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. C collides with A elastically. Then:

• A. The kinetic energy of AB system at maximum compression of the spring is zero.
• B. The kinetic energy of the AB 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 $v\sqrt{\frac{m}{2k}}$.

Answer 1

Answer: (B), (D)

The correct options are
B The kinetic energy of the AB system at maximum compression of the spring is $\frac{m{v}^2}{4}$.
D The maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$.


Since C and A are of equal mass and collision is elastic, C transfers its total momentum to A. Therefore, A has the velocity $v$ immediately after collsion. At maximum compression, the velocity of both the bodies ae equal, say $v^{\prime }$.


Conservation of linear momentum yields
$mv=(m+m)v^{\prime }$
$\Rightarrow v^{\prime }=\left(\frac{v}{2}\right)$.
Initial K.E $=\frac{1}{2}m{v}^2$
$\therefore$ The K.E. of the system at max. compression
$=\frac{1}{2}(m+m)v^{\prime 2}$
$=\frac{1}{2}\left(2m\right)\frac{v^2}{4}=\frac{m{v}^2}{4}$
If the max. compression of spring is $x$,
$\frac{1}{2}k{x}^2=\Delta KE=\frac{m{v}^2}{2}-\frac{m{v}^2}{4}$
$\Rightarrow x=\left(\sqrt{\frac{m}{2k}}\right)v$
Hence B & D are correct options.