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6 May, 16:11

A ball with mass of 0.050 kg is dropped from a height of h1 = 1.5 m. It collides with the floor, then bounces up to a height of h2 = 1.0 m. The Collison takes 0.015 s.

Use part e to Calculate impulse (J) ? Write the formula first.

Use part f to Calculate the average force acting (F) on the ball by the floor during the collision?

*Write the formula

Calculate the change in the kinetic energy (∆K) in the collision. Write formula first.

What happens to the lost kinetic energy?

What type of Collision Is this?

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Answers (1)
  1. 6 May, 16:20
    0
    Impulse of reaction force of floor = change in momentum

    Velocity of impact = √ 2gh₁

    = √ 2 x 9.8 x 1.5 = 5.4 m / s.

    velocity of rebound = √2gh₂

    = √ 2x 9.8 x 1

    = 4.427 m / s.

    Initial momentum =.050 x 5.4 =.27 kg m/s

    Final momentum =.05 x 4.427 =.22 kg. m/s

    change in momentum =.27 -.22 =.05 kg m/s

    Impulse =.05 kg m / s

    Impulse = force x time

    force = impulse / time

    .05 /.015 = 3.33 N.

    kinetic energy = 1/2 m v²

    Initial kinetic energy = 1/2 x. 05 x 5.4²

    = 0.729 J

    Final Kinetic Energy = 1/2 x. 05 x 4.427²

    = 0.489 J

    Change in Kinetic energy = 0.24 J

    Lost kinetic energy is due to conversion of energy into sound light etc.
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