If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is

Options

A. 0.5
B. 1
C. 1.25
D. 2

Answer

To solve the problem, we need to find the percentage increase in kinetic energy when the momentum of a body is increased by 50%.

### Step-by-Step Solution:

1. **Understand the relationship between 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:
\[ KE = \frac{1}{2} mv^2 \]

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 new momentum after a 50% increase**:
- If the initial momentum is \( p \), then a 50% increase in momentum means:
\[ p' = p + 0.5p = 1.5p \]

4. **Calculate the new kinetic energy**:
- Substitute the new momentum into the kinetic energy formula:
\[ KE' = \frac{(p')^2}{2m} = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = \frac{1.125p^2}{m} \]

5. **Calculate the initial kinetic energy**:
- The initial kinetic energy is:
\[ KE = \frac{p^2}{2m} \]

6. **Determine the increase in kinetic energy**:
- The increase in kinetic energy is:
\[ \Delta KE = KE' - KE = \frac{1.125p^2}{m} - \frac{p^2}{2m} \]
- To simplify this, we need a common denominator:
\[ \Delta KE = \frac{1.125p^2}{m} - \frac{0.5p^2}{m} = \frac{(1.125 - 0.5)p^2}{m} = \frac{0.625p^2}{m} \]

7. **Calculate the percentage increase in kinetic energy**:
- The percentage increase in kinetic energy is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta KE}{KE} \right) \times 100 \]
- Substitute the values:
\[ \text{Percentage Increase} = \left( \frac{0.625p^2/m}{(p^2/2m)} \right) \times 100 = \left( \frac{0.625}{0.5} \right) \times 100 = 1.25 \times 100 = 125\% \]

### Final Answer:
The percentage increase in kinetic energy when the momentum of a body is increased by 50% is **125%**.