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

• A. $\sqrt{Mk}\,L$
• B. $\frac{kL^{2}}{2M}$
• C. zero
• D. $\frac{ML^{2}}{k}$

Answer 1

Answer: (A)

$\sqrt{Mk}\,L$

Answer 2

Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted to KE. According to conservation of energy $\frac{1}{2}\,k\,L^{2}-\frac{1}{2}\,Mv^{2}$ $k\,L^{2}=\frac{\left(Mv\right)^{2}}{M}$ $MkL^{2}=p^{2}\,\left(p=Mv\right)$ $\therefore p=L\sqrt{Mk}$