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Case 1- We have two masses of equal size. Mass 1, with positive velocity, collides with Mass 2. This causes Mass 1 to have 0 velocity in the end and Mass 2 to have a velocity.

Case 2- A mass (1) collides with a larger mass (2) that is three times the size. Mass 1 initially has a positive velocity, while mass 2 has 0 velocity. The collision causes the mass 1 to have a negative velocity, while giving mass 2 a velocity that is unknown to us.

In what case would the mass 1 exert a larger impulse on mass 2, if any? Why?

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  • $\begingroup$ "exert a larger impulse". Larger than what? $\endgroup$ – BowlOfRed Nov 1 '15 at 23:23
  • $\begingroup$ In which case would the smaller mass exert a larger impulse on the larger mass? Would the impulse exerted by the smaller mass on the larger mass be larger after impact or before? Or would there not be any larger impulse in this situation? $\endgroup$ – user97298 Nov 1 '15 at 23:28
  • $\begingroup$ I don't understand your question. When you ask "a larger impulse", what would it be larger than? What are you comparing it to? Impulse is exerted during an impact, not before or after. $\endgroup$ – BowlOfRed Nov 1 '15 at 23:30
  • $\begingroup$ Sorry, I only asked part of the question I was intending, I have edited it $\endgroup$ – user97298 Nov 1 '15 at 23:39
  • $\begingroup$ Maybe I'm not understanding ... but in all cases the impulse of Mass 1 on Mass 2 is the same as the impulse of Mass 2 on Mass 1. This is a consequence of Newton's Third Law. $\endgroup$ – garyp Nov 2 '15 at 1:32
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You need to remember that in an elastic collision the equations due to the conservation of linear momentum and kinetic energy are:

$\begin{eqnarray} V_{1\:final}={m_1-m_2\over m_1+m_2}\,V_{1\:initial},\\ V_{2\:final}={m_1\over m_1+m_2}\,V_{1\:initial} \end{eqnarray} $

a) First case: $m_1=m_2$. Here $V_{2\:final}=(V_{1\:initial})/2$. Then, the momentum acquired by the second mass, is

$\begin{equation} \hspace{2cm}{1\over2}\,{m_1V_{1\:initial}} \end{equation}$.

b) Second case: $m_2=3\,m_1$. Here $V_{2\:final}=(V_{1\:initial})/4$. Then, the momentum acquired by the second mass, is

$\begin{equation}\hspace{2cm}{3\over 4}\,{m_1\,V_{1\:initial}}.\end{equation}$

So, is the second case where mass 1 exert a larger impulse on mass 2.

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