A superball with mass m equal to 50 grams is dropped from a height of hi=1.5m .

Question

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

Explanation / Answer

speed of the ball after collision is vf = sqrt(2*g*hf) = sqrt(2*9.8*1) = 4.43 m/sec

speed of the ball before collision is vi = sqrt(2*g*hi) =sqrt(2*9.8*1.5) = 5.42 m/sec

kafter = 0.5*m*vf^2 = 0.5*50*10^-3*4.43^2 = 0.49 J

kbefore = 0.5*m*vi^2 = 0.5*50*10^-3*5.42^2 = 0.734 J

kafter - Kbefore = 0.49-0.734 = -0.244 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.