JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Conservation of Momentum

Two identical blocks

A

and

B

each of mass

m

resting on the smooth horizontal floor are connected by a light spring of natural L and spring constant K.

A

third block

C

of mass

m

moving with a speed

v

along the line joining

A

and

B

collides with

A

. The maximum compression in the spring is

Answer: (D)

Solution

The collision between the identical blocks C and A is elastic, so the two exchange velocities: C comes to rest and A moves off with speed

v

.
The spring is compressed most at the instant A and B move with a common velocity

u

, and momentum conservation for the pair joined by the spring gives

m v = 2 m u \Rightarrow u = \frac{v}{2}

The floor is smooth, so the kinetic energy lost by the pair up to that instant is stored in the spring:

\frac{1}{2} m v^{2} = \frac{1}{2} (2 m) (\frac{v}{2})^{2} + \frac{1}{2} K x^{2}

\frac{m v ^{2}}{4} = \frac{1}{2} K x^{2}

The maximum compression is therefore

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

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option