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Consider an object colliding elastically with an inclined wall (with much bigger mass than the object) as in picuture. The wall is at $45°$ and it is moving.

$(A)$: frame of reference of the wall (the wall is steady)

$(B)$: frame of reference with respect to which the wall is moving towards the right with velocity $V$

enter image description here

The ball collides on the wall in both cases with an angle $\theta_1=45°$ with respect to the normal to the wall. In $(A)$, using conservation of momentum, the ball will be reflected at an angle $\theta_2=\theta_1=45°$.

But what is the value of $\theta_2$ is $(B)$?

My guess is that the conservation of momentum implies the equality of the angles of incidence and reflection only in the frame where the wall is steady, while in the other reference the ball is reflected at an other angle.

But, thinking about it, it seems strange that the ball gets momentum in the direction of the motion of the wall just being reflected on it. So is the situation really like the picture $(B)$, i.e. the angle of reflection is bigger?

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The answer "what is the value of $\theta$ in B" is answered by simple vector summation: you go back to the frame of reference where the wall is stationary, determine the magnitude and direction of the velocity after the impact, then add the velocity of the frame of reference back in.

enter image description here

That's really not so strange. If you imagine the situation where a ball is moving slowing at an angle to a wall that is coming rapidly towards it, you know intuitively that the ball will be moving roughly in the direction of the motion of the wall afterwards (think what happens with a tennis racket and a ball that was tossed in the air to be served... the ball flies across the net, doesn't it?)

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    $\begingroup$ Thanks a lot for the great answer!! If I may ask, in your picture is the velocity $\mathbf{v_1}'$ (the blue vector) equal to $\mathbf{v_1}$ in magnitude (of course the directions are different)? It should be but if I add the green and red vector I get $$|\mathbf{v_1}'|=\sqrt{(|\mathbf{v_1}|-|\mathbf{v_2}|)^2+|\mathbf{v_2}|^2}$$ $\endgroup$
    – Sørën
    Sep 11, 2017 at 19:30
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    $\begingroup$ No - the velocities will not be the same. Just like the ball that is hit with a racket will move faster than before it was hit. Only in the middle picture (frame of reference where the wall is stationary) will the green vector have the same magnitude before and after. In effect, with the wall moving away you absorb some of the energy of the ball (you can convince yourself by putting the wall vertical - 1D "vector addition" is easier...) $\endgroup$
    – Floris
    Sep 11, 2017 at 19:39
  • $\begingroup$ Thank you! I'm asking because in Micheslon Morley experiment (in the "classical" description that assumes aether) the velocity $|\mathbf{v_1}|$ is $c$ and when the light is reflected the velocity $|\mathbf{v_1}'|$ is still c (en.wikipedia.org/wiki/…) and the green vertical vector has magnitude $\sqrt{|\mathbf{v_1}'|^2-|\mathbf{v_2}|^2}=\sqrt{c^2-|\mathbf{v_2}|^2}$. So is in that case $|\mathbf{v_1}'|=c$ an assumption that is made that does not follow from classical mechanics? $\endgroup$
    – Sørën
    Sep 11, 2017 at 20:12
  • $\begingroup$ The velocity of light can never change - but you will get a change in momentum (and thus "color" of the light) when you hit a moving mirror. Doppler effect... yes, in the relativistic limit some of these equations need to be modified (instead of velocities you need momentum - that's the thing that can still be added properly). But you did tag the question "newtonian-mechanics"... $\endgroup$
    – Floris
    Sep 11, 2017 at 20:42

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