ta The Expert TAIHuman-lik x C Physics question Chegg Rashaad X C https://usr05
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ta The Expert TAIHuman-lik x C Physics question Chegg Rashaad X C https://usr05 ar.theexpertta.com/Common/TakeTutorialAssignment.aspx Apps G Google X YouTube to mp3 Co... ValoreBooks.com I.... P chemistry: A Molec... C 18- Kidneys: 10-Me... ADP iPayStatements... b Thunder 08 Account... Audible, try it free Conservation Begin Date: 11/2/2015 12:00:00 AM Due Date: 11/13/2015 11:59:00 PM End Date: 11/13/2015 11:59:00 PM (S%) Problem 11 A cue ball of mass m 0.315 kg is shot at another billiard ball, with mass my 0.535 kg, which is at rest. The cue ball has an initial speed of v 8.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on. Randomized Variables 0.315 kg m 0.535 kg 8.5 Assignment Status Click here for 25% Part (a) Write an expression for the horizontal component of the billiard ball's velocity, v2f after the collision, in terms of the other detailed view variables of the problem. Grade Summary v2f Deductions Problem Status Potential 100% Completed. a B e 7 8 9 HOME Subm. Partial Attempts remaining Completed. per attempt Partial detailed view Completed P t v BACKSPACE DEL CLEAR Partial Submit Hint I Completed. give up! Partial hint. H remaining deduction per Feedback: 0% deduction per feedback. 10 Completed 11 Partial A 25% Part (b) What is this velocity, in meters per second 12 Partial A 25% Part (c) Write an expression for the horizontal component of the cue ball's velocity, v1f after the collision. A 25% Part (d) What is the horizontal component of the cue ball's final velocity. in meters per second? TA, LLC 9:46 PM /13/2015Explanation / Answer
Given that
A cue ball of mass m1 =0.315kg
Another cue of ball m2 =0.535kg
The initial velocity of the first ball u1 =8.5m/s
The initial velocity of the second ball u2 =0m/s
From the conservation of momentum
m1u1+m2u2 =m1v1+mv2
Where v1and v2 are the final velocites of the cue balls after collision
m1u1 =m1v1+m2v2
Therefore, here it is elastic collision, kinetic energy is also conserved then
(1/2)m1u12+(1/2)m2u22 =(1/2)m1v12+(1/2)mv22
From the both the equation by solving we get
u1/v1 =(m1+m2)/(m1-m2)
Therefore the v1 =(m1-m2)/(m1+m2)*u1=(0.315-0.535)/(0.315+0.535)*8.5 =-2.199m/s
v2 =(m1/m2)(u1-v1)=(0.315/0.535)(8.5+2.199) =6.291m/s
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