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 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 elastically with A.

Consider the following statements. Which of the following is/are correct?

(i) The kinetic energy of the A-B system, at maximum compression of the spring, is zero

(ii) The kinetic energy of the A- B system , at maximum compression of the spring, is $\frac{m{v}^2}{4}$

(iii) The maximum compression of the spring, is $v\sqrt{\left(\frac{m}{K}\right)}$

(iv) The maximum compression of the spring, is $v\sqrt{\left(\frac{m}{2K}\right)}$

• A. (ii) & (iv) only
• B. (i) only
• C. (ii) & (iii) only
• D. (i) & (iii) only

Answer 1

The correct option is A (ii) & (iv) only

In situation (i), mass C is moving towards right with velocity v, A and B are at rest.

In situation (ii), which is just after the collision of C with A, C stops and A acquires a velocity v.

When A starts moving towards the right, the spring suffers a compression due to which B also starts moving towards right. The compression of the spring continues till there is a relative velocity between A and B.

Once this relative velocity becomes zero, both A and B move with the same velocity v' and the spring is in a state of maximum compression (situation iii).

Applying momentum conservation in situations (ii) and (iii)

$mv=m{v}^{\prime }=m{v}^{\prime }\Rightarrow v^{\prime }=\frac{v}{2}$

Therefore , KE of the system in situation (iii) is

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

Applying energy conservation, in situations (ii) and (iii) we get

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}K{x}^2$

Solve to get

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