A batsman hits back a ball of mass .0 15kg straight in the direction of the bowler without changing its initial speed of `10 ms^(-1)`. If the ball moves linearly, then the impulse imparted on it (in Ns) is.

Options

A. 3
B. 2
C. 1.5
D. 1.9

Explanation

To find the impulse imparted on the ball by the batsman, we can follow these steps:

### Step 1: Understand the Concept of Impulse
Impulse is defined as the change in momentum of an object. Mathematically, it can be expressed as:
\[ \text{Impulse} (J) = \Delta p = p_f - p_i \]
where \( p_f \) is the final momentum and \( p_i \) is the initial momentum.

### Step 2: Calculate Initial Momentum
The initial momentum of the ball can be calculated using the formula:
\[ p_i = m \cdot v_i \]
where:
- \( m = 0.15 \, \text{kg} \) (mass of the ball)
- \( v_i = -10 \, \text{m/s} \) (initial velocity, negative because it is moving towards the batsman)

Thus, the initial momentum is:
\[ p_i = 0.15 \, \text{kg} \cdot (-10 \, \text{m/s}) = -1.5 \, \text{kg m/s} \]

### Step 3: Calculate Final Momentum
After the batsman hits the ball, it moves back towards the bowler with the same speed but in the opposite direction. Therefore, the final velocity \( v_f \) is:
\[ v_f = 10 \, \text{m/s} \]

Now, we can calculate the final momentum:
\[ p_f = m \cdot v_f = 0.15 \, \text{kg} \cdot 10 \, \text{m/s} = 1.5 \, \text{kg m/s} \]

### Step 4: Calculate the Change in Momentum
Now we can find the change in momentum:
\[ \Delta p = p_f - p_i = 1.5 \, \text{kg m/s} - (-1.5 \, \text{kg m/s}) \]
\[ \Delta p = 1.5 \, \text{kg m/s} + 1.5 \, \text{kg m/s} = 3.0 \, \text{kg m/s} \]

### Step 5: Conclusion
The impulse imparted on the ball is:
\[ J = 3.0 \, \text{Ns} \]

### Final Answer
The impulse imparted on the ball is **3 Ns**.

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