A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebound

Question

A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor? Kg middot m/s A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 52.0 m/s and returns the shot with the ball traveling horizontally at 38.0 m/s in the opposite direction. (Assume the initial direction of the ball is in the -x direction.) (A) What is the impulse delivered to the ball by the racket? N middot s Direction = (b) What work does the racket do on the ball? J

Explanation / Answer

4)

consider the motion of the ball while it is falling down

hi = height from which ball was dropped = 1.25 m

Vi = velocity of ball just before hitting the surface

using conservation of energy

kinetic energy at bottom = initial potential energy at top

(0.5) m Vi2 = mg hi

Vi = sqrt(2g hi) = sqrt(2 x 9.8 x 1.25) = 4.95 m/s

consider the motion of the ball while it is rising up after hitting the surface

hf = height to which ball rises = 0.6 m

Vf = velocity of ball just after hitting the surface

using conservation of energy

kinetic energy at bottom = potential energy at top

(0.5) m Vf2 = mg hf

Vf = sqrt(2g hf) = sqrt(2 x 9.8 x 0.6) = 3.43 m/s

consider up direction as positive and down as negative direction

hence Vi = - 4.95 m/s

Vf = 3.43 m/s

m = mass = 0.120 kg

Impulse is given as

I = change in momentum = m (Vf - Vi) = (0.120) (3.43 - (-4.95)) = 1.01 kgm/s

5)

Vi = initial velocity of ball = - 52 m/s

Vf = final velocity of ball = 38 m/s

m = mass of ball = 0.06 kg

a)

impulse is given as

I = m(Vf - Vi) = (0.06) (38 - (-52)) = 5.4 kgm/s

b)

work done = change in KE = (0.5) m (Vf2 - Vi2) = (0.5) (0.06) ((38)2 - (52)2) = - 37.8 J

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