Two bodies A and B of masses m and 2m respect | vedclass

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(A) Initially,the blocks $A$ and $B$ are at rest and $C$ is moving with velocity $V_0$ to the right. As the masses of $C$ and $A$ are equal and the co

Vedclass · Accepted Answer

$\frac{2}{3} \frac{m V_0^2}{x_0^2}$

Vedclass · Answer

(A) Initially,the blocks $A$ and $B$ are at rest and $C$ is moving with velocity $V_0$ to the right. As the masses of $C$ and $A$ are equal and the collision is elastic,the body $C$ transfers its whole momentum $m V_0$ to body $A$. As a result,body $C$ stops and $A$ starts moving with velocity $V_0$ to the right. At this instant,the spring is uncompressed and body $B$ is still at rest.The momentum of the system at this instant is $P = m V_0$.Now,the spring is compressed and body $B$ comes into motion. After time $t_0$,the compression of the spring is $x_0$ and the common velocity of $A$ and $B$ is $v$. Since the external force on the system is zero,the law of conservation of linear momentum gives:$m V_0 = m v + (2m) v$$m V_0 = 3m v$$v = \frac{V_0}{3}$The law of conservation of energy gives:$\frac{1}{2} m V_0^2 = \frac{1}{2} m v^2 + \frac{1}{2} (2m) v^2 + \frac{1}{2} k x_0^2$$\frac{1}{2} m V_0^2 = \frac{3}{2} m v^2 + \frac{1}{2} k x_0^2$Substituting $v = \frac{V_0}{3}$:$\frac{1}{2} m V_0^2 = \frac{3}{2} m \left( \frac{V_0}{3} \right)^2 + \frac{1}{2} k x_0^2$$\frac{1}{2} m V_0^2 = \frac{3}{2} m \left( \frac{V_0^2}{9} \right) + \frac{1}{2} k x_0^2$$\frac{1}{2} m V_0^2 = \frac{1}{6} m V_0^2 + \frac{1}{2} k x_0^2$$\frac{1}{2} k x_0^2 = \frac{1}{2} m V_0^2 - \frac{1}{6} m V_0^2$$\frac{1}{2} k x_0^2 = \frac{3-1}{6} m V_0^2 = \frac{2}{6} m V_0^2 = \frac{1}{3} m V_0^2$$k = \frac{2}{3} \frac{m V_0^2}{x_0^2}$.

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