Problem 4: A golf ball of mass m-0.22 kg is dropped from a height h. It interact

Question

Problem 4: A golf ball of mass m-0.22 kg is dropped from a height h. It interacts with the floor for t-0.155 s, and applies a force ofF= 17 N to the floor when it elastically collides with it. Randomized Variables m = 0.22 kg t= 0.155 s F= 17N Part (a) Write an expression for the ball's velocity, v, just after it rebounds from the floor. (Hint: The fact that the collision is elastic is important when solving this problem.) Expression: 2 Select from the variables below to write your expression. Note that all variables may not be required. , , 0, d, F, g, h, i, j, k, m, P, r, S, t Part (b) What is the ball's velocity v, in meters per second right after it rebounds? Numeric :A numeric value is expected and not an expression. N . 155s 08-i Part (c) How high h, in meters, will the ball travel on the rebound? Numeric : A numeric value is expected and not an expression.

Explanation / Answer

part a ) impulse = change in momentum

F * t = mv - (-mv)

F* t = 2mv

v = Ft / 2m

b) v = (17*0.155) / (2*0.22) = 5.98 m/s

c) from conservation of energy

mgh = (1/2) mv2

h = v2 / 2g = 5.982 / (2*9.8) = 1.829 m

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