Question

Two blocks A and B each of mass $m$ are connected by a massless spring of neutral length L and spring constant K. The blocks are initially resting on a smooth horizontal floor 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 maximum compression of the spring is zero then kinetic energy of the system A-B at maximum compression of the spring is zero.
• B. the maximum compression of the spring is zero.The kinetic energy of the A-B system at maximum compression of the spring is $\frac{m{v}^2}{4.}$
  
• C. the maximum compression of the spring is $v\sqrt{\frac{m}{K}}$ and kinetic energy at this is $\frac{m{v}^2}{4}$ .
  
• D. the maximum compression of the spring is $v\sqrt{\frac{m}{2K}}$ and kinetic energy at this is $\frac{m{v}^2}{\mathrm{4\; .}}$

Answer 1

The correct option is D the maximum compression of the spring is $v\sqrt{\frac{m}{2K}}$ and kinetic energy at this is $\frac{m{v}^2}{\mathrm{4\; .}}$

Let the velocity acquired by A and B be V, then
mv = mV + mV
$\Rightarrow V=\frac{v}{2}$
Also, $\frac{1}{2}m{v}^2=\frac{1}{2}m{V}^2+\frac{1}{2}m{V}^2+\frac{1}{2}k{x}^2$
where $x$ is the maximum compression of the spring.
On solving the above equations, we get
$x=v\sqrt{\frac{m}{2K}}$
At maximum compression, kinetic energy of the A-B system is = $\frac{1}{2}m{V}^2+\frac{1}{2}m{V}^2=m{V}^2=\frac{m{v}^2}{4}$