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

Options

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

Explanation

To solve the problem, we will calculate the change in momentum of the cricket ball step by step.

### Step 1: Convert mass from grams to kilograms
The mass of the cricket ball is given as 150 grams. To convert this to kilograms, we use the conversion factor \(1 \text{ kg} = 1000 \text{ g}\).

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

**Hint:** Always convert mass to kilograms when using SI units.

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

**Hint:** Pay attention to the signs when dealing with vector components.

### Step 3: Calculate initial momentum
The initial momentum \(\bar{P_i}\) is given by the formula:
\[ \bar{P_i} = m \cdot \bar{u} \]
Substituting the values:
\[ \bar{P_i} = 0.15 \text{ kg} \cdot (3 \hat{i} - 4 \hat{j}) = (0.15 \cdot 3) \hat{i} + (0.15 \cdot -4) \hat{j} = 0.45 \hat{i} - 0.6 \hat{j} \text{ kg m/s} \]

**Hint:** Momentum is a vector quantity; calculate each component separately.

### Step 4: Calculate final momentum
The final momentum \(\bar{P_f}\) is calculated similarly:
\[ \bar{P_f} = m \cdot \bar{v} \]
Substituting the values:
\[ \bar{P_f} = 0.15 \text{ kg} \cdot (-3 \hat{i} + 4 \hat{j}) = (0.15 \cdot -3) \hat{i} + (0.15 \cdot 4) \hat{j} = -0.45 \hat{i} + 0.6 \hat{j} \text{ kg m/s} \]

**Hint:** Ensure to apply the correct signs for the final velocity components.

### Step 5: Calculate change in momentum
The change in momentum \(\Delta \bar{P}\) is given by:
\[ \Delta \bar{P} = \bar{P_f} - \bar{P_i} \]
Substituting the values:
\[ \Delta \bar{P} = (-0.45 \hat{i} + 0.6 \hat{j}) - (0.45 \hat{i} - 0.6 \hat{j}) \]
\[ \Delta \bar{P} = (-0.45 - 0.45) \hat{i} + (0.6 + 0.6) \hat{j} = -0.9 \hat{i} + 1.2 \hat{j} \text{ kg m/s} \]

**Hint:** When subtracting vectors, subtract the corresponding components.

### Final Answer
The change in momentum of the cricket ball is:
\[ \Delta \bar{P} = -0.9 \hat{i} + 1.2 \hat{j} \text{ kg m/s} \]