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

Question

As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height ofhi=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 positivey 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.

Explanation / Answer

mgh = .5mv^2

v is proportional to sqrt(h)

v1 = sqrt(2gh) = 5.42 downwards

momentum just before collision= - mv1 = -0.271 kg m/s

v2 = sqrt(2*9.8*1) = 4.4272 m/s upwards

momentum just after collision = mv2 = 0.22 kg m/s

impulse = change in momentum / time = .22+.271 / .015 = 32.824 kg m/s^2

Favg,y = 32.824 N

Kafter = .5mv2^2 = 0.49

Kbefore = .5mv1^2 = .73441

change in kE = -0.2444

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