Linear momentum of particle is increased by
(a) 100% (b) 1%
without changing its mass. Find percentage increase in its kinetic energy in both cases.

Explanation

To solve the problem, we will analyze the relationship between linear momentum (P) and kinetic energy (K) of a particle. The formulas we will use are:

1. The relationship between kinetic energy and momentum:
\[ K = \frac{P^2}{2m} \]

Where:
- \( K \) is the kinetic energy,
- \( P \) is the momentum,
- \( m \) is the mass of the particle.

### Step 1: Analyze the case when momentum is increased by 100%

1. **Initial Momentum**: Let the initial momentum be \( P_i \).
2. **Final Momentum**: If the momentum is increased by 100%, the final momentum \( P_f \) will be:
\[ P_f = P_i + 100\% \text{ of } P_i = P_i + P_i = 2P_i \]
3. **Initial Kinetic Energy**: The initial kinetic energy \( K_i \) is:
\[ K_i = \frac{P_i^2}{2m} \]
4. **Final Kinetic Energy**: The final kinetic energy \( K_f \) is:
\[ K_f = \frac{P_f^2}{2m} = \frac{(2P_i)^2}{2m} = \frac{4P_i^2}{2m} = \frac{2P_i^2}{m} \]
5. **Percentage Increase in Kinetic Energy**: The change in kinetic energy \( \Delta K \) is:
\[ \Delta K = K_f - K_i = \frac{2P_i^2}{m} - \frac{P_i^2}{2m} = \frac{4P_i^2}{2m} - \frac{P_i^2}{2m} = \frac{3P_i^2}{2m} \]
The percentage increase in kinetic energy is:
\[ \text{Percentage Increase} = \frac{\Delta K}{K_i} \times 100 = \frac{\frac{3P_i^2}{2m}}{\frac{P_i^2}{2m}} \times 100 = 3 \times 100 = 300\% \]

### Step 2: Analyze the case when momentum is increased by 1%

1. **Initial Momentum**: Let the initial momentum be \( P_i \).
2. **Final Momentum**: If the momentum is increased by 1%, the final momentum \( P_f \) will be:
\[ P_f = P_i + 1\% \text{ of } P_i = P_i + 0.01 P_i = 1.01 P_i \]
3. **Final Kinetic Energy**: The final kinetic energy \( K_f \) is:
\[ K_f = \frac{P_f^2}{2m} = \frac{(1.01 P_i)^2}{2m} = \frac{1.0201 P_i^2}{2m} \]
4. **Percentage Increase in Kinetic Energy**: The change in kinetic energy \( \Delta K \) is:
\[ \Delta K = K_f - K_i = \frac{1.0201 P_i^2}{2m} - \frac{P_i^2}{2m} = \frac{(1.0201 - 1)P_i^2}{2m} = \frac{0.0201 P_i^2}{2m} \]
The percentage increase in kinetic energy is:
\[ \text{Percentage Increase} = \frac{\Delta K}{K_i} \times 100 = \frac{\frac{0.0201 P_i^2}{2m}}{\frac{P_i^2}{2m}} \times 100 = 0.0201 \times 100 = 2.01\% \]

### Final Answers:
- For a 100% increase in momentum, the percentage increase in kinetic energy is **300%**.
- For a 1% increase in momentum, the percentage increase in kinetic energy is approximately **2.01%**.