Law of conservation of energy:
From law of conservation of energy,
mghi + 0 = 0 + (1/2)mvi2
ghi = (1/2)vi2
vi = sqrt (2ghi) = sqrt[(2)(9.8)(1.5)] = 5.42 m/s
As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height of hi =1.5 m. It collides with a table, then bounces up to a height of hf = 1.0 m. 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 = 9.8 m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance.
a. Find the y component of the momentum, pbefore,y, of the ball immediately before the collision.
b. Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.
c. Find Jy, the y component of the impulse imparted to the ball during the collision.
d. Find the y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball.
e. Find Kafter - Kbefore, the change in the kinetic energy of the ball during the collision, in joules.