If the kinetic energy of a body increases by `125%`, the percentage increases in its momentum is

Options

A. `50%`
B. `62.5%`
C. `250%`
D. `200%`

Explanation

To solve the problem, we need to find the percentage increase in momentum when the kinetic energy of a body increases by 125%. Let's break down the solution step by step.

### Step 1: Understand the increase in kinetic energy
The initial kinetic energy (KE_initial) is denoted as \( KE_i \). If the kinetic energy increases by 125%, the final kinetic energy (KE_final) can be expressed as:
\[ KE_{final} = KE_i + 1.25 \times KE_i = 2.25 \times KE_i \]

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum (P) using the formula:
\[ KE = \frac{P^2}{2m} \]
where \( m \) is the mass of the body.

### Step 3: Set up the equation for final kinetic energy
Using the relationship between kinetic energy and momentum, we can write:
\[ KE_{final} = \frac{P_{final}^2}{2m} \]
Substituting the expression for \( KE_{final} \):
\[ 2.25 \times KE_i = \frac{P_{final}^2}{2m} \]

### Step 4: Set up the equation for initial kinetic energy
Similarly, the initial kinetic energy can be expressed as:
\[ KE_i = \frac{P_{initial}^2}{2m} \]
Substituting this into the equation gives:
\[ 2.25 \times \frac{P_{initial}^2}{2m} = \frac{P_{final}^2}{2m} \]

### Step 5: Cancel common terms
The \( 2m \) terms cancel out:
\[ 2.25 \times P_{initial}^2 = P_{final}^2 \]

### Step 6: Solve for the ratio of final and initial momentum
Taking the square root of both sides:
\[ P_{final} = \sqrt{2.25} \times P_{initial} = \frac{3}{2} \times P_{initial} \]

### Step 7: Calculate the percentage increase in momentum
The percentage increase in momentum can be calculated using the formula:
\[ \text{Percentage Increase} = \frac{P_{final} - P_{initial}}{P_{initial}} \times 100 \]
Substituting \( P_{final} \):
\[ \text{Percentage Increase} = \frac{\frac{3}{2} P_{initial} - P_{initial}}{P_{initial}} \times 100 \]
This simplifies to:
\[ \text{Percentage Increase} = \frac{\frac{3}{2} P_{initial} - \frac{2}{2} P_{initial}}{P_{initial}} \times 100 = \frac{\frac{1}{2} P_{initial}}{P_{initial}} \times 100 = \frac{1}{2} \times 100 = 50\% \]

### Final Answer
Thus, the percentage increase in momentum is **50%**.

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