Q. Two identical blocks $A$ and $B$, each of mass ‘$m$’ resting on smooth floor are connected by a light spring of natural length $L$ and spring constant $K$, with the spring at its natural length. A third identical block ‘$C$’ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. The maximum compression in the spring is
A) $v\sqrt{\frac{m}{2k}}$
B) $m\sqrt{\frac{v}{2k}}$
C) $\sqrt{\frac{mv}{k}}$
D) $\frac{mv}{2k}$
Explanation
(a) [image:6_image86.png]
Initial momentum of the system (block $C) = mv$
After striking with $A$, the block $C$ comes to rest and now both block $A$ and $B$ moves with velocity $V$, when compression in spring is maximum
By the law of conservation of linear momentum
$mv = (m + m)V \Rightarrow V = \frac{v}{2}$
By the law of conservation of energy
K.E. of block $C =$ K.E. of system + P.E. of system
$\frac{1}{2}\ mv^{2} = \frac{1}{2}(2m)V^{2} + \frac{1}{2}kx^{2}$
$\Rightarrow \frac{1}{2}mv^{2} = \frac{1}{2}(2m)\left( \frac{v}{2} \right)^{2} + \frac{1}{2}kx^{2} \Rightarrow kx^{2} = \frac{1}{2}mv^{2}$
$\Rightarrow x = v\sqrt{\frac{m}{2k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.