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# M2 question watch

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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.
2. hmm, is this an Edexcel question?
3. Yup. It's from the 2nd review exercise.
4. (Original post by shift3)
Yup. It's from the 2nd review exercise.
question no? .... ah q20 working on it
5. 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?
6. 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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