If the increase in the kinetic energy of a body is 22%, then the increase in the momentum will be

Options

A. 0.22
B. 0.44
C. 0.1
D. 3

Answer

To solve the problem of finding the increase in momentum when the kinetic energy of a body increases by 22%, we can follow these steps:

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

Momentum (p) is given by:
\[ p = mv \]

### Step 2: Express kinetic energy in terms of momentum
From the momentum equation, we can express velocity as:
\[ v = \frac{p}{m} \]
Substituting this into the kinetic energy formula gives:
\[ K.E = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{p^2}{2m} \]

### Step 3: Set up the equations for initial and final kinetic energy
Let the initial kinetic energy be \( K_1 \) and the final kinetic energy after a 22% increase be \( K_2 \):
\[ K_1 = \frac{p_1^2}{2m} \]
\[ K_2 = K_1 + 0.22 K_1 = 1.22 K_1 \]

### Step 4: Relate the increase in kinetic energy to momentum
Using the expression for kinetic energy in terms of momentum:
\[ K_2 = \frac{p_2^2}{2m} \]
Setting the two expressions for \( K_2 \) equal gives:
\[ 1.22 K_1 = \frac{p_2^2}{2m} \]
Substituting \( K_1 \):
\[ 1.22 \left(\frac{p_1^2}{2m}\right) = \frac{p_2^2}{2m} \]
This simplifies to:
\[ 1.22 p_1^2 = p_2^2 \]

### Step 5: Solve for the ratio of momenta
Taking the square root of both sides:
\[ p_2 = \sqrt{1.22} p_1 \]
Calculating \( \sqrt{1.22} \):
\[ \sqrt{1.22} \approx 1.104 \]
Thus:
\[ p_2 \approx 1.104 p_1 \]

### Step 6: Find the increase in momentum
The increase in momentum is:
\[ \Delta p = p_2 - p_1 = (1.104 p_1 - p_1) = 0.104 p_1 \]

### Step 7: Express the increase as a fraction of the initial momentum
To express this increase as a percentage:
\[ \text{Percentage increase} = \left(\frac{\Delta p}{p_1}\right) \times 100 = \left(0.104\right) \times 100 = 10.4\% \]

### Final Answer
The increase in momentum is approximately \( 0.104 p_1 \) or \( 10.4\% \).