Class Management 1I Help hapter 7-9 HW Begin Date: 11/21/2017 10:00:00 AM -Due D
ID: 1795281 • Letter: C
Question
Class Management 1I Help hapter 7-9 HW Begin Date: 11/21/2017 10:00:00 AM -Due Date: 11/30/2017 11:59:00 PM End Date: 12/15/2017 12:00:00 AM (13%) Problem 8: A baseball of mass m 0.26 kg is thrown at another ball hanging from the ceiling by a length of string L-1.95 m. The second ball m2- 0.72 kg is initially at rest while the baseball has an initial horizontal velocity of V 4.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity.) Randomized Variables m1 = 0.26 kg m2 = 0.72 kg L=1.95 m Vi 4.5 m/s m, 2Otheexp 50% Part (a) Select an express on for the magnitude ofthe closest distance from the ceiling the second ball will each d. 50% Part (b) what is the angle that the string makes with the vertical at the highest point of travel in degrees? = Grade Su Deductions Potential sinO | cos() | tan() | | (117| 8 cotan0 asin acos0 atan0 acotan0 sinh0 cosh0 tanh coa0+ Submissior Attempts ne (4% per att detailed via | 4 | 5 | 6 ODegrees Radians Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: deduction per feedback. _Explanation / Answer
8. given m1 = 0.26 kg
L = 1.95 m
m2 = 0.72 kg
V1 = 4.5 m/s
after collision baseball falls down
a. let after collision the spedd of m1 be u
then from conservation of momentum
m1V1 = m2*u
u = 1.625 m/s
from conservation of energy
0.5*m2*u^2 = m2gL(1 - cos(theta))
also, d = Lcos(theta)
hence
d = Lcos(theta) = (L - 0.5u^2/g)
d = L - 0.5(m1^2V1^2/m2^2*g)
d = 1.815 m
b. 1.95cos(theta) = 1.815
theta = 21.411 deg