Question

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} $

Answer 1

Answer: (A)

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
$m v=(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} m v^{2}=\frac{1}{2}(2 m) V^{2}+\frac{1}{2} k x^{2} $
$\Rightarrow \frac{1}{2} m v^{2}=\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^{2}+\frac{1}{2} k x^{2} $
$\Rightarrow k x^{2}=\frac{1}{2} m v^{2} $
$\Rightarrow x=v \sqrt{\frac{m}{2 k}}$