# M2 question

Watch
This discussion is closed.
#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.
0
15 years ago
#2
hmm, is this an Edexcel question?
0
#3
Yup. It's from the 2nd review exercise.
0
15 years ago
#4
(Original post by shift3)
Yup. It's from the 2nd review exercise.
question no? .... ah q20 working on it
0
15 years ago
#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?
0
#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)
0
X
new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

What I expected (116)
25.78%
Better than expected (93)
20.67%
Worse than expected (241)
53.56%