# Relative frame of motion issue

If an object normally bounces off a surface that has zero velocity with a velocity $v_\text{a}$ (elastically), how fast will the velocity be if the surface is moving with velocity $v_\text{b}$ (also elastic)?

My reasoning was that from the surface's frame of reference, the collision must preserve the balls velocity since it was elastic, so if it was incoming at a perceived speed of $v_\text{a},$ it must exit at that speed. This means that from an outside frame, it would be a velocity of $v_\text{a} + v_\text{b}.$

Is this reasoning correct?

• Seems correct to me. – Internet Guy Jun 13 '18 at 4:00

## 1 Answer

No your reasoning is wrong. Firstly I assume that the surface incoming has infinite mass. Then according to the frame of reference on the surface the ball is incoming at $\left(v_\text{a}+v_\text{b}\right)$ which means, with respect to it, the ball bounces too with $\left(v_\text{a}+v_\text{b}\right).$ But taking into account the velocity of the plank the velocity w.r.t. earth becomes $\left(v_\text{a}+2v_\text{b}\right).$

• This reasoning must be done at advanced stages of linear momentum. – Manas Singh Jun 13 '18 at 3:57
• Normal reasoning says to make normal linear momentum equations by takung surface of finite mass and on getting the result take the limit to infinity – Manas Singh Jun 13 '18 at 3:58
• Another possible way is making impulse equations – Manas Singh Jun 13 '18 at 3:59
• Gotcha, I forgot to account for the surface's body when calculating the perceived speed of the object relative to the the surface. Thanks! – user1939991 Jun 13 '18 at 4:36