If the linear momentum is increased by 50%, then KE will be increased by :

Options

A. `50%`
B. `100%`
C. `125%`
D. `25%`

Explanation

To solve the problem of how much the kinetic energy (KE) will increase when the linear momentum is increased by 50%, we can follow these steps:

### Step 1: Understand the relationship between momentum and kinetic energy
The linear momentum \( P \) of an object is given by the formula:
\[ P = m \cdot v \]
where \( m \) is the mass and \( v \) is the velocity of the object.

The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]

### Step 2: Define the initial momentum and kinetic energy
Let the initial momentum be \( P_{\text{initial}} = m \cdot v \).
The initial kinetic energy can be expressed as:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]

### Step 3: Calculate the new momentum after a 50% increase
If the momentum is increased by 50%, the new momentum \( P_{\text{final}} \) will be:
\[ P_{\text{final}} = P_{\text{initial}} + 0.5 \cdot P_{\text{initial}} = 1.5 \cdot P_{\text{initial}} \]
Thus,
\[ P_{\text{final}} = 1.5 \cdot (m \cdot v) = m \cdot v' \]
where \( v' \) is the new velocity.

### Step 4: Relate the new velocity to the initial velocity
From the equation \( P_{\text{final}} = m \cdot v' \), we can express the new velocity:
\[ m \cdot v' = 1.5 \cdot (m \cdot v) \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ v' = 1.5 \cdot v \]

### Step 5: Calculate the new kinetic energy
Now, we can find the new kinetic energy \( KE_{\text{final}} \):
\[ KE_{\text{final}} = \frac{1}{2} m (v')^2 = \frac{1}{2} m (1.5v)^2 \]
Calculating \( (1.5v)^2 \):
\[ (1.5v)^2 = 2.25v^2 \]
Thus,
\[ KE_{\text{final}} = \frac{1}{2} m (2.25v^2) = \frac{2.25}{2} m v^2 = 1.125 m v^2 \]

### Step 6: Find the percentage increase in kinetic energy
Now we can find the percentage increase in kinetic energy:
1. Initial kinetic energy:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
2. Final kinetic energy:
\[ KE_{\text{final}} = 1.125 \cdot \frac{1}{2} m v^2 \]
3. The increase in kinetic energy:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 1.125 \cdot \frac{1}{2} m v^2 - \frac{1}{2} m v^2 \]
\[ \Delta KE = (1.125 - 1) \cdot \frac{1}{2} m v^2 = 0.125 \cdot \frac{1}{2} m v^2 \]

4. Percentage increase:
\[ \text{Percentage Increase} = \left( \frac{\Delta KE}{KE_{\text{initial}}} \right) \times 100 \]
\[ = \left( \frac{0.125 \cdot \frac{1}{2} m v^2}{\frac{1}{2} m v^2} \right) \times 100 = 12.5\% \]

### Final Answer
The kinetic energy will increase by **12.5%** when the linear momentum is increased by 50%.

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