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(D) When block $C$ collides with block $A$,they stick together (assuming perfectly inelastic collision) or transfer momentum. Since $C$ and $A$ have e

Vedclass · Accepted Answer

Both $(b)$ and $(c)$.

Vedclass · Answer

(D) When block $C$ collides with block $A$,they stick together (assuming perfectly inelastic collision) or transfer momentum. Since $C$ and $A$ have equal mass $m$,after the collision,$C$ stops and $A$ moves with velocity $v$. Now,consider the system of blocks $A$ and $B$ connected by the spring. The initial momentum of the system is $mv$. Let $V$ be the common velocity of blocks $A$ and $B$ at the moment of maximum compression. By conservation of linear momentum: $mv = (m + m)V \Rightarrow V = \frac{v}{2}$. At maximum compression $x$,the kinetic energy of the system is converted into potential energy of the spring and the remaining kinetic energy of the center of mass. Using conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}(m+m)V^2 + \frac{1}{2}Kx^2$ $\frac{1}{2}mv^2 = \frac{1}{2}(2m)(\frac{v}{2})^2 + \frac{1}{2}Kx^2$ $\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx^2$ $\frac{1}{4}mv^2 = \frac{1}{2}Kx^2 \Rightarrow x^2 = \frac{mv^2}{2K} \Rightarrow x = v\sqrt{\frac{m}{2K}}$. At maximum compression,the kinetic energy of the $A-B$ system is: $K.E. = \frac{1}{2}(2m)V^2 = m(\frac{v}{2})^2 = \frac{mv^2}{4}$. Thus,both statements $(b)$ and $(c)$ are correct.

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