10.-/1.5 pointsBUElemPhys1 7.P.102. A rubber ball, with a mass of 50.0 grams is

Question

10.-/1.5 pointsBUElemPhys1 7.P.102. A rubber ball, with a mass of 50.0 grams is dropped from rest from a height of 1.60 m above the floor. It hits the floor, and then reaches a maximum height of 80.0 cm when it comes back up again. In this problem, use g 10.0 m/s2. (a) The collision with the floor causes some mechanical energy to be lost (this energy generally ends up as thermal energy). How much mechanical energy is lost in this case? (b) What is the ball's speed, just as it leaves the floor on its way up? m/s (c) What is the magnitude of the impulse experienced by the ball for the entire time it is in contact with the floor? kg m/s

Explanation / Answer

(a)

Initial potential energy of rubber ball, Ui = mghi = (50*10-3)(10)(1.6) = 0.8 J

Initial kinetic energy of rubber ball, Ki = 0.5mvi2= (0.5)(50*10-3)(0)2 = 0 J

Initial total energy of rubber ball, Ei = 0.8 + 0 = 0.8 J

Final potential energy of rubber ball, Uf = mghf = (50*10-3)(10)(0.8) = 0.4 J

Final kinetic energy of rubber ball, Kf = 0.5mvf2= (0.5)(50*10-3)(0)2 = 0 J

Final total energy of rubber ball, Ei = 0.4 + 0 = 0.4 J

Hence mechanical energy lost during collision, = 0.8 – 0.4 = 0.4 J

(b)

Let at the beginning of upward motion, velocity is v. Then

Kinetic energy at the beginning of the upward motion = 0.5mv2

Potential energy at the beginning of upward motion = 0 J

Total energy at the beginning of upward motion = 0.5mv2

Total energy at the end of upward motion = 0.4 J

So as per energy conservation,

0.5mv2 = 0.4

v2 = 0.4/(0.5m)

v2 = 0.4/(0.5*(50*10-3))

v = 4 m/s

(c)

At the beginning of the downward motion, total energy, Ei = 0.8 J

Let at the end of downward motion, velocity is v. Then

Kinetic energy at the end of the downward motion = 0.5mv2

Potential energy at the end of downward motion = 0 J

Total energy at the end of downward motion = 0.5mv2

So as per energy conservation,

0.5mv2 = 0.8

v2 = 0.8/(0.5m)

v2 = 0.8/(0.5*(50*10-3))

v = 5.657 m/s        

As at the beginning of upward motion, v = 4 m/s.

So change in momentum and hence impulse = (50*10-3)(5.657 – 4) = 0.08285 kg m/s.

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