As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is

Question

As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height of hi=1.5m . It collides with a table, then bounces up to a height of hf=1.0m . The duration of the collision (the time during which the superball is in contact with the table) is tc=15ms . In this problem, take the positive y direction to be upward, and use g=9.8m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance.

Part A

Find the y component of the momentum, pbefore,y , of the ball immediately before the collision.

Part B

Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.

Part C

Find Jy , the y component of the impulse imparted to the ball during the collision.

Part D

Find the y component of the time-averaged force Favg,y , in newtons, that the table exerts on the ball

Part E

Find KafterKbefore , the change in the kinetic energy of the ball during the collision, in joules.

Explanation / Answer

use the kinematic equations:

v^2 = vo^2 + 2ay

1. v^2 = 0 + 2(9.8)(1.5)

    v = 5.42 m/s  

    mv = 0.27 kgm/s

2. 0 = vo^2 + 2(9.8)(-1.0)

    vo = 4.43 m/s

    mv = -0.22 kgm/s

3. impulse = Ft = mv

   F = (0.27 + 0.22...) / 0.015= 32.8N

4. Impulse = Ft = 32.8(0.015) = 0.492 Ns

5. (1/2)(0.05)(4.43)2 -(1/2)(0.05)(5.42)2 =- 0.245J

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