When the momentum of a body decreases by 10%, its K.E. decreases by

Options

A. 0.2
B. 0.4
C. 0.36
D. none of these

Answer

To solve the problem, we need to determine how much the kinetic energy (K.E.) of a body decreases when its momentum decreases by 10%.

### 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 \( K.E. \) is given by the formula:
\[ K.E. = \frac{1}{2} mv^2 \]

2. **Express kinetic energy in terms of momentum**:
- We can express velocity \( v \) in terms of momentum:
\[ v = \frac{p}{m} \]
- Substituting this into the kinetic energy formula gives:
\[ K.E. = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{p^2}{2m} \]
- This shows that kinetic energy is directly proportional to the square of momentum.

3. **Determine the new momentum after a 10% decrease**:
- If the momentum decreases by 10%, the new momentum \( p' \) is:
\[ p' = p - 0.1p = 0.9p \]

4. **Calculate the new kinetic energy with the decreased momentum**:
- Using the new momentum in the kinetic energy formula:
\[ K.E.' = \frac{(p')^2}{2m} = \frac{(0.9p)^2}{2m} = \frac{0.81p^2}{2m} \]

5. **Determine the decrease in kinetic energy**:
- The initial kinetic energy was:
\[ K.E. = \frac{p^2}{2m} \]
- The decrease in kinetic energy \( \Delta K.E. \) can be calculated as:
\[ \Delta K.E. = K.E. - K.E.' = \frac{p^2}{2m} - \frac{0.81p^2}{2m} \]
- Simplifying this gives:
\[ \Delta K.E. = \frac{p^2}{2m} (1 - 0.81) = \frac{p^2}{2m} \cdot 0.19 \]

6. **Express the decrease in kinetic energy as a percentage**:
- The percentage decrease in kinetic energy is:
\[ \text{Percentage Decrease} = \frac{\Delta K.E.}{K.E.} \times 100 = \frac{0.19 \cdot \frac{p^2}{2m}}{\frac{p^2}{2m}} \times 100 = 19\% \]

### Final Answer:
The kinetic energy decreases by **19%** when the momentum of the body decreases by 10%.