Question

The kinetic energy of a body increases by 300% due to change in its velocity. The momentum increases by

• A. $100\%$
• B. $200\%$
• C. $\sqrt{300}\%$
• D. $400\%$

Answer 1

The correct option is A $100\%$

when the kinetic energy increasing by $300$%, new kinetic energy t will be

$K^{\prime }=k+300$% k

$K^{\prime }=K+3K$

$K^{\prime }=4K...........\left(i\right)$

We know that $K.E=K=\frac{p^2}{2m}$($m$ is the mass of object)

$\therefore$ ,equation (i) become

$\begin{array}{c}=\frac{p{}^{\prime 2}}{2m}=4\frac{p^2}{2m}\\ =p{}^{\prime 2}=4p^2\\ p^{\prime }=\sqrt{4}{p}^2\\ p^{\prime }=2p..........\left(ii\right)\end{array}$

% change in momentum $=\frac{p^{\prime }-p}{p}\times 100$

$=\frac{2p-p}{p}\times 100$$=$1 \times 100=100$

$\therefore$ increases in momentum is $100$%.