A hammer weighing 3 kg strikes the head of a nail with a speed of `2ms^(-1)` drives it by 1 cm into the wall. The impulse imparted to the wall is

Options

A. 6Ns
B. 3Ns
C. 2Ns
D. 12Ns

Explanation

To solve the problem, we need to calculate the impulse imparted to the wall by the hammer. We can do this by determining the change in momentum of the hammer as it strikes the nail.

### Step 1: Calculate the initial momentum of the hammer
The initial momentum (p_initial) of the hammer can be calculated using the formula:
\[ p_{\text{initial}} = m \cdot v \]
where:
- \( m = 3 \, \text{kg} \) (mass of the hammer)
- \( v = 2 \, \text{m/s} \) (initial speed of the hammer)

Substituting the values:
\[ p_{\text{initial}} = 3 \, \text{kg} \cdot 2 \, \text{m/s} = 6 \, \text{kg m/s} \]

### Step 2: Calculate the final momentum of the hammer
When the hammer strikes the nail and drives it into the wall, it comes to a stop. Therefore, the final momentum (p_final) of the hammer is:
\[ p_{\text{final}} = 0 \, \text{kg m/s} \]

### Step 3: Calculate the change in momentum of the hammer
The change in momentum (Δp) of the hammer is given by:
\[ \Delta p = p_{\text{final}} - p_{\text{initial}} \]
Substituting the values:
\[ \Delta p = 0 - 6 \, \text{kg m/s} = -6 \, \text{kg m/s} \]

### Step 4: Determine the impulse imparted to the wall
According to the impulse-momentum theorem, the impulse (J) imparted to the wall is equal to the change in momentum of the hammer:
\[ J = \Delta p \]
Thus:
\[ J = -6 \, \text{kg m/s} \]

Since impulse is a magnitude, we can express it as:
\[ J = 6 \, \text{kg m/s} \]

### Final Answer
The impulse imparted to the wall is **6 kg m/s**.

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