Question

The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ______%

Answer 1

Correct Answer: 125

The relationship between momentum (p) and kinetic energy (KE) of a body is given by the equations: \( p = mv \) and \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. We need to find the percentage increase in kinetic energy when the momentum is increased by 50%.

If the initial momentum is \( p \), then a 50% increase results in a new momentum \( p' = 1.5p \). Assuming the mass \( m \) remains constant, we can express the initial and new kinetic energies as \( KE_1 = \frac{p^2}{2m} \) and \( KE_2 = \frac{(1.5p)^2}{2m} \) respectively.

Calculating \( KE_2 \):

\( KE_2 = \frac{2.25p^2}{2m} = \frac{2.25p^2}{2m} = 1.125 \cdot \frac{p^2}{2m} = 1.125 \cdot KE_1 \)

The increase in kinetic energy \( ΔKE = KE_2 - KE_1 = 1.125KE_1 - KE_1 = 0.125KE_1 \)

Thus, the percentage increase in kinetic energy is:

\(\frac{ΔKE}{KE_1} \times 100\% = \frac{0.125KE_1}{KE_1} \times 100\% = 12.5\%\)

Therefore, the kinetic energy of the body increases by 125%, which is within the specified range of 125% to 125%.