# Angle of reflection of an object colliding on a moving wall

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 $V$

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?

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.
• 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 $$|\mathbf{v_1}'|=\sqrt{(|\mathbf{v_1}|-|\mathbf{v_2}|)^2+|\mathbf{v_2}|^2}$$ Sep 11, 2017 at 19:30
• 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? Sep 11, 2017 at 20:12