A ball of 0.1kg strikes a wall at right angle with a speed of 6 m/s and rebounds along its original path at 4 m/s. The change in momentum in Newton- sec is-

Options

A. `10^(3)`
B. `10^(2)`
C. 10
D. 1

Answer

To solve the problem of finding the change in momentum of the ball, we can follow these steps:

### Step 1: Understand the given data
- Mass of the ball (m) = 0.1 kg
- Initial velocity (u) = 6 m/s (towards the wall)
- Final velocity (v) = -4 m/s (rebound direction, opposite to the initial direction)

### Step 2: Calculate the initial momentum
The initial momentum (p_initial) can be calculated using the formula:
\[ p_{\text{initial}} = m \cdot u \]
Substituting the values:
\[ p_{\text{initial}} = 0.1 \, \text{kg} \cdot 6 \, \text{m/s} = 0.6 \, \text{kg m/s} \]

### Step 3: Calculate the final momentum
The final momentum (p_final) can be calculated using the formula:
\[ p_{\text{final}} = m \cdot v \]
Substituting the values:
\[ p_{\text{final}} = 0.1 \, \text{kg} \cdot (-4 \, \text{m/s}) = -0.4 \, \text{kg m/s} \]

### Step 4: Calculate the change in momentum
The change in momentum (Δp) is given by:
\[ \Delta p = p_{\text{final}} - p_{\text{initial}} \]
Substituting the values:
\[ \Delta p = -0.4 \, \text{kg m/s} - 0.6 \, \text{kg m/s} \]
\[ \Delta p = -1.0 \, \text{kg m/s} \]

### Step 5: Express the change in momentum in Newton-seconds
Since 1 kg m/s is equivalent to 1 Newton-second, we can express the change in momentum:
\[ \Delta p = -1.0 \, \text{N s} \]

### Final Answer
The change in momentum is **1 N s** (considering the magnitude).

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