Question

If momentum of a body is increased by 20%, then its kinetic energy increases by

• A. 36%
• B. 40%
• C. 44%
• D. 48%

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between momentum and kinetic energy. Let's break it down step-by-step:

1. **Formula for Momentum and Kinetic Energy**: The momentum (\(p\)) of a body is given by the formula:

\(p = mv\) 

where \(m\) is the mass of the body and \(v\) is its velocity.

The kinetic energy (\(KE\)) of a body is given by the formula:

\(\text{KE} = \frac{1}{2}mv^2\)

2. **Relating Momentum and Kinetic Energy**: We can express kinetic energy in terms of momentum. By substituting \(v = \frac{p}{m}\) (from the momentum formula) into the kinetic energy equation, we have:

\(\text{KE} = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}\)

3. **Calculate the Increase in Kinetic Energy**: Given that the momentum of the body increases by 20%, the new momentum is:

\(p_{\text{new}} = p + 0.2p = 1.2p\)

The new kinetic energy based on the increased momentum is:

\(\text{KE}_{\text{new}} = \frac{{(1.2p)}^2}{2m} = \frac{1.44p^2}{2m}\)

4. **Determine the Percentage Increase in Kinetic Energy**: The initial kinetic energy was \(\frac{p^2}{2m}\). The percentage increase in kinetic energy can be calculated as:

\(\text{Percentage Increase} = \left(\frac{\text{KE}_{\text{new}} - \text{KE}}{\text{KE}}\right) \times 100%\)

\(= \left(\frac{\frac{1.44p^2}{2m} - \frac{p^2}{2m}}{\frac{p^2}{2m}}\right) \times 100%\)

\(= \left(\frac{1.44p^2 - p^2}{p^2}\right) \times 100%\)

\(= \left(0.44\right) \times 100%\)

\(= 44%\)

Hence, when the momentum of a body is increased by 20%, its kinetic energy increases by 44%.

Answer 2

\(K=\frac{p^2}{2m}\)
\(K=\frac{(1.2p)^2}{2m}\)
\(⇒\frac{K−K}{K}=(1.2)^2−1=0.44\)
⇒ Kinetic energy increases by 44% 
So, the correct option is (C): 44%