Class Management I Help Homework 8 Begin Date: 3/29/2018 12:00:00 AM-Due Date: 4

Question

Class Management I Help Homework 8 Begin Date: 3/29/2018 12:00:00 AM-Due Date: 49/2018 11:59:00 PM End Date: 4/9/2018 11-59:00 PM (8%) Problem 1: A cue ball of mass m.-0.395 kg is shot at another billiard ball, with mass m2 = 0.555 kg, which is at rest The cue ball has an initial speed of v-S.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on A 256 Part (o) Write an expression for the horizontal component of the billiard ball's velocity, Va, afer the collision, in terms of the other variables of the problem. Grade Summary Potential 100% Submissions 0% Attempts remaining 7 per attempt) etailed view my Submit Hint I give up Hints: 2% deduction per hint. Hints remaining Feedback: 2% deduction per feedback. 25% Part (b) what is this velocity, in meters per second? i i à 25% Part (c) Wrte an expression for the horzontal component of the cue ball's velocity, v after the collision. - 25% Part (d) what is the horizontal component of the cue bars final velocity, in meters per second?

Explanation / Answer

Now it is elastic collision, so
using momentum conservation

Pi = Pf
m1u1 + m2u2 = m1v1f + m2v2f

u2 = 0

u1 = v

m1v = m1v1f + m2v2f

another condition for elastic collision
u1 - u2 = v2f - v1f
u2 = 0

v2f = v + v1f

m1v = m1v1f + m2v2f

m1v = m1*v1f + m2*(v + v1f)

v1f*(m1 + m2) = (m1 - m2)v

Solving above equation

v1f = v*(m1 - m2)/(m1 + m2)

v2f = v + v1f

v2f = v + v*(m1 - m2)/(m1 + m2)

v2f = 2*m1*v/(m1+ m2)

Part B

v2f = 2*0.395*5.5/(0.395 + 0.555)

v2f = 4.574 m/sec

Part C

from above solution

v1f = v*(m1 - m2)/(m1 + m2)

Part D

v1f = 5.5*(0.395 - 0.555)/(0.395 + 0.555)

v1f = -0.926 m/sec

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