Question

If the momentum of a particle is increased by 60%, then percentage increase in its kinetic energy, is
(1)125%
(2)148%
(3)156%
(4)200%

Answer 1

Dear Student,


$$\begin{gathered} lettheinitialmomentumbeP=mv \\ momentumcanbeincreasedvelocity \\ letthenewvelocitybev\text{'} \\ mv\text{'}=1.6mv \\ v\text{'}=1.6v \\ initialkineticenergy \\ K=\frac{m{v}^2}{2} \\ newKE \\ K\text{'}=\frac{mv{\text{'}}^2}{2}=\frac{2.56m{v}^2}{2}=2.56K \\ percentageincrease=\frac{2.56K-K}{K}*100=156percentincreasein \\ REgards \end{gathered}$$