Question

The momentum of a body increases by $20$%. Find the percentage increase in its kinetic energy.

Answer 1

The relation between kinetic energy and momentum is

$KE=\frac{p^2}{2m}........\left(1\right)$ Where, p is the momentum

When momentum is increases by $20$%, new momentum is $p^{\prime }$

$p^{\prime }=p+0.2p$

$p^{\prime }=1.2p$

New kinetic energy

$K{E}^{\prime }=\frac{p{{}^{\prime }}^2}{2m}$

$K{E}^{\prime }=\frac{{\left(1.2p\right)}^2}{2m}$

$K{E}^{\prime }=\frac{1.44p^2}{2m}$

$K{E}^{\prime }=1.44KE$ ( from equation 1 )

Change in kinetic energy

$\Delta KE=\frac{K{E}^{\prime }-KE}{KE}\times 100$

$\Delta KE=\frac{1.44KE-KE}{KE}\times 100$

$\Delta KE=44$%

So, kinetic energy is increased by 44 %