If the linear momentum of a body is increased by 5%, by what percentage shall the kinetic energy increase (approx.)?

Options

A. 0.1
B. 0.2
C. 0.25
D. 0.5

Explanation

To solve the problem of how much the kinetic energy increases when the linear momentum of a body is increased by 5%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between momentum and kinetic energy**:
- The linear momentum \( p \) of a body is given by:
\[ p = mv \]
where \( m \) is the mass and \( v \) is the velocity.
- The kinetic energy \( KE \) of the body is given by:
\[ KE = \frac{1}{2} mv^2 \]

2. **Express kinetic energy in terms of momentum**:
- We can express kinetic energy in terms of momentum:
\[ KE = \frac{p^2}{2m} \]

3. **Calculate the final momentum after a 5% increase**:
- If the initial momentum is \( p_i \), then after a 5% increase, the final momentum \( p_f \) is:
\[ p_f = p_i + 0.05 p_i = 1.05 p_i \]

4. **Calculate the initial and final kinetic energy**:
- The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{p_i^2}{2m} \]
- The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{p_f^2}{2m} = \frac{(1.05 p_i)^2}{2m} = \frac{1.1025 p_i^2}{2m} \]

5. **Find the change in kinetic energy**:
- The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_f - KE_i = \frac{1.1025 p_i^2}{2m} - \frac{p_i^2}{2m} = \frac{(1.1025 - 1)p_i^2}{2m} = \frac{0.1025 p_i^2}{2m} \]

6. **Calculate the percentage increase in kinetic energy**:
- The percentage change in kinetic energy is given by:
\[ \text{Percentage Change} = \left(\frac{\Delta KE}{KE_i}\right) \times 100 = \left(\frac{\frac{0.1025 p_i^2}{2m}}{\frac{p_i^2}{2m}}\right) \times 100 \]
- Simplifying this gives:
\[ \text{Percentage Change} = 0.1025 \times 100 = 10.25\% \]

### Final Answer:
The kinetic energy increases by approximately **10.25%** when the linear momentum is increased by 5%.