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shift3
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#1
Report Thread starter 15 years ago
#1
A light inextensible string of length 4a has a particle A, of mass m, attached at one end and a particle B, of mass m, attached at the other end. The string passes through a small smooth ring which is fixed at a point O at a distance 3a above a horizontal table. The system is hanging in equilibrium with OB = 2a when a smooth bead of mass 2m, which is threaded on the string between O and B, is released from rest at O. The bead falls under gravity until it collides with and adheres to the particle B to form a composite particle C.
Given that the string remains taut, show that the speed of C immediately after the collision is sqrt(ga).


I can solve this, but I'm not really content with my reasoning. I did it in two ways, once considering energy and once using the standard equations of motion, but I think I'm doing something wrong... So I wanted to see how you guys would approach it.
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KingAS
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#2
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hmm, is this an Edexcel question?
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shift3
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#3
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Yup. It's from the 2nd review exercise.
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KingAS
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(Original post by shift3)
Yup. It's from the 2nd review exercise.
question no? .... ah q20 working on it
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KingAS
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hehe i cant do it! Noooo... better revise that chapter... got about 2.5 hours before my exams!! so can you post wht u got so far?
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shift3
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#6
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Before C is released:
PE=mgh=(2m)g(3a)=6mga
KE=0

When C joins B:
PE=mgh=(m+m+2m)ga=4mga
KE=(1/2)mv²=(1/2)(m+m+2m)v²=2mv²

so....
2mv²=2mga
v²=sqrt(ga)
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