Question

Two blocks of masses $m$ and $M$ are moving with speeds $v_{1}and{v}_2(v_1>v_2)$ in the same direction on the frictionless surface respectively, $M$ being ahead of $m$. An ideal spring of force constant $k$ is attached to the backside of $M$ as shown. The maximum compression of the spring when the block collides is:

• A. $v_1\sqrt{\frac{m}{k}}$
• B. $v_2\sqrt{\frac{M}{k}}$
• C. $(v_1-v_2)\sqrt{\frac{mM}{(M+m)k}}$
• D. $Noneofaboveiscorrect.$

Answer 1

Answer: (C)

$(v_1-v_2)\sqrt{\frac{mM}{(M+m)k}}$

The initial kinetic energy of the system is given by:
$T_{initial}=E_{total}=\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2$

When the system has minimum kinetic energy, the two bodies would be moving at the same velocity.
Using momentum conservation.
$m{v}_1+M{v}_2=(m+M)v$
$v=\frac{m{v}_1+M{v}_2}{m+M}$

The kinetic energy at this point would be:
$T_{min}=\frac{1}{2}(m+M)v^2=\frac{1}{2}\frac{(m{v}_1+M{v}_2{)}^2}{m+M}$
Using $T_{initial}-T_{min}=\frac{1}{2}k{x}_{max}^2$
We get:
$x_{max}=(v_1-v_2)\sqrt{\frac{mM}{(M+m)k}}$
Hence option C is correct.