The head of a hammer (m = 1.5 kg) moving at 4.5 m/s strikes a nail and bounces back with the same speed after an elastic collision lasting 0.075 s. What is the magnitude of the average force the hammer exerts on the nail?

Options

A. 6.8 N
B. 90 N
C. 60 N
D. 180 N

Answer

To solve the problem of finding the magnitude of the average force the hammer exerts on the nail during an elastic collision, we can follow these steps:

### Step 1: Understand the problem
The hammer has a mass \( m = 1.5 \, \text{kg} \) and is moving at a speed of \( u = 4.5 \, \text{m/s} \) before striking the nail. After the collision, it bounces back with the same speed but in the opposite direction.

### Step 2: Determine the initial and final velocities
- Initial velocity of the hammer before the collision:
\[ u = -4.5 \, \text{m/s} \quad (\text{downward direction}) \]
- Final velocity of the hammer after the collision:
\[ v = 4.5 \, \text{m/s} \quad (\text{upward direction}) \]

### Step 3: Calculate the change in momentum
The change in momentum (\( \Delta p \)) of the hammer can be calculated using the formula:
\[ \Delta p = m(v - u) \]
Substituting the values:
\[ \Delta p = 1.5 \, \text{kg} \times (4.5 \, \text{m/s} - (-4.5 \, \text{m/s})) \]
\[ \Delta p = 1.5 \, \text{kg} \times (4.5 + 4.5) \, \text{m/s} \]
\[ \Delta p = 1.5 \, \text{kg} \times 9 \, \text{m/s} = 13.5 \, \text{kg m/s} \]

### Step 4: Calculate the average force
The average force (\( F_{\text{avg}} \)) exerted by the hammer on the nail can be calculated using the formula:
\[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \]
Where \( \Delta t = 0.075 \, \text{s} \). Substituting the values:
\[ F_{\text{avg}} = \frac{13.5 \, \text{kg m/s}}{0.075 \, \text{s}} \]
\[ F_{\text{avg}} = 180 \, \text{N} \]

### Final Answer
The magnitude of the average force the hammer exerts on the nail is \( 180 \, \text{N} \).

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