A 8.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.320 m along the surface before stopping. What was the initial speed of the bullet?

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

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