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A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered straight back with a speed of 15.0 m/s. Draw a diagram of this situation. Represent every number appropriately. Your diagram should have two parts, before the collision and after the collision. We will estimate that the momentum of the system before the collision is approximately equal to the collision after the collision. Use this estimate to approximate the velocity of the bowling ball after the collision.

Answer :

Final answer:

The approximate velocity of the bowling ball after the collision is 11.32 m/s.

Explanation:

Collision between a Bowling Ball and a Bowling Pin

When a 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered straight back with a speed of 15.0 m/s, we can use the principle of conservation of momentum to approximate the velocity of the bowling ball after the collision.

Before the collision, the momentum of the system is given by the sum of the momenta of the bowling ball and the bowling pin:

Momentum before collision = (mass of bowling ball * velocity of bowling ball) + (mass of bowling pin * velocity of bowling pin)

Substituting the given values:

  1. Mass of bowling ball = 5.50 kg
  2. Velocity of bowling ball = 9.00 m/s
  3. Mass of bowling pin = 0.850 kg
  4. Velocity of bowling pin = 15.0 m/s

Using the principle of conservation of momentum, we can assume that the momentum before the collision is approximately equal to the momentum after the collision:

Momentum before collision = Momentum after collision

Since the bowling pin is scattered straight back, its velocity after the collision is in the opposite direction with the same magnitude:

Velocity of bowling pin after collision = -15.0 m/s

Let's denote the velocity of the bowling ball after the collision as v. Using the principle of conservation of momentum, we can write:

(mass of bowling ball * velocity of bowling ball) + (mass of bowling pin * velocity of bowling pin) = (mass of bowling ball * velocity of bowling ball after collision) + (mass of bowling pin * velocity of bowling pin after collision)

Substituting the given values:

  1. Mass of bowling ball = 5.50 kg
  2. Velocity of bowling ball = 9.00 m/s
  3. Mass of bowling pin = 0.850 kg
  4. Velocity of bowling pin = 15.0 m/s
  5. Velocity of bowling pin after collision = -15.0 m/s

Simplifying the equation:

(5.50 kg * 9.00 m/s) + (0.850 kg * 15.0 m/s) = (5.50 kg * v) + (0.850 kg * -15.0 m/s)

Solving for v:

49.5 kg·m/s + 12.75 kg·m/s = 5.50 kg·v - 12.75 kg·m/s

62.25 kg·m/s = 5.50 kg·v

v = 62.25 kg·m/s / 5.50 kg

v ≈ 11.32 m/s

Answer:

The approximate velocity of the bowling ball after the collision is 11.32 m/s.

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