards me with a velocity ##-u##. What will be the ball's velocity after a perfectly elastic collision?

From the bus driver's point of view, who is at rest, the ball's initial velocity is ##u+v##. After the collision, its velocity has to have the same value, but an opposite direction, so ##-(u+v)##. So that means that relative to me standing on the ground at rest, the ball's new velocity is ##-(v+2u)##.

I understand this, but how to derive the new velocity from my point of view? I still can't explain why it is ##+2u## instead of ##+u##, I mean that the new velocity isn't##-(u+v)##. How to explain it? It is clear that it is because of the moving bus, but why in particular?

 

[Dale]

You can just write down the conservation of momentum $$mv-Mu=mv_f-Mu_f$$ and the conservation of kinetic energy $$\frac{1}{2}mv^2+\frac{1}{2}Mu^2=\frac{1}{2}mv_f^2+\frac{1}{2}Mu_f^2$$ That is two equations in two unknowns, so you can solve it to get $$u_f=\frac{Mu-m(u+2v)}{m+M}$$$$v_f=\frac{mv-M(2u+v)}{m+M}$$ Then we simply take the limit as ##M\rightarrow \infty## to get $$u_f=u$$$$v_f=-2u-v$$

 

[Frabjous]

You can do the first case in your head because you know the solution for an object hitting a stationary infinite mass. You do not know the solution for a moving object hitting a moving infinite mass, so you have to go back to first principles and use conservation of momentum and energy for the second case. This is an example of why choosing a “smart” frame of reference can make solving a problem easier.

 

[A.T.]

I understand this, but how to derive the new velocity from my point of view? I still can't explain why it is ##+2u## instead of ##+u##, I mean that the new velocity isn't##-(u+v)##.

Can you explain why it should be ##-(u+v)##? It doesn't make any sense, because for ##v=0## the ball would just remain stuck to the bus, which isn't an elastic collision.

To check solutions it often helps to consider limiting / extreme cases.

 

[Juanda]

Not very long ago there was a heated thread about gravity-assisted maneuvers where something very similar came up. This might help you.

From the article

Another analogy, illustrated by the cartoon above, involves a moving railroad train that represents Jupiter, moving along its track about the Sun. The kid in the propeller beanie throws a tennis ball that represents a spacecraft. It encounters the train, which transfers its momentum into the ball.

It's interesting to note the speeds in the cartoon. The propeller-beanie kid sees his tennis ball moving away from him at 30 miles per hour. So does the Sun, sitting on the stationary platform. The engineer driving the train sees the ball coming at about 80 MPH, since the train is moving 50 MPH with respect to the ground. The train and ball interact at 80 MPH. The ball rebounds from the front of the train at nearly the same 80 MPH, which can be added to the 50 MPH speed of the train, because it acquired it from the train. The result approaches a total of 130 MPH.

https://solarsystem.nasa.gov/basics/primer/
Credit to @A.T. who originally shared the source. Well, and credit to the NASA guys who wrote it of course.