(11%) Problem 6: A golf ball is dropped from a height of hi-3.95 m above the gro
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(11%) Problem 6: A golf ball is dropped from a height of hi-3.95 m above the ground. After it bounces, it only reaches a height h 2.24 m above the ground. The golf ball has mass ml = 0.1985 kg Randomized Variables h1 = 3.95 m h2= 2.24 m m-0.1985 kg 33% Part (a) What is the magnitude of the impulse 1 in kilogram meters per second, the golf ball experienced during the bounce? Grade Summary Deductions Potential I0.432 0% 100% Submissions Attempts remaining: 4 0% per attempt) 789HOME tan acosO sinh cosO Sin cotanO asin()a atan)acotan() coshO tanh)cotanh(O 45 6- detailed view 0% 0% 0% END o Degrees Radians BACKSPACE ?DEL? CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback Submission History Hints Feedback Note: Feedback not accessed Answer 1 = 0.43 I0.43 I--0.43 I-04322 I-0.4322 Totals 0% 0% 0% 0% 0% 0% 0% 0% 0% 0% 0% 0% Totals 0% 0% 0% 0% 33% Part (b) If the golf ball was in contact with the ground for t 0.131 s, what was the magnitude of the constant force F acting on it, in newtons? 33% Part (c) How much energy, in Joules, did the golf ball transfer to the environment during the bounce?Explanation / Answer
a) vf = sqrt [2 g h] = sqrt [2 * 9.8 * 2.24] = 6.63 m/s
vi = sqrt [2 g h] = sqrt [2 * -9.8 * 3.95] = -8.79 m/s
delta p = m (vf - vi) = 0.1985 (6.63 + 8.79)
magnitude of impulse = 3.06 kg.m/s
b) impulse = F t
F = 3.06 / 0.131
magnitude of constant force = 23.36 N
c) delta E = 1/2 * m * (vf2 - vi2)
= 1/2 * 0.1985 * (6.632 - (-8.792)) = -3.30 J
amount of energy = 3.30 J
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