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
$mv=(m+m)V$
$\Rightarrow V=\frac{v}{2}$
By the law of conservation of energy
K E of block $C=KE$ of system $+PE$ of system
$\frac{1}{2}m{v}^2=\frac{1}{2}(2m)V^2+\frac{1}{2}k{x}^2$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}(2m){\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x}^2$
$\Rightarrow k{x}^2=\frac{1}{2}m{v}^2$
$\Rightarrow x=v\sqrt{\frac{m}{2k}}$