In the previous problem `3` the magnitude of the momentum transferred during the hit is .

Options

A. zero
B. `0.75 Kg-ms^(-1)`
C. `1.5 kg-ms^(-1)`
D. `14 kg ms^(-1)`

Explanation

To solve the problem of finding the magnitude of the momentum transferred during the hit of a cricket ball, we can follow these steps:

### Step 1: Identify the given values
- Mass of the cricket ball, \( m = 150 \) grams = \( 0.15 \) kg (since 1 kg = 1000 grams)
- Initial velocity, \( \vec{V_i} = 3 \hat{i} + 4 \hat{j} \) m/s
- Final velocity, \( \vec{V_f} = -3 \hat{i} + 4 \hat{j} \) m/s

### Step 2: Calculate the change in velocity
The change in velocity \( \Delta \vec{V} \) can be calculated as:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} \]
Substituting the values:
\[ \Delta \vec{V} = (-3 \hat{i} + 4 \hat{j}) - (3 \hat{i} + 4 \hat{j}) = -3 \hat{i} - 3 \hat{j} \]

### Step 3: Calculate the change in momentum
Using the formula for momentum \( \vec{P} = m \vec{V} \), the change in momentum \( \Delta \vec{P} \) is given by:
\[ \Delta \vec{P} = m \Delta \vec{V} \]
Substituting the mass and change in velocity:
\[ \Delta \vec{P} = 0.15 \text{ kg} \cdot (-3 \hat{i} - 3 \hat{j}) = -0.45 \hat{i} - 0.45 \hat{j} \text{ kg m/s} \]

### Step 4: Calculate the magnitude of the change in momentum
The magnitude of the momentum \( |\Delta \vec{P}| \) can be calculated using the formula:
\[ |\Delta \vec{P}| = \sqrt{(-0.45)^2 + (-0.45)^2} \]
Calculating this gives:
\[ |\Delta \vec{P}| = \sqrt{0.2025 + 0.2025} = \sqrt{0.405} \approx 0.636 \text{ kg m/s} \]

### Step 5: Final answer
The magnitude of the momentum transferred during the hit is approximately \( 0.636 \) kg m/s.

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