When momentum of a body increases by `200%` its KE increases by

Options

A. `200%`
B. `300%`
C. `400%`
D. `800%`

Answer

To solve the problem of how much the kinetic energy (KE) increases when the momentum of a body increases by 200%, we can follow these steps:

### Step 1: Understand the relationship between momentum and kinetic energy
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the object.

Momentum (P) is defined as:
\[ P = mv \]
From this, we can express velocity in terms of momentum:
\[ v = \frac{P}{m} \]

### Step 2: Express kinetic energy in terms of momentum
Substituting the expression for \( v \) into the kinetic energy formula, we get:
\[ KE = \frac{1}{2} m \left(\frac{P}{m}\right)^2 = \frac{P^2}{2m} \]

### Step 3: Define initial momentum and kinetic energy
Let the initial momentum be \( P_0 \). The initial kinetic energy \( KE_0 \) can be expressed as:
\[ KE_0 = \frac{P_0^2}{2m} \]

### Step 4: Calculate the final momentum
If the momentum increases by 200%, the final momentum \( P_f \) is:
\[ P_f = P_0 + 200\% \text{ of } P_0 = P_0 + 2P_0 = 3P_0 \]

### Step 5: Calculate the final kinetic energy
Now, substituting \( P_f \) into the kinetic energy formula:
\[ KE_f = \frac{P_f^2}{2m} = \frac{(3P_0)^2}{2m} = \frac{9P_0^2}{2m} \]

### Step 6: Relate final kinetic energy to initial kinetic energy
Now we can relate the final kinetic energy to the initial kinetic energy:
\[ KE_f = 9 \cdot \frac{P_0^2}{2m} = 9 \cdot KE_0 \]

### Step 7: Calculate the percentage increase in kinetic energy
The percentage increase in kinetic energy can be calculated as:
\[ \text{Percentage Increase} = \frac{KE_f - KE_0}{KE_0} \times 100\% \]
Substituting the values:
\[ \text{Percentage Increase} = \frac{9KE_0 - KE_0}{KE_0} \times 100\% = \frac{8KE_0}{KE_0} \times 100\% = 800\% \]

### Conclusion
Thus, when the momentum of a body increases by 200%, its kinetic energy increases by **800%**.

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