0
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ Seems correct to me. $\endgroup$
    – User
    Commented Jun 13, 2018 at 4:00

1 Answer 1

Reset to default
0
$\begingroup$

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 $\left(v_\text{a}+2v_\text{b}\right).$

Improve this answer
$\endgroup$
4
  • $\begingroup$ This reasoning must be done at advanced stages of linear momentum. $\endgroup$
    – Manas Singh
    Commented Jun 13, 2018 at 3:57
  • $\begingroup$ 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 $\endgroup$
    – Manas Singh
    Commented Jun 13, 2018 at 3:58
  • $\begingroup$ Another possible way is making impulse equations $\endgroup$
    – Manas Singh
    Commented Jun 13, 2018 at 3:59
  • $\begingroup$ Gotcha, I forgot to account for the surface's body when calculating the perceived speed of the object relative to the the surface. Thanks! $\endgroup$
    – user1939991
    Commented Jun 13, 2018 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.