A 0.145 kg ball is dropped from a height of 8.0 m, and it bounces back to a heig

Question

A 0.145 kg ball is dropped from a height of 8.0 m, and it bounces back to a height of 7.0 m. Assuming there is no air friction in the path of the ball, use conservation of energy method to find the followings [(a), (b), and (c)]. PLEASE SHOW YOUR WORK.

(a) The speed of the ball in its decent at 4.0 m above the ground level.

Using conversation of energy (KE1 + PE1 = KE2 + PE2), I know this answer is 8.85 m/s. That is correct, as it was graded correct.

(b) The speed of the ball just before hitting the ground.

(c) The speed of the ball just after hitting the ground.

(d) What is the impulse on the ball from the ground?

Explanation / Answer

here,

mass of the ball , m = 0.145 kg

initial height , hi = 8 m

final height , hf = 7 m

(b)

let The speed of the ball just before hitting the ground be u

u = sqrt(2*g*hi)

u = 12.52 m/s

The speed of the ball just before hitting the ground is 12.52 m/s

(c)

The speed of the ball just after hitting the ground be v

v = sqrt(2*g*hf)

v = 11.71 m/s

The speed of the ball just after hitting the ground is 11.71 m/s

(d)

the impulse on the ball from the ground , I = m * ( v - u)

I = 0.145 * ( 11.71 - ( - 12.52))

I = 3.51 kg.m/s

the impulse on the ball from the ground is 3.51 kg.m/s

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