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Hello fellow stackexchangers, this is my first post, so sorry if this is too vague or violates guidelines.

I am studying Physics and this problem came up, I will type it verbatim

A small object collides with a large object and sticks. Which object experiences the larger magnitude of momentum change -the small object -the large object-both objects experience the same magnitude of momentum change

I posted this on the math site because I wish to see a proof for the general situation. Note that this situation occurs in one dimension, that conservation of momentum applies, and that the two objects become one after the collison. The collision is inelastic and kinetic energy is not conserved.

If this is more suitable for the Physics site, please tell me.

Onion_Kurian
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  • I believe the equation for momentum is like mivi + mi2vi2 = (mi + mi2)vf, where (mi + mi2) substitutes mf. Is this correct? I haven't taken physics in a while. I can work with this equation though. – Riley Carney Feb 04 '16 at 23:40

1 Answers1

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Since the initial momentum is $m_1v_1+m_2v_2$ and
and the final momentum is $(m_1+m_2)v_f$, we have:

$m_1v_1 + m_2v_2 = (m_1+m_2)v_f$

or

$v_f\to \frac{m_1 v_1+m_2 v_2}{m_1+m_2}$

To find the change in object 1's momentum, we compute

$m_1 \left(v_1-v_f\right)$

Substituting, we get:

$m_1 \left(v_1-\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\right)$

which simplifies to:

$\frac{m_1 m_2 \left(v_1-v_2\right)}{m_1+m_2}$

Similarly, for object 2, we compute:

$m_2 \left(v_2-v_f\right)$

after substitution:

$m_2 \left(v_2-\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\right)$

and simplifying:

$\frac{m_1 m_2 \left(v_2-v_1\right)}{m_1+m_2}$

which is the exact negative of the answer for object 1, so the absolute changes in momentum are identical.