Question

The momentum of a body is increased by $50$%. The kinetic energy of the body will be increased by

• A. $50$%
• B. $125$%
• C. $330$%
• D. $400$%

Answer 1

The correct option is B $125$%

Kinetic energy K and linear momentum P are related as

$K=\frac{P^2}{2m}$.....(1)

The new momentum is P+50% 0f P =(3/2)P.

If the new kinetic energy is K’ then

$K\text{'}=\frac{\left(\right(3/2)P{)}^2}{2m}=\frac{9}{4}\frac{P^2}{2m}$………(2)

Divide equation (2) by Equation (1) to get

K′=(9/4)K

Now the % change in K is

$\frac{K\text{'}-K}{K}\times 100=\frac{\left(9/4\right)K-K}{K}\times 100=125$%

Hence the correct option is B