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 at its natural length as shown in Fig. 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 AB system at maximum compression of the spring is zero.
• B. The KE of the AB system at maximum compression of the spring is $\left(1/4\right)m{v}^2$
• 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

Answer: (B), (D)

The correct options are
B The KE of the AB system at maximum compression of the spring is $\left(1/4\right)m{v}^2$
D The maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$.

[Ref. image 1, 2]

If two same masses collide elastically then their velocity get exchange.

At maximum compression, relative velocity of A and B will be zero ( w.r.t each other ) [Ref. image 2]

Applying momentum conservation at situations (2) and (3)

$\Rightarrow mv+0=m{v}_1+m{v}_1$

$\Rightarrow v_1=\frac{v}{2}$

KE of AB system at maximum compression situation -

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

Applying energy conservation between (2) and (3) situations -

$\Rightarrow \frac{1}{2}m{v}^2+0=\frac{1}{4}m{v}^2+\frac{1}{2}k{x}^2$

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

$\Rightarrow \frac{1}{2}k{x}^2=\frac{1}{4}m{v}^2$

$\Rightarrow x^2=\frac{m{v}^2}{2k}$

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

Hence, option B and D correct.