A 1.5 kg hammer moving with a velocity of 10 m/s strikes a nail for 0.005 s. The average force exerted on the nail is

Options

A. 5000 N
B. 3000 N
C. 1500 N
D. 750 N

Answer

To find the average force exerted on the nail by the hammer, we can use the relationship between momentum, force, and time. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the hammer (m) = 1.5 kg
- Initial velocity of the hammer (v1) = 10 m/s
- Final velocity of the hammer (v2) = 0 m/s (since it comes to rest)
- Time duration (t) = 0.005 s

### Step 2: Calculate the change in momentum
The change in momentum (Δp) can be calculated using the formula:
\[ \Delta p = m(v2 - v1) \]
Substituting the values:
\[ \Delta p = 1.5 \, \text{kg} \times (0 \, \text{m/s} - 10 \, \text{m/s}) = 1.5 \, \text{kg} \times (-10 \, \text{m/s}) = -15 \, \text{kg m/s} \]

### Step 3: Relate change in momentum to average force
According to the impulse-momentum theorem:
\[ \Delta p = F_{\text{avg}} \times t \]
Where \(F_{\text{avg}}\) is the average force. Rearranging the formula gives:
\[ F_{\text{avg}} = \frac{\Delta p}{t} \]

### Step 4: Substitute the values to find average force
Substituting the change in momentum and time into the equation:
\[ F_{\text{avg}} = \frac{-15 \, \text{kg m/s}}{0.005 \, \text{s}} = -3000 \, \text{N} \]
The negative sign indicates that the force is in the opposite direction of the hammer's initial motion, but we are interested in the magnitude of the force:
\[ F_{\text{avg}} = 3000 \, \text{N} \]

### Conclusion
The average force exerted on the nail is **3000 N**.

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