A cat with mass m_c is sitting at rest on the | vedclass

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(B) Let the left plank be $P_1$ and the right plank be $P_2$. Initially,the system is at rest. When the cat leaps from $P_1$ to $P_2$ with speed $v_{c

Vedclass · Accepted Answer

$\frac{2m_c v_{cg}}{m_s}$

Vedclass · Answer

(B) Let the left plank be $P_1$ and the right plank be $P_2$. Initially,the system is at rest. When the cat leaps from $P_1$ to $P_2$ with speed $v_{cg}$ (to the right),by conservation of momentum for the system (cat + $P_1$): $0 = m_c v_{cg} + m_s v_{P1} \implies v_{P1} = -\frac{m_c v_{cg}}{m_s}$ (to the left). When the cat lands on $P_2$,the cat and $P_2$ move together. By conservation of momentum for the system (cat + $P_2$): $m_c v_{cg} = (m_c + m_s) v_{P2} \implies v_{P2} = \frac{m_c v_{cg}}{m_c + m_s}$. When the cat leaps back from $P_2$ to $P_1$ with speed $v_{cg}$ (to the left),let the final velocity of $P_2$ be $v'_{P2}$. By conservation of momentum for the system (cat + $P_2$): $(m_c + m_s) v_{P2} = m_c (-v_{cg}) + m_s v'_{P2}$. Substituting $v_{P2}$: $(m_c + m_s) \left( \frac{m_c v_{cg}}{m_c + m_s} \right) = -m_c v_{cg} + m_s v'_{P2}$. $m_c v_{cg} = -m_c v_{cg} + m_s v'_{P2} \implies 2m_c v_{cg} = m_s v'_{P2} \implies v'_{P2} = \frac{2m_c v_{cg}}{m_s}$.

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