As shown in the figure , a superball with mass m equal to 50 grams is dropped fr

Question

As shown in the figure , a superball with mass m equal to 50 grams is dropped from a height of h = 1.5 m} . It collides with a table, then bounces up to a height of h_{ m f}= 1.0; { m m} . The duration of the collision (the time during which the superball is in contact with the table) is = 15 . In this problem, take the positive y direction to be upward, and use g = 9.8; { m m/s^2} for the magnitude of the acceleration due to gravity. Neglect air resistance.


Find the y component of the momentum, p_{{ m before}{,}y}, of the ball immediately before the collision.
Express your answer numerically, to two significant figures.
y} =



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.
p_{{ m after}{,}y} =



Part C
Find the y component of the time-averaged force F_{{ m avg}{,}y}, in newtons, that the table exerts on the ball.
Express your answer numerically, to two significant figures.
F_{{ m avg}_y} =



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


Part E
Find K_{ m after}-K_{ m before}, the change in the kinetic energy of the ball during the collision, in joules.
Express your answer numerically, to two significant figures.
K_{ m after}-K_{ m before} =

Explanation / Answer

A. Find the y component of the momentum, p_{{ m 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.Find the y component of the time-averaged force F_{{ m avg}{,}y}, in newtons, that the table exerts on the ball.= [0.22 -(- 0.27)]/0.015 = 32.7 N D. J_y, the y component of the impulse imparted to the ball during the collision=0.22+0.27=0.49 N E. Find K_{ m after}-K_{ m 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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