- Mass of the bullet (m1) = 8g = 0.008 Kg
- Velocity of the bullet (v1) = 420 m/s
- We know, kinetic energy = [tex] \frac{1}{2}m {v}^{2} [/tex]
- Therefore, the kinetic energy of the bullet
[tex] = \frac{1}{2} \times 0.008 \times 420J \\ = 1.68J[/tex]
- So, the kinetic energy of the bullet is 1.68 J. It is said that the man's kinetic energy should be same as that of the bullet.
- Mass of the man (m2) = 75 Kg
- Let the velocity of the man be v2.
- Therefore,
[tex]1.68J = \frac{1}{2} \times 75 \: kg \times v2 \\ = > v2 = \frac{1.68 \times 2}{75} m {s}^{ - 1} \\ = 0.0448 \: m \: {s}^{ - 1} [/tex]
Answer:
0.0448 m/s
Hope you could get an idea from here.
Doubt clarification - use comment section.
