Answer:
The initial speed of the bullet is 169.12 m/s
Explanation:
Given;
mass of bullet, m₁ = 8 g = 0.008 kg
mass of wooden block, m₂ = 1.20-kg
coefficient of kinetic friction, μ = 0.20
distance moved by the block, d = 0.320 m
Step 1:
frictional force on the wooden block after the collision
Fk = μm₂g
Neglect mass of the bullet since it is small compared to mass of the block
Fk = 0.2 x 1.2 x 9.8
Fk = 2.352 N
Step 2:
acceleration of the block after the impact
From Newton's second law of motion;
F = ma
Fk = m₂a
a = Fk / m₂
a = 2.352 / 1.2
a = 1.96 m/s²
Step 3:
velocity of the block after the impact
v² = u² + 2as
where;
s is the distance moved by the block = d
v² = 0² + 2(1.96 x 0.32)
v² = 1.2544
v = √1.2544
v = 1.12 m/s
Step 4:
velocity of the bullet before the collision
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₂ is speed of the wooden block before collision = 0, since it was resting ....
0.008u₁ + 1.2 x 0 = 1.12(0.008 + 1.2)
0.008u₁ = 1.35296
u₁ = 1.35296 / 0.008
u₁ = 169.12 m/s
Therefore, the initial speed of the bullet is 169.12 m/s