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. A third identical block C, also of mass m, moving on the floor with a speed along the line joining A and B,collides with A (see figure) Then

• A. The kinetic energy of A – B system at maximum compression of the spring is zero
• B. The kinetic energy of A – B system at maximum compression of the spring is $\frac{m{v}^2}{4}$
• C. The maximum compression of the spring is $v\sqrt{\frac{m}{k}}$
• D. The maximum compression of the spring is $v\sqrt{\frac{2m}{k}}$
  Let x be the maximum compression of the spring. From the law of conservation of energy, we have $\frac{1}{2}m{v}^2=m{u}^2+\frac{1}{2}k{x}^2$

Answer 1

The correct option is B The kinetic energy of A – B system at maximum compression of the spring is $\frac{m{v}^2}{4}$

Since the spring in massless, the momentum imparted to the system A – B by block C is equally shared between A and B. Since A and B have the same mass, they will have the same velocity; let it be u. from the law of conservation of momentum, we have $mv=mu+mu=2mu$ Which gives u =$\frac{v}{2}$ Let x be the maximum compression of the spring. From the law of conservation of energy, we have $\frac{1}{2}m{v}^2=m{u}^2+\frac{1}{2}k{x}^2$ $or{v}^2=2u^2+\frac{k{x}^2}{m}=2{\left(\frac{v}{2}\right)}^2+\frac{k{x}^2}{m}Or\frac{k{x}^2}{m}=\frac{v^2}{2}$ $orx=v\sqrt{\frac{m}{2k}}$ (1) The kinetic energy is equally shared between masses A and B and the spring is at maximum compression, i.e $\frac{1}{2}k{x}^2=\frac{1}{2}m{u}^2+\frac{1}{2}m{u}^2=m{u}^2$ Thus, kinetic energy of A – B maximum compression $m{u}^2=m{\left(\frac{v}{2}\right)}^2=\frac{m{v}^2}{4}$ Using (i) notice that $\frac{1}{2}k{x}^2=\frac{m{v}^2}{4}$ . Hence the correct choice is (b)