The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is _______ %. [8-Apr-2023 shift 1]

JEE Mains - Increase in Kinetic Energy

Question: The momentum of a body is increased by 50%. What is the percentage increase in the kinetic energy of the body?
[8-Apr-2023 Shift 1]

Select the correct answer:

(A) 50%

(B) 100%

(C) 125%

(D) 150%

Solution

Let the initial momentum be \( p \) and the mass be \( m \). Then the initial kinetic energy is: \[ KE_1 = \frac{p^2}{2m}. \]

After increasing the momentum by 50%, the new momentum is: \[ p' = 1.5p. \] The new kinetic energy becomes: \[ KE_2 = \frac{{p'}^2}{2m} = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m}. \]

The percentage increase in kinetic energy is: \[ \frac{KE_2 - KE_1}{KE_1} \times 100\% = \frac{\frac{2.25p^2}{2m} - \frac{p^2}{2m}}{\frac{p^2}{2m}} \times 100\%. \] Simplify: \[ \frac{2.25 - 1}{1} \times 100\% = 1.25 \times 100\% = 125\%. \]

Therefore, the kinetic energy increases by 125%.