When the momentum of a body increases by 100%, its K.E. increases by :

Options

A. 4
B. 1
C. 3
D. none

Answer

To solve the problem of how much the kinetic energy (K.E.) increases when the momentum of a body increases by 100%, we can follow these steps:

### Step-by-Step Solution:

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

2. **Express kinetic energy in terms of momentum**:
- We can express \( K.E. \) in terms of momentum. From the momentum formula, we can solve for \( v \):
\[ v = \frac{p}{m} \]
- Substituting \( v \) into the kinetic energy formula:
\[ K.E. = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{1}{2} m \frac{p^2}{m^2} = \frac{p^2}{2m} \]

3. **Determine the initial kinetic energy**:
- Let the initial momentum be \( p \). The initial kinetic energy \( K.E_1 \) can be expressed as:
\[ K.E_1 = \frac{p^2}{2m} \]

4. **Calculate the new momentum after a 100% increase**:
- If the momentum increases by 100%, the new momentum \( p' \) becomes:
\[ p' = p + 100\% \text{ of } p = p + p = 2p \]

5. **Calculate the new kinetic energy**:
- The new kinetic energy \( K.E_2 \) with the new momentum \( p' \) is:
\[ K.E_2 = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = \frac{2p^2}{m} \]

6. **Find the increase in kinetic energy**:
- The increase in kinetic energy \( \Delta K.E. \) is given by:
\[ \Delta K.E. = K.E_2 - K.E_1 = \frac{2p^2}{m} - \frac{p^2}{2m} \]
- To combine these, we need a common denominator:
\[ K.E_1 = \frac{p^2}{2m} \implies \Delta K.E. = \frac{2p^2}{m} - \frac{p^2}{2m} = \frac{4p^2}{2m} - \frac{p^2}{2m} = \frac{3p^2}{2m} \]

7. **Calculate the percentage increase in kinetic energy**:
- The percentage increase in kinetic energy can be calculated as:
\[ \text{Percentage Increase} = \frac{\Delta K.E.}{K.E_1} \times 100 = \frac{\frac{3p^2}{2m}}{\frac{p^2}{2m}} \times 100 = 3 \times 100 = 300\% \]

### Conclusion:
When the momentum of a body increases by 100%, its kinetic energy increases by **300%**.