KE of a body is increased by `44%.` What is the percent increse in the momentum ?

Options

A. `10%`
B. `20%`
C. `30%`
D. `44%`

Answer

To solve the problem, we need to find the percentage increase in momentum when the kinetic energy of a body is increased by 44%.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy and Momentum**:
- The kinetic energy (KE) of a body is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- The momentum (p) of a body is given by the formula:
\[ p = mv \]

2. **Initial Kinetic Energy**:
- Let the initial kinetic energy be \( KE_0 = k_0 \).

3. **Final Kinetic Energy**:
- The kinetic energy is increased by 44%, so the final kinetic energy is:
\[ KE_f = k_0 + 0.44 k_0 = 1.44 k_0 \]

4. **Relating Kinetic Energy to Momentum**:
- We can express kinetic energy in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
- Therefore, the initial momentum \( p_0 \) can be expressed as:
\[ p_0 = \sqrt{2m k_0} \]

5. **Final Momentum**:
- The final kinetic energy can also be expressed in terms of momentum:
\[ KE_f = \frac{p_f^2}{2m} \]
- Setting this equal to the final kinetic energy:
\[ 1.44 k_0 = \frac{p_f^2}{2m} \]
- Rearranging gives:
\[ p_f^2 = 2m \cdot 1.44 k_0 \]
- Thus, the final momentum \( p_f \) is:
\[ p_f = \sqrt{2m \cdot 1.44 k_0} = \sqrt{1.44} \cdot \sqrt{2m k_0} = 1.2 p_0 \]

6. **Calculating Percentage Increase in Momentum**:
- The percentage increase in momentum is calculated as:
\[ \text{Percentage Increase} = \frac{p_f - p_0}{p_0} \times 100\% \]
- Substituting the values:
\[ \text{Percentage Increase} = \frac{1.2 p_0 - p_0}{p_0} \times 100\% = \frac{0.2 p_0}{p_0} \times 100\% = 20\% \]

### Final Answer:
The percentage increase in momentum is **20%**.

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