Question

When momentum of a body increases by $200\%$, its $KE$ increases by

• A. $200\%$
• B. $300\%$
• C. $400\%$
• D. $800\%$

Answer 1

The correct option is D $800\%$

Momentum , $P=mv$

Kinetic Energy, $KE=\frac{1}{2}m{v}^2$

$\Rightarrow$ $mKE=\frac{1}{2}(mv{)}^2=\frac{p^2}{2}$

$\Rightarrow KE=\frac{P^2}{2m}$

Now $P$ is increases by $200\%$ , $P_1=3P$

$\Rightarrow K{E}_1=\frac{(3P{)}^2}{2m}$

So $KE$ increases by $=\frac{8P^2}{2m}$

$\Rightarrow \Delta KE\%=\frac{\frac{9P^2}{2m}-\frac{p^2}{2m}}{\frac{p^2}{2m}}\times 100=800\%$