A bullet of mass 25gm strikes a target with a velocity of 400m/s . Calculate the K.E when it strikes the target,work done by the bullet,opposing force of the target if the bullet penetrates to a depth of 12.5 cm.
Answers
Mass of the bullet, m = 25 g = 0.025 kg.
Initial speed of the bullet, u = 400 m/s.
Kinetic energy of the bullet when it strikes the target,
→ K_i = mu²/2
→ K_i = 0.025 × 400² / 2
→ K_i = 2000 J
→ K_i = 2 kJ
The bullet comes to rest after penetrating through the target. SO the final speed, v = 0 m/s.
So is the kinetic energy.
→ K_f = 0 J
Hence work done by the bullet is, by work - energy theorem,
→ W = K_f - K_i
→ W = -2 kJ
The bullet comes to rest after travelling a distance, s = 12.5 cm = 0.125 m.
The retardation of the bullet is, using third equation of motion,
→ a = (v² - u²) / 2s
→ a = (0² - 400²) / (2 × 0.125)
→ a = - 640000 m/s²
→ a = - 640 km/s²
Hence the opposing force of the target is, using Third Law,
→ F = - ma
→ F = - 0.025 × - 640
→ F = 16 kN
Answer:
16N is the answer based on the above numerical