A 3g bullet travelling at 300m/s hits and embeds itself in a 1-kg wooden block resting on a frictionless surface.
How far will it slide after being hit by a bullet that exerts 3N friction force
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Final velocity of block and bullet is 0.897 m/s and the distance traveled by the block is 0.135 m
Explanation:
Data given:
Mass of bullet "m1" = 3 g = 0.003 kg
Speed of bullet "v1" = 300 m/s
Mass of wooden block "m2" = 1 kg
Speed of block "v2" = 0 m/s (at rest)
Frictional force "f" = 3 N
Solution:
According to law of conservation of momentum:
m1 × v1 + m2 × v2 = (m1 + m2) × v
⇒ 0.003 × 300 + 1 × 0 = (0.003 + 1) × v
⇒ 0.9 = 1.003 × V
⇒ v = 0.9 / 1.003
⇒ v = 0.897 m/s
(b) Final velocity of block = 0
Initial velocity of block = 0.897 m/s
Using 3rd equation of motion.
v² = u² + 2as
0 = (0.897)^2 + 2as
- 2as = (0.897)^2
a = -0.805/2s
a = -0.805/2 s ( -ve sign indicates deceleration)
According to Newton's second law:
F = ma
3 = 1.003 × 0.805/ 2 s
S = 1.003 × 0.805/6
S = 0.135 m
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