Class Management Help HW#8 Begin Date: 7/14/2017 5:00:00 PM-Due Date: 7/18/2017
ID: 1632237 • Letter: C
Question
Class Management Help HW#8 Begin Date: 7/14/2017 5:00:00 PM-Due Date: 7/18/2017 11:59:00 PM End Date: 7/18/2017 11:59:00 PM (8%) Problem 3: A block of mass m1 = 13.5 kg slides along a horizontal surface (with friction, ,-0.24) a distance d= 2.2 m before striking a second block of mass m2- 6.25 kg. The first block has an initial velocity of v-8.75 m/s. Randomized Variables m 13.5 kg m2- 6.25 kg k= 0.24 d= 2.2 m v= 8.75 m/s ©theexpertta.com 50% Part (a) Assuming that block one stops after it collides with block two, what is block two s velocity after impact in ms? Grade Summary Deductions Potential 090 100% sin) tan() Submissions Attempts remaining: 10 (0% per attempt) - cos cotansi acos(0 atan)acotanOsinhO cosh tnh cotanh0 123 0 detailed view ODegrees O Radians EL CLEAR Submit Hint I give up! Hints: 100 deduction per hint. Hints remaining. S Feedback: 3% deduction per feedback. 50% Part (b) How far does block two travel, in meters, before coming to rest after the collision?Explanation / Answer
a) Only force acting on block along the direction of motion is friction, f = µkmg = ma
acceleration, a = µk*g = 0.24 * -9.8m/s² = - 2.352 m/s2
Velocity of block 1 just before the impact: v2 = u2 + 2as = (8.75)2 + 2*(- 2.352)*2.2 = 66.214
v = 8.137 m/s impact velocity
Now conserve momentum:
13.5kg * 8.137 + 6.25kg * 0 = 13.5kg * 0 + 6.25 * v2
v2 = 17.576 m/s
(b) 02 = (v2)2 + 2ad2 = (17.576)2 + 2*(- 2.352)*d2
d2 = 65.67 m