M2 Collisions Watch

martinjolly
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Report Thread starter 12 years ago
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Ive tried this question over and over again and have got nowhere with it so any help would be much appreciated thanks:

Two small balls, A and B of equal radius and mass 4kg and 5kg respectively lie at rest on a smooth horizontal floor. Ball A is projected with speed u and collides with ball B. Following this collision ball B then strikes a smooth vertical wall normally. After rebounding from the wall, ball B agian collides wit ball A. Given that ball B is brought to rest by the second collison with A.

a) show that 2e³ - 3e² - 3e + 2 = 0 where e is the coefficient of restitution between the two balls and between ball B and the wall.

b) verify that e = 0.5 is the only practical solution for the equation
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Malik
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M2 Excercise 4D Question 7

Diagram:


a)
1st Collision:

using c.of.m=> 4u=4v1+5v2 (1)
e=(v2-v1)/u=> ue=v2-v1(2)
(2)x4=> 4ue=4v2-4v1 (3)
(1)+(3)=> 4ue+4u=9v2 => v2=[4u(e+1)]/9
.:. v1= [u(4-5e)]/9 (by substituting v2 into (1))

2nd Collision:
V=eu
.:. v3=e(v2) => v3= [4ue(e+1)]/9

Final Collision:
using c.of.m:
4u(4-5e)]/9 -20ue(1+e)/9 = -4v4
4u(4-5e)-20ue(1+e)=-36v4
-[u(4-5e)-5ue(1+e)]=9v4
-4u+5eu+5eu+5e²u=9v4 -----(4)

e=v4/(v1+v3)
e=9v4/(u(4-5e)+4ue(1+e)
eu(4-5e)+4e²u(1+e)=9v4
4eu-5e²u+4e²u+4e³u=9v4
4e³u-e²u+4eu=9v4 --------(5)

(4)=(5) 9v4=9v4
-4u +5eu +5eu +5e²u =4e³u -e²u +4eu (u cancels out)
-4 +5e +5e +5e² =4e³ -e² +4e
4e³-6e²-6e+4=0
2e³-3e²-3e+2=0

b) if e=1/2 then (2e-1) is a factor of 2e³-3e²-3e+2=0
Divide 2e³-3e²-3e+2=0 by (2e-1) to get=> e²-e-2 which factorises to (e-2)(e+1)

2e³-3e²-3e+2=(2e-1)(e-2)(e+1)
0«e«1
.:. e=1/2 is the only practical value.

I've answered this before its, a tough question which you wont expect to find on a real exam paper. Make sure you open the diagram first to understand it properly.
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