Class Management Help h 6 test Begin Date: 10/27/2017 12:05:00 AM- -Due Date: 10
ID: 1790313 • Letter: C
Question
Class Management Help h 6 test Begin Date: 10/27/2017 12:05:00 AM- -Due Date: 10/28/2017 11:59:00 PM End Date: 10/29/2017 1:00:00 ANM (8%) Problem 9: A stone is dropped from a height of hi-3.65 m, above the ground. After it bounces, it only makes it to a height h2 = 2.3 m above the ground. The stone has mass m = 0.1685 kg. Randomized Variables h1= 3.65 m h2 = 2.3 m m 0.1685 kg 33% Part (a) What is the magnitude of the impulse 1 in kilogram meters per second, the stone experienced during the bounce? Grade Summary Deductions Potential 100% 096 tan() | ( cosO cotan acos0 Submissions Attempts remaining: 3 60s per attempt) sin0 1 23 atan0 acotanO sinh0 cosh(). tanho detailed view cotan Degrees Radians Submit Hist I give up Hints:deduction per hint Hints remaining:- Feedback: deduction per feedback 33% Part (b) If the stone was 1n contact with the ground for 0.18 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 stone transfer to the environment during the bounce?Explanation / Answer
(A) Applying energy conservation to find speed just before hitting floor,
m g h1 = m v1^2 / 2
v1 = sqrt(2 x 9.81 x 3.65) = 8.46 m/s downward
after the collision,
v2 = sqrt(2 x 9.81 x 2.3) = 6.72 m/s upward
v1 = - 8.46 m/s and v2 = 6.72 m/s
(A) Impulse = change in momentum = 0.1685(6.72 + 8.46)
= 2.56 kg m/s
(B) impulse = F t
2.56 = F (0.187)
F =13.8 N
(C) change in energy = 0.1685 x 9.81 (3.65 - 2.3)
= 2.23 J