Consider a one-dimensional collision involvin | vedclass

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(D) From the conservation of linear momentum: $m_1 v_0 = m_1 v_1 + m_2 v_2$,so $v_2 = \frac{m_1}{m_2}(v_0 - v_1)$.For a collision,the coefficient of r

Vedclass · Accepted Answer

$0 < v_1 < v_0$ ; $-v_0 < v_1 < v_0/2$

Vedclass · Answer

(D) From the conservation of linear momentum: $m_1 v_0 = m_1 v_1 + m_2 v_2$,so $v_2 = \frac{m_1}{m_2}(v_0 - v_1)$.For a collision,the coefficient of restitution $e$ satisfies $0 \le e \le 1$,where $e = \frac{v_2 - v_1}{v_0}$.Substituting $v_2$: $e = \frac{\frac{m_1}{m_2}(v_0 - v_1) - v_1}{v_0} = \frac{m_1 v_0 - (m_1 + m_2)v_1}{m_2 v_0}$.Solving for $v_1$: $v_1 = \frac{m_1 - e m_2}{m_1 + m_2} v_0$.Case $(A)$: If $m_1 > m_2$,the minimum $v_1$ occurs at $e=1$ $(v_1 = \frac{m_1 - m_2}{m_1 + m_2} v_0 > 0)$ and maximum at $e=0$ $(v_1 = \frac{m_1}{m_1 + m_2} v_0 Case $(B)$: If $m_1 < m_2$,the minimum $v_1$ occurs at $e=1$ $(v_1 = \frac{m_1 - m_2}{m_1 + m_2} v_0 < 0)$ and maximum at $e=0$ $(v_1 = \frac{m_1}{m_1 + m_2} v_0 < v_0)$. Thus,$\frac{m_1 - m_2}{m_1 + m_2} v_0 < v_1 < v_0$. Given the options,the range $-v_0 < v_1 < v_0/2$ is the most appropriate fit.

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