Question

Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring in its natural length, as shown in figure.

A third identical block C, also of mass m moves on the floor with a speed v along the line joining A and B collides with A, elastically. Then,

• A. The kinetic energy of AB system at maximum compression of the spring is zero.
• B. The kinetic energy of the AB system at maximum compression of the spring is mv².
• C. The maximum compression of the spring is.
• D. The maximum compression of the spring is.

Answer 1

The correct option is D

The maximum compression of the spring is.

Due to equal mass, C transfers its total momentum to A. Therefore A has the velocity v. At max. compression the velocity of both the bodies are equal say, v'. Conservation of linear momentum yields

$mv=(m+m)v^{\prime }$

$\Rightarrow v^{\prime }=\left(\frac{v}{2}\right)$

$\therefore$ The K.E. of the system

$=\frac{1}{2}(m+m)v^{\prime 2}$

$=\frac{1}{2}\left(2m\right)\frac{v^2}{4}=\frac{m{v}^2}{4}$

To calculate the maximum compression x,

$\frac{1}{2}k{x}^2=△KE=\frac{m{v}^2}{2}-\frac{m{v}^2}{4}$

$\Rightarrow x=(\sqrt{\frac{m}{2k}})$ v.