Question

Two identical blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor are connected by a light spring of natural length $L$ and spring constant $K$. A third block $C$ of mass $m$ moving with a speed $v$ along the line joining $A$ and $B$ collides elastically with $A$. The maximum compression in the spring is

• A. $\sqrt{\frac{mv}{2K}}$
• B. $\sqrt{\frac{m}{2K}}$
• C. $\sqrt{\frac{mv}{K}}$
• D. $v\sqrt{\frac{m}{2K}}$

Answer 1

The correct option is D $v\sqrt{\frac{m}{2K}}$

Let $v_1$ be the common velocity of blocks $A$ and $B$ at the maximum compression of the spring.

From conservation of momentum:
$mv=m{v}_1+m{v}_1$
$v_1=\frac{v}{2}$
And from energy conservation:
$\frac{1}{2}m{v}^2=(\frac{1}{2}\times m{\left(\frac{v}{2}\right)}^2)\times 2+\frac{1}{2}K{x}^2$
$\frac{m{v}^2}{2}-\frac{m{v}^2}{4}=\frac{1}{2}K{x}^2$
$\frac{m{v}^2}{4}=\frac{1}{2}K{x}^2$
Then the maximum compression in the spring is
$x=\sqrt{\frac{m{v}^2}{2K}}$
$x=v\sqrt{\frac{m}{2K}}$