QJEEM23Ph5a9d: JEE Main 2023 Physics - Class 11 JEEM23Ph5a9d Solution: The momentum of a body is increased by . The percentage…

The momentum (p) and kinetic energy (K) of a body are related by the equations:

$p = mv$ ,

$K = \frac{1}{2}mv^2$ ,

where m is the mass and v is the velocity of the body.

We can express v in terms of p and m:

$v = \frac{p}{m}$ ,

and substitute this into the equation for K to get:

$K = \frac{p^2}{2m}$ .

So, the kinetic energy is proportional to the square of the momentum.

If the momentum is increased by 50%, the new momentum is 1.5p, and the new kinetic energy is:

$K' = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25K$ .

The percentage increase in the kinetic energy is then:

$\frac{K'-K}{K} \times 100 = \frac{2.25K - K}{K} \times 100 = 1.25 \times 100 = 125\%$ .

So, the percentage increase in the kinetic energy of the body is 125%.