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Two blocks of mass $M_1$ and $M_2$ moving along a 1-dimensional straight line with velocities $V_1$ and $V_2$, respectively, collide elastically. After the collision they move with respective velocities $U_1$ and $U_2$. What is the ratio $M_1/M_2$ if the velocities are to be interchanged, ie, $U_1=V_2$ and $U_2=V_1$?

I'm very confused as to where to begin in regards to this question. I've tried rearranging the conversation of momentum equation ($m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2$) to express it in terms of $m_1$ and $m_2$ and then dividing them into each other. However this leaves me with $m_1$ and $m_2$ on both sides which is definitely incorrect. Is there an easier way to find the ratio?

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closed as off-topic by John Rennie, Bill N, dmckee Apr 21 '16 at 1:40

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Considering momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ Therefore, $$m_1u_1-m_1v_1=m_2v_2-m_2u_2$$ Factorising gives us: $$m_1(u_1-v_1)=m_2(v_2-u_2)$$ Allowing us to rearrange to: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-u_{2}}{u_{1}-v_{1}}$$ Using the fact that you say the velocities can be interchanged, we obtain a final answer of: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-v_{1}}{v_{2}-v_{1}}$$ This final step gives a ratio of 1 between the two masses.

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    $\begingroup$ The correct ratio is one. Think about what happens when one billiard ball hits another completely head-on without any spin effects. $\endgroup$ – Jerry Schirmer Apr 20 '16 at 19:37
  • $\begingroup$ @JerrySchirmer Of course - I was overlooking the fact that the velocities are interchanged, so was confused by the apparent value of 1. This makes sense now, I will edit the answer! $\endgroup$ – Noah P Apr 20 '16 at 19:39
  • $\begingroup$ Its a bit confusing how this can be correct math since you have Kg = ms^-1 as an equation. Its a bit quirky. $\endgroup$ – WDUK Nov 28 '18 at 1:03

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