JEE Physics 2023 Question with Solution

Given: The momentum of the body increases by $50\%$.

Find: The percentage increase in kinetic energy.

From the relation between kinetic energy and momentum,

$$K = \frac{P^2}{2m}$$

where $P$ is momentum and $m$ is mass.

Let the initial momentum be $P_1$. Then the initial kinetic energy is

$$K_1 = \frac{P_1^2}{2m}$$

The new momentum is

$$P_2 = P_1 + 0.5P_1 = 1.5P_1$$

So the new kinetic energy becomes

$$K_2 = \frac{P_2^2}{2m} = \frac{(1.5P_1)^2}{2m} = \frac{2.25P_1^2}{2m} = 2.25K_1$$

Now the percentage increase in kinetic energy is

$$\frac{K_2 - K_1}{K_1} \times 100 = \frac{2.25K_1 - K_1}{K_1} \times 100 = 125\%$$

Therefore, the percentage increase in kinetic energy is $125\%$.

Given: Momentum increases by $50\%$.

Find: The percentage increase in kinetic energy.

Since

$$K = \frac{P^2}{2m}$$

for constant mass, kinetic energy is proportional to the square of momentum:

$$K \propto P^2$$

If momentum becomes $1.5P$, then kinetic energy becomes

$$(1.5)^2 = 2.25$$

Times the initial value.

So the increase is

$$(2.25 - 1) \times 100 = 125\%$$

Therefore, the correct numerical answer is $125$.