If momentum of an object is increased by 10%, then is kinetic energy will increase by

Options

A. 0.2
B. 0.21
C. 0.4
D. 0.19

Answer

To find out how much the kinetic energy of an object increases when its momentum is increased by 10%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formulas**:
- The momentum \( p \) of an object is given by:
\[ p = mv \]
where \( m \) is mass and \( v \) is velocity.
- The kinetic energy \( KE \) can be expressed in terms of momentum as:
\[ KE = \frac{p^2}{2m} \]

2. **Initial Kinetic Energy**:
- Let the initial momentum be \( p \). The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{p^2}{2m} \]

3. **Increase in Momentum**:
- If the momentum is increased by 10%, the new momentum \( p' \) becomes:
\[ p' = p + 0.1p = 1.1p \]

4. **Final Kinetic Energy**:
- The final kinetic energy \( KE_f \) with the new momentum \( p' \) is:
\[ KE_f = \frac{(p')^2}{2m} = \frac{(1.1p)^2}{2m} \]
- Expanding this gives:
\[ KE_f = \frac{1.21p^2}{2m} \]

5. **Change in Kinetic Energy**:
- The change in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_f - KE_i \]
- Substituting the expressions for \( KE_f \) and \( KE_i \):
\[ \Delta KE = \frac{1.21p^2}{2m} - \frac{p^2}{2m} \]
- Factoring out \( \frac{p^2}{2m} \):
\[ \Delta KE = \frac{p^2}{2m}(1.21 - 1) = \frac{p^2}{2m} \times 0.21 \]

6. **Conclusion**:
- The increase in kinetic energy is:
\[ \Delta KE = 0.21 \frac{p^2}{2m} \]
- This means the kinetic energy increases by 21% of the initial kinetic energy.

### Final Answer:
The kinetic energy will increase by 21% when the momentum is increased by 10%.