Question

please answer me Ques No. 10 whose answer is 50 percent

Answer 1

Hope it helps you.... :)...

Answer 2

Kinetic energy is increased by 125 % .
Let initial kinetic energy is K
then, final kinetic energy will be K + 125% of K
= K + 1.25K = 2.25K

we know, kinetic energy is directly proportional to square of linear momentum.
e.g., $K.E\propto P^2$
so, $frac{K.E_1}{K.E_2}=\frac{P_1^2}{P_2^2}$

$\frac{K}{2.25K}=\frac{P_1^2}{P_2^2}$

$\frac{10^2}{15^2}=\frac{P_1^2}{P_2^2}$

so, $\frac{P_1}{P_2}=\frac{10}{15}$

or, $\frac{P_2}{P_1}=\frac{15}{10}$

or,$\frac{P_2-P_1}{P_1}\times100=\frac{15-10}{10}\times100=50$

hence, percentage change in momentum is 50%