If the K.E. of a body is increased by 300%, its momentum will increase by

Options

A. 100 %
B. 15 %
C. `sqrt300%`
D. 17.5 %

Explanation

To solve the problem, we need to understand the relationship between kinetic energy (K.E.) and momentum (P).

### Step-by-Step Solution:

1. **Understanding Kinetic Energy and Momentum**:
The kinetic energy (K.E.) of a body is given by the formula:
\[ K.E. = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the body. The momentum (P) is given by:
\[ P = mv \]

2. **Initial Kinetic Energy**:
Let the initial kinetic energy be \( K_1 \). Thus,
\[ K_1 = \frac{1}{2} mv_1^2 \]
where \( v_1 \) is the initial velocity.

3. **Increase in Kinetic Energy**:
The problem states that the kinetic energy is increased by 300%. This means the new kinetic energy \( K_2 \) is:
\[ K_2 = K_1 + 3K_1 = 4K_1 \]

4. **Relating Kinetic Energy to Momentum**:
We can express the kinetic energy in terms of momentum. From the momentum formula, we can express velocity as:
\[ v = \frac{P}{m} \]
Substituting this into the kinetic energy formula gives:
\[ K.E. = \frac{1}{2} m \left(\frac{P}{m}\right)^2 = \frac{P^2}{2m} \]

5. **Initial Momentum**:
Let the initial momentum be \( P_1 \). Then, the initial kinetic energy can be expressed as:
\[ K_1 = \frac{P_1^2}{2m} \]

6. **New Momentum**:
The new kinetic energy \( K_2 \) can be expressed in terms of the new momentum \( P_2 \):
\[ K_2 = \frac{P_2^2}{2m} \]
Since \( K_2 = 4K_1 \), we have:
\[ \frac{P_2^2}{2m} = 4 \left(\frac{P_1^2}{2m}\right) \]
Simplifying gives:
\[ P_2^2 = 4P_1^2 \]

7. **Finding New Momentum**:
Taking the square root of both sides, we find:
\[ P_2 = 2P_1 \]

8. **Calculating Percentage Increase in Momentum**:
The increase in momentum is:
\[ \Delta P = P_2 - P_1 = 2P_1 - P_1 = P_1 \]
To find the percentage increase:
\[ \text{Percentage Increase} = \left(\frac{\Delta P}{P_1}\right) \times 100 = \left(\frac{P_1}{P_1}\right) \times 100 = 100\% \]

### Final Answer:
The momentum will increase by **100%**.