Question

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is?

• A. Zero
• B. $\frac{M{L}^2}{K}$
• C. $\sqrt{MK}L$
• D. $\frac{K{L}^2}{2M}$

Answer 1

The correct option is C $\sqrt{MK}L$

We know that, total mechanical energy remains constant, if conservative forces act on the system only.

$\therefore$ K.E+ P.E= E (Constant)

$\Delta K+\Delta U=0$

$\Delta K=-\Delta U$

If the initial velocity is v, then kinetic energy is $\frac{1}{2}m{v}^2$

Final potential energy due to spring is= $\frac{1}{2}k{x}^2$

From conservation of energy, K.E=P.E

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$v=\left(\sqrt{\frac{k}{m}}\right)x$, where x= compression= $L$

Therefore, $v=L\sqrt{\frac{k}{m}}$

Hence, maximum momentum, P= mv

=$L\sqrt{Mk}$