Two identical blocks A and B,each of mass m,r | vedclass

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(A) $1$. When block $C$ collides with block $A$,it sticks to it (assuming perfectly inelastic collision as no other information is given). By conserva

Vedclass · Accepted Answer

$v\sqrt{\frac{m}{2K}}$

Vedclass · Answer

(A) $1$. When block $C$ collides with block $A$,it sticks to it (assuming perfectly inelastic collision as no other information is given). By conservation of linear momentum for the system $(C+A)$: $mv = (m+m)V_A \Rightarrow V_A = v/2$.$2$. Now,we have a system where block $A$ (with $C$ attached) moves with velocity $v/2$ towards block $B$,which is initially at rest.$3$. The maximum compression in the spring occurs when both blocks $(A+C)$ and $B$ move with the same velocity $V_{cm}$ relative to the center of mass frame.$4$. By conservation of linear momentum for the whole system $(C+A+B)$: $mv = (m+m+m)V_{cm} \Rightarrow V_{cm} = v/3$.$5$. By conservation of mechanical energy: $\frac{1}{2}(2m)(v/2)^2 = \frac{1}{2}(3m)(v/3)^2 + \frac{1}{2}Kx_{max}^2$.$6$. $\frac{1}{2}(2m)(v^2/4) = \frac{1}{2}(3m)(v^2/9) + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{mv^2}{4} = \frac{mv^2}{6} + \frac{1}{2}Kx_{max}^2$.$7$. $\frac{1}{2}Kx_{max}^2 = \frac{mv^2}{4} - \frac{mv^2}{6} = \frac{3mv^2 - 2mv^2}{12} = \frac{mv^2}{12}$.$8$. $x_{max}^2 = \frac{mv^2}{6K} \Rightarrow x_{max} = v\sqrt{\frac{m}{6K}}$.*Note: If the collision is elastic,$C$ comes to rest and $A$ moves with velocity $v$. Then $mv = (m+m)V_{cm} \Rightarrow V_{cm} = v/2$. Energy: $\frac{1}{2}mv^2 = \frac{1}{2}(2m)(v/2)^2 + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx_{max}^2 \Rightarrow \frac{1}{2}Kx_{max}^2 = \frac{1}{4}mv^2 \Rightarrow x_{max} = v\sqrt{\frac{m}{2K}}$. This matches option $A$.*

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