A cricket ball of mass 150 g has an intial velocity u = `(3 hati + 4 hatj)ms^(-1)` and a final velocity `v = -(3hati + 4 hatj) ms^(-1)` , after being hit. The change in momentum (final momentum - initial momentum ) is (in `Kg ms^(1)`)

Options

A. zero
B. `-(0.45 hati + 0.6 hatj)`
C. `-(0.9 hatj + 1.2 hatj)`
D. `-5(hati + hatj ) hati`

Explanation

To find the change in momentum of the cricket ball, we will follow these steps:

### Step 1: Convert mass from grams to kilograms
The mass of the cricket ball is given as 150 grams. We need to convert this to kilograms because the standard unit of mass in the SI system is kilograms.

\[ m = 150 \text{ g} = \frac{150}{1000} \text{ kg} = 0.15 \text{ kg} \]

### Step 2: Identify the initial and final velocities
The initial velocity \( \mathbf{u} \) and final velocity \( \mathbf{v} \) are given as:
\[ \mathbf{u} = (3 \hat{i} + 4 \hat{j}) \text{ m/s} \]
\[ \mathbf{v} = -(3 \hat{i} + 4 \hat{j}) \text{ m/s} = (-3 \hat{i} - 4 \hat{j}) \text{ m/s} \]

### Step 3: Calculate the change in velocity
The change in velocity \( \Delta \mathbf{v} \) is given by the final velocity minus the initial velocity:
\[ \Delta \mathbf{v} = \mathbf{v} - \mathbf{u} \]
Substituting the values:
\[ \Delta \mathbf{v} = (-3 \hat{i} - 4 \hat{j}) - (3 \hat{i} + 4 \hat{j}) \]
\[ \Delta \mathbf{v} = -3 \hat{i} - 4 \hat{j} - 3 \hat{i} - 4 \hat{j} \]
\[ \Delta \mathbf{v} = -6 \hat{i} - 8 \hat{j} \]

### Step 4: Calculate the change in momentum
The change in momentum \( \Delta \mathbf{p} \) can be calculated using the formula:
\[ \Delta \mathbf{p} = m \cdot \Delta \mathbf{v} \]
Substituting the mass and the change in velocity:
\[ \Delta \mathbf{p} = 0.15 \text{ kg} \cdot (-6 \hat{i} - 8 \hat{j}) \]
\[ \Delta \mathbf{p} = 0.15 \cdot -6 \hat{i} + 0.15 \cdot -8 \hat{j} \]
\[ \Delta \mathbf{p} = -0.9 \hat{i} - 1.2 \hat{j} \]

### Step 5: Express the final answer
The change in momentum in vector form is:
\[ \Delta \mathbf{p} = -0.9 \hat{i} - 1.2 \hat{j} \text{ kg m/s} \]
This can also be expressed as:
\[ \Delta \mathbf{p} = 0.9 \hat{i} + 1.2 \hat{j} \text{ kg m/s} \quad \text{(taking the negative sign out)} \]

### Final Answer
The change in momentum is:
\[ \Delta \mathbf{p} = 0.9 \hat{i} + 1.2 \hat{j} \text{ kg m/s} \]