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.

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

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

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

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

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

Explanation / Answer

(a) Velocity of ball before hitting the table, u = -(2ghi)1/2 = -(2 * 9.8 * 1.5)1/2 = -5.42 m/s

So, momentum before collision, pi = mu = 0.05 * (-5.42) = -0.27 kg-m/s

(b) Velocity of ball after collision, v = (2ghf)1/2 = (2 * 9.8 * 1)1/2 = 4.43 m/s

(c) Impulse imparted, I = pf - pi = (0.05 * 4.43) - (-0.27) = 0.49 kg-m/s

(d) Time-averaged force, Favg = I/tc = 0.49/(15 * 10-3) = 32.67 N

(e) Change in KE = KEf - KEi = m(v2 - u2)/2 = 0.05 * [4.432 - (-5.42)2] / 2 = -0.244 J

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