Q.
The blocks $A$ and $B$, each of mass $m$, are connected by 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 $B$, and collides with $A$. Then
A) The KE o the $A - B$ system, at maximum compression of the spring, is zero
B) The KE of the $A - B$ system, at maximum compression of the spring is $\frac{1}{4}mv^{2}$
C) The maximum compression of the spring is $v\sqrt{\frac{m}{k}}$
D) The maximum compression of the spring is $v\sqrt{\frac{m}{2k}}$
Explanation
(d) The compression of spring is maximum when velocities of both blocks $A$ and $B$ is same. Let it be $v_{0}$, then from conservation law of momentum
$mv = mv_{0} + mv_{0} = 2mv_{0} \Rightarrow v_{0} = \frac{v}{2}$
$\therefore$ kinetic energy of $A - B$ system at that stage
$= \frac{1}{2}(m + m) \times \left( \frac{v}{2} \right)^{2} = \frac{mv^{2}}{4}$
Further loss in KE> = gain in elastic potential energy
$ie$, $\frac{1}{2}mv^{2} - \frac{1}{4}mv^{2} = \frac{1}{4}mv^{2} = \frac{1}{2}kx^{2}$
$\Rightarrow \ x = v\sqrt{\frac{m}{2k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.