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Question

(A) The initial momentum of the system is zero,i.e.,$P_i = 0$.
Since the spring force is an internal force,the linear momentum of the system is conse

Vedclass · Accepted Answer

$\left( \frac{m_2}{m_1} \right)^2$

Vedclass · Answer

(A) The initial momentum of the system is zero,i.e.,$P_i = 0$.
Since the spring force is an internal force,the linear momentum of the system is conserved during the release.
Let the velocities acquired by masses $m_1$ and $m_2$ immediately after release be $v_1$ and $v_2$. By conservation of linear momentum: $m_1 v_1 = m_2 v_2$,which implies $\frac{v_1}{v_2} = \frac{m_2}{m_1}$.
When the blocks move on the surface,the work done by friction equals the initial kinetic energy of each block.
For block $1$: $\mu m_1 g x_1 = \frac{1}{2} m_1 v_1^2 \Rightarrow x_1 = \frac{v_1^2}{2 \mu g}$.
For block $2$: $\mu m_2 g x_2 = \frac{1}{2} m_2 v_2^2 \Rightarrow x_2 = \frac{v_2^2}{2 \mu g}$.
Taking the ratio: $\frac{x_1}{x_2} = \frac{v_1^2}{v_2^2}$.
Substituting $\frac{v_1}{v_2} = \frac{m_2}{m_1}$,we get $\frac{x_1}{x_2} = \left( \frac{m_2}{m_1} \right)^2$.

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