The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates int

Question

The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates into the wood for a distance of 12 cm is:

Accepted Answer

1. **Identify Given Values:**
   - Mass ($m$) = $20 	ext{ g} = 0.02 	ext{ kg}$
   - Initial velocity ($u$) = $250 	ext{ m/s}$
   - Final velocity ($v$) = $0 	ext{ m/s}$ (stops)
   - Distance ($s$) = $12 	ext{ cm} = 0.12 	ext{ m}$

2. **Method 1: Work-Energy Theorem**
   The work done by the resistive force ($F$) equals the change in kinetic energy [Source 93, 95].
   $$ W = \Delta K = K_f - K_i $$
   $$ -F \cdot s = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 $$
   $$ F \cdot s = \frac{1}{2}mu^2 $$
   $$ F = \frac{mu^2}{2s} $$

3. **Calculation:**
   $$ F = \frac{0.02 	imes (250)^2}{2 	imes 0.12} $$
   $$ F = \frac{0.02 	imes 62500}{0.24} $$
   $$ F = \frac{1250}{0.24} \approx 5208.33 	ext{ N} $$

4. **Scientific Notation:**
   $$ F \approx 5.2 	imes 10^3 	ext{ N} $$
   (This approach aligns with Example 4.2 in Source 79, which uses kinematic equations and Newton's second law to solve a similar bullet-block problem).

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