Question #83c7e

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Question #83c7e

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Answer 1 · 2015-10-14T14:16:28

I found that depends on the ratio of the two masses only.

Explanation:

I tried considerng Conservation of Momentum $p$ $p$ (along $x$ $x$ ):
Before: $0$ $0$
After: $m_{1} v_{1} + m_{2} v_{2}$ $m_1v_1+m_2v_2$
and together:
$0 = m_{1} v_{1} + m_{2} v_{2}$ $0=m_1v_1+m_2v_2$

so: $m_{1} v_{1} = - m_{2} v_{2}$ $m_1v_1=-m_2v_2$

squaring both and rearranging:

$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{m_{2}^{2}}{m_{1}^{2}}$ $v_1^2/v_2^2=m_2^2/m_1^2$

Kinetic Energy is $K = \frac{1}{2} m v^{2}$ $K=1/2mv^2$
So:
$K_1/K_2=(cancel(1/2)m_1v_1^2)/(cancel(1/2)m_2v_2^2)$ ,
substituting: $v_1^2/v_2^2=m_2^2/m_1^2$
$K_1/K_2=(cancel(m_1))/(cancel(m_2))*m_2^cancel(2)/m_1^cancel(2)=m_2/m_1$

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Question #83c7e

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