± A Superball Collides Inelastically with a Table As shown in the figure (Figure

Question

± A Superball Collides Inelastically with a Table

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. Express your answer numerically, to two significant figures.

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. Express your answer numerically, to two significant figures.

Part C Find Jy, the y component of the impulse imparted to the ball during the collision. Express your answer numerically, to two significant figures.

Part D Find the y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball. Express your answer numerically, to two significant figures.

Part E Find KafterKbefore, the change in the kinetic energy of the ball during the collision, in joules. Express your answer numerically, to two significant figures.

Please show your work in detail, I'm posting because I need to know how to get to the answer, I'm studying for an exam.

Explanation / Answer

A. Find the y component of the momentum, p before y of the ball immediately before the collision.
= 0.05*sq rt(2*9.8*1.5) = 0.27 kg m s^-1

B.Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table. = 0.05*sq rt(2*9.8*1.0) = 0.22 kg m s^-1

C)

J_y, the y component of the impulse imparted to the ball during the collision=0.22+0.27=0.49 N

D)

Find the y component of the time-averaged force Favg,y in newtons, that the table exerts on the ball.= [0.22 -(- 0.27)]/0.015 = 32.7 N

E. Find K after - K before, the change in the kinetic energy of the ball during the collision, in joules. (0.27^2 - 0.22^2)/(2*0.05) = 0.25 J

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