Bullet Retardation Simulation
Example 5.2: Calculating average resistive force
Example
Question:
A bullet of mass \(0.04\,\mathrm{kg}\) moving with a speed of \(90\,\mathrm{m\,s^{-1}}\) enters a heavy wooden block and is stopped after a distance of \(60\,\mathrm{cm}\). What is the average resistive force exerted by the block on the bullet?
Solution:
The retardation \(a\) of the bullet (assumed constant) is given by:
\[
a = \frac{-u^2}{2s} = \frac{-90^2}{2 \times 0.6}\,\mathrm{m\,s^{-2}} = -6750\,\mathrm{m\,s^{-2}}
\]
The retarding force, by the second law of motion, is:
\[
F = m a = 0.04\,\mathrm{kg} \times 6750\,\mathrm{m\,s^{-2}} = 270\,\mathrm{N}
\]
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.
Results:
Initial velocity:
90 m/s
Retardation:
-6750 m/s²
Stopping distance:
60 cm
Average resistive force:
270 N
Physics Explanation:
This simulation demonstrates how to calculate the average resistive force acting on a bullet as it penetrates a wooden block.
Given Data:
- Bullet mass (m) = 0.04 kg
- Initial velocity (u) = 90 m/s
- Final velocity (v) = 0 m/s
- Stopping distance (s) = 60 cm = 0.6 m
Calculations:
1. Calculate retardation using kinematic equation:
\[ v^2 = u^2 + 2as \]
\[ 0 = 90^2 + 2a(0.6) \]
\[ a = -\frac{90^2}{2 \times 0.6} = -6750 \, \text{m/s}^2 \]
2. Calculate average resistive force using Newton's Second Law:
\[ F = ma = 0.04 \, \text{kg} \times 6750 \, \text{m/s}^2 = 270 \, \text{N} \]
Note:
The actual resistive force may vary during penetration. This calculation gives the average resistive force over the stopping distance.



