Question

A light spring of spring constant $k$ is kept compressed between two blocks of masses $m$ and $M$ on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions.
The spring loses contact with the blocks when it acquires natural length. If the spring was initially compressed through a distance $x$, find the final speed of the block of mass $M$.

• A. ${\left[\frac{kM}{M(M+m)}\right]}^{1/2}x$
• B. ${\left[\frac{km}{M(M+m)}\right]}^{1/2}x$
• C. ${\left[\frac{M}{M(M+m)}\right]}^{1/2}x$
• D. ${\left[\frac{m}{M(M+m)}\right]}^{1/2}x$

Answer 1

The correct option is B ${\left[\frac{km}{M(M+m)}\right]}^{1/2}x$

Answer is A.
Consider the two blocks plus the spring to be the system. No external force acts on this system in horizontal direction. Hence, the linear momentum will remain constant.

Suppose, the block of mass $M$ moves with a speed $v_1$ and the other block with a speed $v_2$ after losing contact with the spring.

From conservation of linear momentum in horizontal direction we have,

$M{v}_1-m{v}_2=0$ or $v_1=\frac{m}{M}{v}_2$... (i)

Initially, the energy of the system $=\frac{1}{2}k{x}^2$

Finally, the energy of the system $=\frac{1}{2}m{v}_2^2+\frac{1}{2}M{v}_1^2$

As there is no friction, mechanical energy will remain conserved.

Therefore, $\frac{1}{2}m{v}_2^2+\frac{1}{2}M{v}_1^2=\frac{1}{2}k{x}^2$

Solving Eqs. (i) and (ii); we get

or, $v_2={\left[\frac{kM}{m(M+m)}\right]}^{1/2}x$ and $v_1={\left[\frac{km}{M(M+m)}\right]}^{1/2}x$.

Hence, the final speed of the block of mass M is ${\left[\frac{km}{M(M+m)}\right]}^{1/2}x$.