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2) the figure below shows a bullet of mass 200 g traveling horizontally towards the east with speed 400 m/s, which strikes a block of mass 1.5 kg that is initially at rest on a frictionless table. after striking the block, the bullet is embedded in the block and the block and the bullet move together as one unit. (a) what is the magnitude and direction of the velocity of the block/bullet combination immediately after the impact? (b) what is the magnitude and direction of the impulse by the block on the bullet? (c) what is the magnitude and direction of the impulse from the bullet on the block? (d) if it took 3.0 ms for the bullet to change the speed from 400 m/s to the final speed after impact, what is the average force on the block and the bullet during this time?

User Tomas M
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1 Answer

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Final answer:

The post-impact velocity of the block/bullet system is approximately 47.06 m/s east. The impulse on the bullet by the block is about -70.59 Ns west, and the impulse on the block by the bullet is 70.59 Ns east. The average force experienced during impact is around 23530 N east.

Step-by-step explanation:

Answer to Bullet and Block Collision Problem

To solve this physics problem focused on projectile motion and impulse, we use the conservation of momentum and concepts from Newton's laws of motion. Here's the detailed solution:

  • a. Velocity of the block/bullet combination: We use conservation of momentum since no external forces are acting horizontally. Before collision, the total momentum is the momentum of the bullet alone (since the block is at rest).
  • m_bullet * v_bullet_initial = (m_bullet + m_block) * v_final
  • (0.2 kg * 400 m/s) = (0.2 kg + 1.5 kg) * v_final
  • v_final = 80 m/s / 1.7 kg ≈ 47.06 m/s
  • The final velocity is approximately 47.06 m/s towards the east.
  • b. Impulse on the bullet by the block: Impulse is a change in momentum, which is the same as the magnitude of force multiplied by the time over which it acts.
  • Impulse = m_bullet * (v_final - v_bullet_initial)
  • Impulse = 0.2 kg * (47.06 m/s - 400 m/s)
  • Impulse ≈ -70.59 Ns (west)
  • c. Impulse from the bullet on the block: According to Newton's third law, the impulse exerted by the bullet on the block will be equal in magnitude and opposite in direction to the impulse on the bullet by the block.
  • Impulse ≈ 70.59 Ns (east)
  • d. Average force during the impact: The average force is found by dividing the impulse by the time during which the force acts.
  • Average force = Impulse / Time
  • Average force = 70.59 Ns / 0.003 s
  • Average force ≈ 23530 N directed towards the east.

User DavidJBerman
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