Ask Question
30 November, 14:46

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?

+1
Answers (1)
  1. 30 November, 16:34
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers