A block C of mass m is moving with velocity v | vedclass

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(D) Step $1$: Collision between $C$ and $A$. Since the collision is elastic and both have mass $m$,they exchange velocities. Block $C$ comes to rest,a

Vedclass · Accepted Answer

$\frac{2}{3} \frac{mv_0^2}{x_0^2}$

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(D) Step $1$: Collision between $C$ and $A$. Since the collision is elastic and both have mass $m$,they exchange velocities. Block $C$ comes to rest,and block $A$ starts moving with velocity $v_0$.Step $2$: After the collision,block $A$ moves towards block $B$ and compresses the spring. When the spring is compressed by $x_0$,both $A$ and $B$ move with the same velocity $v$.Step $3$: Apply conservation of linear momentum for the system $(A+B)$:$m v_0 = (m + 2m) v$$m v_0 = 3m v$$v = \frac{v_0}{3}$Step $4$: Apply conservation of mechanical energy for the system $(A+B)$:Initial kinetic energy of $A$ = Final kinetic energy of $(A+B)$ + Potential energy of the spring$\frac{1}{2} m v_0^2 = \frac{1}{2} (m + 2m) v^2 + \frac{1}{2} k x_0^2$$\frac{1}{2} m v_0^2 = \frac{1}{2} (3m) \left(\frac{v_0}{3}\right)^2 + \frac{1}{2} k x_0^2$$m v_0^2 = 3m \frac{v_0^2}{9} + k x_0^2$$m v_0^2 = \frac{m v_0^2}{3} + k x_0^2$$k x_0^2 = m v_0^2 - \frac{m v_0^2}{3} = \frac{2}{3} m v_0^2$$k = \frac{2 m v_0^2}{3 x_0^2}$

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