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 spring at its natural length as shown in the 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 and collides with A. Then,

• A. The KE of the A-B system at maximum compression of the spring is zero
• B. The KE of the AB system at maximum compression of the spring is $\frac{m{v}^2}{2}$
• C. The maximum compression of the spring is $v\sqrt{\frac{m}{k}}$
• D. The maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$

Answer 1

The correct option is D

The maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$

Let $v_1$ be the velocity acquired by A and B.

According to the law of conservation of momentum,

$mv=m{v}_1+m{v}_1$ or $v_1=\frac{v}{2}$

According to law of conservation of energy,

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_1^2+\frac{1}{2}k{x}^2$,

where x is the compression.

$\therefore$ $m{v}^2=\frac{m{v}^2}{4}+\frac{m{v}^2}{4}+k{x}^2\Rightarrow x=v\sqrt{\frac{m}{2k}}$

At maximum compression, KE of the AB system will be,

$\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_1^2=m{v}_1^2=\frac{m{v}^2}{4}$