The momentum of a body is increased by 25%. The kinetic energy is increased by about

Options

A. `25%`
B. `5%`
C. `56%`
D. `38%`

Explanation

To solve the problem of how much the kinetic energy increases when the momentum of a body is increased by 25%, we can follow these steps:

### Step-by-Step Solution:

1. **Define Initial Momentum and Kinetic Energy**:
- Let the initial momentum \( p_i \) of the body be given by the formula:
\[ p_i = mv \]
- The initial kinetic energy \( KE_1 \) is given by:
\[ KE_1 = \frac{1}{2} mv^2 \]

2. **Express Kinetic Energy in Terms of Momentum**:
- We can express the kinetic energy in terms of momentum:
\[ KE_1 = \frac{p_i^2}{2m} \]

3. **Calculate Final Momentum**:
- The momentum is increased by 25%, so the final momentum \( p_f \) is:
\[ p_f = p_i + 0.25p_i = 1.25p_i \]

4. **Calculate Final Kinetic Energy**:
- The final kinetic energy \( KE_2 \) can be expressed as:
\[ KE_2 = \frac{p_f^2}{2m} = \frac{(1.25p_i)^2}{2m} = \frac{1.5625p_i^2}{2m} \]

5. **Calculate the Increase in Kinetic Energy**:
- The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_2 - KE_1 = \frac{1.5625p_i^2}{2m} - \frac{p_i^2}{2m} \]
- Simplifying this gives:
\[ \Delta KE = \frac{(1.5625 - 1)p_i^2}{2m} = \frac{0.5625p_i^2}{2m} \]

6. **Calculate the Percentage Increase in Kinetic Energy**:
- The percentage increase in kinetic energy is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta KE}{KE_1}\right) \times 100 \]
- Substituting the values:
\[ \text{Percentage Increase} = \left(\frac{\frac{0.5625p_i^2}{2m}}{\frac{p_i^2}{2m}}\right) \times 100 = 0.5625 \times 100 = 56.25\% \]

### Final Answer:
The kinetic energy is increased by approximately **56.25%**.

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