Question

$KE$ of a body is increase by $44$%. What is the percent increase in the momentum?

• A. $11$%
• B. $20$%
• C. $30$%
• D. $44$%

Answer 1

The correct option is A $11$%

Momentum $P=mv$

KInetic energy $KE=\frac{1}{2}m{v}^2$ $\Rightarrow mKE=\frac{1}{2}(mv{)}^2$ $\Rightarrow KE=\frac{P^2}{2m}$ $\Rightarrow P=\sqrt{2mKE}$

$P_1=\sqrt{2mKE}$ , $P_2=\sqrt{2m\times 144KE}$

$\Rightarrow \frac{P_2}{P_1}=12$ $\Rightarrow P_2=12P_1$

Percent increase in momentum $=\frac{P_2-P_1}{P_1}\%=11\%$