Question

A plank of mass $5kg$ placed on a frictionless horizontal plane. Further a block of mass $1kg$ is placed over the plank. A massless spring of natural length $2m$ is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring's natural length. They system is now released from the rest. What is the velocity of the plank when block leaves the plank? (The stiffness constant of spring is $100N/m)$.

Answer 1

Given,

Mass of plank, $M=5kg$

Mass of block, $m=1kg$
Compressed length $L=1m$ (which is half of natural length)

Sprung stiffness, $K=100N/m$

From conservation of momentum

Final momentum = initial momentum.

$MV+mv=0$

$v=\frac{-MV}{m}=\frac{-5V}{1}=-5V$

Conservation of kinetic energy

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

$\frac{1}{2}100\times 1^2=\frac{1}{2}5V^2+\frac{1}{2}1\times {(-5V)}^2$

$30V^2=100$

$V=\sqrt{\frac{10}{3}}m/s$

Velocity of plank is $\sqrt{\frac{10}{3}}m/s$