The increase in kinetic energy for a body whose increase in linear momentum is 50% will be

Options

A. ` 100%`
B. ` 125%`
C. `300%`
D. `175% `

Explanation

To solve the problem of finding the increase in kinetic energy when the linear momentum of a body increases by 50%, we can follow these steps:

### 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 and \( v \) is the velocity.
- The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]

2. **Define the initial momentum and kinetic energy**:
- Let the initial momentum be \( p = mv \).
- The initial kinetic energy is:
\[ KE = \frac{1}{2} mv^2 \]

3. **Calculate the new momentum after a 50% increase**:
- An increase of 50% in momentum means:
\[ p' = p + 0.5p = 1.5p \]
- Substituting for \( p \):
\[ p' = 1.5(mv) = m(1.5v) \]
- This indicates that the new velocity \( v' \) is:
\[ v' = 1.5v \]

4. **Calculate the new kinetic energy with the increased velocity**:
- The new kinetic energy \( KE' \) is:
\[ KE' = \frac{1}{2} m(v')^2 = \frac{1}{2} m(1.5v)^2 \]
- Simplifying this:
\[ KE' = \frac{1}{2} m(2.25v^2) = \frac{9}{8} mv^2 \]

5. **Find the increase in kinetic energy**:
- The increase in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE' - KE = \frac{9}{8} mv^2 - \frac{1}{2} mv^2 \]
- To combine these, convert \( \frac{1}{2} mv^2 \) to a fraction with a denominator of 8:
\[ \Delta KE = \frac{9}{8} mv^2 - \frac{4}{8} mv^2 = \frac{5}{8} mv^2 \]

6. **Calculate the percentage increase in kinetic energy**:
- The percentage increase is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta KE}{KE}\right) \times 100 = \left(\frac{\frac{5}{8} mv^2}{\frac{1}{2} mv^2}\right) \times 100 \]
- Simplifying this:
\[ \text{Percentage Increase} = \left(\frac{5/8}{1/2}\right) \times 100 = \left(\frac{5}{8} \times \frac{2}{1}\right) \times 100 = \left(\frac{10}{8}\right) \times 100 = 125\% \]

### Final Answer:
The increase in kinetic energy when the linear momentum increases by 50% is **125%**.