A `5.0g` bullet moving at `100m//s` strikes a log. Assume that the bullet undergoes uniform deceleration and stops in `6.0cm`. Find (a) the time taken for the bullet to stop, (b) the impulse on the log and (c) the average force experienced by the log.

Explanation

Let's solve the problem step by step.

### Given Data:
- Mass of the bullet, \( m = 5.0 \, \text{g} = 5.0 \times 10^{-3} \, \text{kg} \)
- Initial velocity of the bullet, \( u = 100 \, \text{m/s} \)
- Final velocity of the bullet, \( v = 0 \, \text{m/s} \) (since it stops)
- Displacement, \( s = 6.0 \, \text{cm} = 6.0 \times 10^{-2} \, \text{m} \)

### (a) Time taken for the bullet to stop

We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Since the bullet comes to a stop, \( v = 0 \):
\[ 0 = (100)^2 + 2a(6.0 \times 10^{-2}) \]
Rearranging gives:
\[ 2a(6.0 \times 10^{-2}) = -10000 \]
\[ a = \frac{-10000}{2 \times 6.0 \times 10^{-2}} = \frac{-10000}{0.12} = -83333.33 \, \text{m/s}^2 \]

Now, we can calculate the time taken to stop using:
\[ v = u + at \]
Substituting the known values:
\[ 0 = 100 + (-83333.33)t \]
\[ t = \frac{100}{83333.33} \approx 0.0012 \, \text{s} = 1.2 \times 10^{-3} \, \text{s} \]

### (b) Impulse on the log

Impulse is defined as the change in momentum:
\[ \text{Impulse} = \Delta p = m(v - u) \]
Substituting the values:
\[ \Delta p = 5.0 \times 10^{-3} \, \text{kg} \times (0 - 100) = -5.0 \times 10^{-3} \times 100 = -0.5 \, \text{kg m/s} \]
The magnitude of the impulse is:
\[ \text{Impulse} = 0.5 \, \text{Ns} \]

### (c) Average force experienced by the log

The average force can be calculated using the impulse and the time taken:
\[ F_{\text{avg}} = \frac{\text{Impulse}}{t} \]
Substituting the values:
\[ F_{\text{avg}} = \frac{0.5}{1.2 \times 10^{-3}} \approx 416.67 \, \text{N} \]

### Final Answers:
- (a) Time taken for the bullet to stop: \( t \approx 1.2 \times 10^{-3} \, \text{s} \)
- (b) Impulse on the log: \( \text{Impulse} = 0.5 \, \text{Ns} \)
- (c) Average force experienced by the log: \( F_{\text{avg}} \approx 416.67 \, \text{N} \)