Q.
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 collides is
A) $\sqrt{MkL}$
B) $\frac{kL^{2}}{2M}$
C) Zero
D) $\frac{ML^{2}}{k}$
Explanation
(a) Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE.
According to conservation of energy
$\frac{1}{2}kL^{2} = \frac{1}{2}Mv^{2}$
Or $kL^{2} = \frac{(Mv)^{2}}{M}$
$MKL^{2} = p^{2}$ $(p = Mv)$
$\therefore p = L\sqrt{MK}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.