Kinetic energy of a particle is increased by
(a) 50% (b) 1%
Find percentage change in linear momentum.

Explanation

To solve the problem, we need to find the percentage change in linear momentum when the kinetic energy of a particle is increased by either 50% or 1%. Let's break it down step by step.

### Step 1: Understand the relationship between kinetic energy and momentum
The kinetic energy (K) of a particle is related to its momentum (P) by the equation:
\[ K = \frac{P^2}{2m} \]
From this, we can express momentum in terms of kinetic energy:
\[ P = \sqrt{2mK} \]

### Step 2: Calculate the initial momentum
Let the initial kinetic energy be \( K_i \). Therefore, the initial momentum \( P_i \) can be expressed as:
\[ P_i = \sqrt{2mK_i} \]

### Step 3: Calculate the final momentum for a 50% increase in kinetic energy
If the kinetic energy is increased by 50%, the final kinetic energy \( K_f \) becomes:
\[ K_f = K_i + 0.5K_i = 1.5K_i \]
Now, substituting this into the momentum equation gives:
\[ P_f = \sqrt{2mK_f} = \sqrt{2m(1.5K_i)} = \sqrt{1.5} \cdot \sqrt{2mK_i} = \sqrt{1.5} \cdot P_i \]

### Step 4: Calculate the percentage change in momentum
The percentage change in momentum is given by:
\[ \text{Percentage Change} = \frac{P_f - P_i}{P_i} \times 100 \]
Substituting \( P_f \):
\[ \text{Percentage Change} = \frac{\sqrt{1.5} P_i - P_i}{P_i} \times 100 = (\sqrt{1.5} - 1) \times 100 \]
Calculating \( \sqrt{1.5} \) gives approximately 1.2247, so:
\[ \text{Percentage Change} = (1.2247 - 1) \times 100 \approx 22.47\% \]
Thus, the percentage change in momentum when kinetic energy is increased by 50% is approximately **22%**.

### Step 5: Calculate the final momentum for a 1% increase in kinetic energy
For a 1% increase in kinetic energy, the final kinetic energy \( K_f \) becomes:
\[ K_f = K_i + 0.01K_i = 1.01K_i \]
Now substituting this into the momentum equation gives:
\[ P_f = \sqrt{2mK_f} = \sqrt{2m(1.01K_i)} = \sqrt{1.01} \cdot \sqrt{2mK_i} = \sqrt{1.01} \cdot P_i \]

### Step 6: Calculate the percentage change in momentum for the 1% increase
Using the same formula for percentage change:
\[ \text{Percentage Change} = \frac{P_f - P_i}{P_i} \times 100 \]
Substituting \( P_f \):
\[ \text{Percentage Change} = \frac{\sqrt{1.01} P_i - P_i}{P_i} \times 100 = (\sqrt{1.01} - 1) \times 100 \]
Calculating \( \sqrt{1.01} \) gives approximately 1.00499, so:
\[ \text{Percentage Change} = (1.00499 - 1) \times 100 \approx 0.499\% \]
Thus, the percentage change in momentum when kinetic energy is increased by 1% is approximately **0.5%**.

### Final Answers:
- For a 50% increase in kinetic energy, the percentage change in momentum is approximately **22%**.
- For a 1% increase in kinetic energy, the percentage change in momentum is approximately **0.5%**.