(3)1 (62m)d ma = mand Question 12 Before During After A ball with a mass of 0.52

Question

(3)1 (62m)d ma = mand Question 12 Before During After A ball with a mass of 0.52 kg is dropped on a table. The initial height of the ball is 0.71 meters above the table. The time that the ball is in contact with the table is 23 ms. During the collision with the table the ball loses 27% of its kinetic energy (a 10 pts) Calculate the change in the momentum of the ball during its collision with the table. (b 5 pts) Calculate the average force acting on the ball during its collision with the table. (C 10pts) Calculate the height to which the ball will rise after the collision. PE m 3 b483ke 2 m

Explanation / Answer

(a) Intial velocity of the ball before striking the table, v1 = sqrt{2*g*h] ] sqrt[2*9.81*0.71] = 3.73 m/s.

Initial kinetic energy of the ball before striking the table, KE1 = (1/2)*m*v1^2

= 0.5*0.52*3.73^2 = 3.62 J

Final kinetic energy after striking the table, KE2 = 0.73*3.62 = 2.64 J

so, final velocity of the ball v2 = sqrt[(2*2.64)/0.52] = 3.19 m/s

So, change in momentum = mv1 - mv2 = m(v1 - v2) = 0.52(3.73 + 3.19) = 13.31 kg-m/s (please note that direction of final velocity is opposite to that of the initial velocity that is why we have added the velocities above)

(b) Average force acting on the wall = Rate of change of momentum = 13.31 / (23x10^-3) = 578.6 N

(c) height raised by the ball after collision, h2 = v2^2 / 2g = 3.19^2 / (2*9.81) = 0.52 m

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