Class Management | Help HW 5 Begin Date: 10/1/2017 12:00:00 AM- Due Date: 10/10/
ID: 1777297 • Letter: C
Question
Class Management | Help HW 5 Begin Date: 10/1/2017 12:00:00 AM- Due Date: 10/10/2017 11:59:00 PM End Date: 12/16/2017 12:00:00 AM (696) Problem 14: A baseball of mass m,-0.4-kg is thrown at another ball hanging from the ceiling by a length of string L- 1.9m. The second ball m-0.9 kg is initially at rest while the baseball has an initial horizontal velocity of Vi-3.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity. Randomized Variables mt1 = 0.47k m2 = 0.91 kg L= 1.9 m 3.5 m/s 2Otheexpertta.com 50% Part (a) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d. Grade Summary Dechictions 0% Potential 100% (mi)2 2m29 2m2g 2g Submissions Atteropts remairing (33% per attempt) detailed view 3 --2g Submit Hint I give upt Hints: 1% deduction per hint Hints remaining2 Feedback: 1% deduction per feedback. 50% Part (b) What is the angle that the string makes with the vertical at the highest point of travel in degrees?Explanation / Answer
distance d from ceiling(d)
d = L – h
d = L – (0.5mv12/mg)
d = L – (v12/2g)
(b)
h = L – L * cos
h = 1.9 – 1.9*cos
cos = 1 – h/1.9
h = v12/2g
h = 3.52/19.6
h = 0.625 m
For the 1st ball, horizontal momentum = 0.47 * 3.5 = 1.645 m-s
final horizontal momentum = 0
For the 2nd ball, horizontal momentum = 0.91 * v2
0.91 * v2 = 1.645
v2 = 1.8 m/s
h = (1.82/19.6)
h = 0.166 meter
0.166 = 1.9 – 1.9*cos
= 24o