Velocity of a ball bounced off by a train I saw this question Velocity of a ball rebounding off a moving train, but I would like to emphasize a different case, which is quite close though:
A bouncy ball is at rest on the railroad, and a train will hit it at $200 \ \text{mph}$ mph. What is the final velocity of the ball?
I'd have said it's also $200 \ \text{mph}$, but in my exercise book, it is given with $400 \ \text{mph}$ which, for me, doesn't make sense.
   
 A: The speed is $200 \ \text{mph}$ relative to the train, and since the train is travelling at $200 \ \text{mph}$, the speed relative to the tracks is $400 \ \text{mph}$.
If the speed of the ball were $200 \ \text{mph}$ relative to the tracks, then it would have to be stuck to the front of the train.
A: In a collision which conserves energy, in the center of mass system, the two masses must leave the center of mass with the same speed that they had while approaching.  Assuming the center of mass pretty much moves with the train, then the ball approaches with a relative speed of (v) and must leave with the same relative speed.  Add that to the speed of the center of mass and you get a speed of (2v) relative to the ground.
A: The collision between the ball and the train can be considered to be a one-dimensional elastic collision. In an elastic collision, both the momentum and the kinetic energy are conserved.
Let $m$, $u$ and $v$ represent the mass, initial velocity and final velocity.
For an elastic collision, the final velocity of the body is given by:
$$v_1 = \frac{(m_1 - m_2) \ u_1}{m_1+m_2}+\frac{2m_2 u_2}{m_1+m_2}$$
Where subscripts 1 and 2 are for body 1 and body 2.
Now using the same equation for determining the velocity of the ball.
$$v_b = \frac{(m_b - m_t) \ u_b}{m_b+m_t}+\frac{2m_t u_t}{m_b+m_t}$$
Where the subscripts $b$ and $t$ are for the ball and train, respectively.
After collision, the velocity of the train remains almost the same. The initial velocity of the ball was zero. $m_b$ can be ignored compared to $m_t$.
$$v_b = 0 + \frac{2m_t u_t}{m_t}$$
$$v_b = {2 u_t}$$
A: Throw an idealized ball at 200 mph at a stationary train and it will bounce off at 200 mph. Of course this assumes the balls mass is insignificant when compared to the trains mass.1
If you can accept that then understand that it's no different if you throw the train at the ball. Why? Because neither the train nor the ball care what the ground is doing since the ground isn't hitting either of them.
The problem is requiring you to change your perspective. We tend to root our minds in what the ground is doing. And indeed the velocities given here are measured against the ground. The truth is all that matters here is how fast the ball and the train are coming together. So long as that is happening at 200 mph it doesn't matter how fast anything is moving relative to the ground.
The only reason the ground becomes part of the problem is the inputs and outputs are measured against it. You don't need it while working out the collision. You only need it when expressing your answer in the same form as the question: velocity relative to the ground.
1 If you wish to consider toy trains or wrecking balls see Wikipedia's Elastic Collision.
