Question

Two blocks A and B each of mass m, are connected by a massless spring of natural length Land 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. Then

• A. the kinetic energy of the A-B system, at maximum compression of the spring, is zero
• B. the kinetic energy of the A-B system, at maximum compression of the spring, is $mv^2/4$
• C. the maximum compression of the spring is $v\sqrt{(m/k)}$
• D. the maximum compression of the spring is v

Answer 1

The Correct Option is D

After collision between C and A, C stops while A moves with speed of C i.e. v [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, A and B will move with same speed v/2 (From conservation of linear momentum).

Let x be the maximum compression in this position.

$\therefore$ KE of A-B system at maximum compression

$ \hspace15mm =\frac{1}{2}(2m)\bigg(\frac{v}{2}\bigg)^2$

or $ \hspace10mm =K_{max}=mv^2/4$

From conservation of mechanical energy in two positions shown in above figure

or $ \hspace15mm =\frac{1}{2}mv^2=\frac{1}{4}mv^2+\frac{1}{2}kx^2$
$ \hspace20mm =\frac{1}{2}kx^2=\frac{1}{4}mv^2 \, \, \Rightarrow \, \, \therefore \, \, x = v+\sqrt{\frac{m}{2k}}$