A 3g bullet travelling at 300m/s hits and embeds itself in a 1-kg wooden block resting on a frictionless surface.
a) What is the final speed of the block and bullet ?
b) If the block after being hit slides on a surface that exerts 3N friction force on the block, how far will it slide ?
Answers
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 = 3 N
To Find:
(a) Final speed of the block and bullet.
(b) Distance to which the box slides.
Calculation:
(a) Using law of conservation of momentum, we have:
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
v² = u² + 2as
⇒ 0 = (0.897)² + 2as
⇒ a = -0.805/2s
⇒ a = -0.805/2s
Negative sign shows deceleration
As F = ma
⇒ 3 = 1.003 × 0.805/2s
⇒ s = 1.003 × 0.805/6
⇒ s = 0.135 m
- So the final velocity of block and bullet is 0.897 m/s and the distance travelled will be 0.135 m
Final velocity of block and bullet is 0.897 m/s and the distance travelled by the block is 0.135 m
Explanation:
We are given that:
- 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:
(a) 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
The final velocity of block and bullet is 0.897 m/s and the distance travelled by the block is 0.135 m