Question

When the mass of a body is increased by 100% and velocity of a body is decreased by 50%. What is the percentage change in its kinetic energy?

Answer 1

Kinetic energy, $K=\frac{1}{2}m{v}^2$

According to question, mass $m$ becomes $m^{\prime }=m+m\left(100/100\right)=2m$ and the velocity $v$ becomes $v^{\prime }=v-v\left(50/100\right)=v/2$,

Thus, the kinetic energy becomes, $K^{\prime }=\frac{1}{2}{m}^{\prime }{v}^{\prime 2}=\frac{1}{2}\left(2m\right)(v/2{)}^2=K/2$

Thus, % change in kinetic energy $=\frac{K-K^{\prime }}{K}\times 100=\frac{K-K/2}{K}\times 100=50$%