# Collisions in 2-D problem

1. Jul 4, 2012

### Ashu2912

1. The problem statement, all variables and given/known data
See the attachment for the diagram. 'A' is a fixed point at a height 'H' above a smooth horizontal plane. An ideal string of length 'L' (>H) has one end attached to 'A' and the other to a particle. The particle is held horizontally (the string is taut) and released. The collision between the particle and wall is inelastic. We have to find the height when the particle is at momentary rest again.
I assumed the following unknowns :
'$\theta$' = The angle shown in the diagram.
'h' = Answer
'v' = Velocity of particle wrt groung just before the collision.
g= Acceleration due to gravity at the earth's surface.
Frame of reference = ground (inertial)

2. Relevant equations
v = (2gH)1/2
The velocity along the common normal just after the collision is 0 and the velocity along the common tangent is vsin$\theta$, just after the collision. Hence the magnitude of velocity at this point is vsin$\theta$.
Applying conservation of energy to particle + earth system just after collision to max. height point,
h = (v sin$\theta$)2/2g
[External forces do no work]

3. The attempt at a solution
The answer comes out at H3/L2, but the answer given is H5/L4.

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2. Jul 4, 2012

### gnulinger

Do you think you could type up the problem description exactly as it was posted? I am having trouble picturing what this looks like.

3. Jul 5, 2012

### Ashu2912

See the images......
One of them has the question and the other has the diagram....

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4. Jul 5, 2012

5. Jul 5, 2012

### Ashu2912

It means that the collision between the particle and the surface is perfectly inelastic.

6. Jul 5, 2012

### lewando

Assuming no energy loss, I also get H3/L2. Hmm...another typo?

7. Jul 5, 2012

### Yukoel

Hello Ashu2912
The string is inextensible right? So check for the velocity of the rising ball after the collision.From what your picture looks like I assume the ball travels for some distance on the ground with its velocity constrained to be parallel to the ground because the string has slacked.Then as you work out in your attempts it rises up yet again.While rising up resolve the velocities along and perpendicular to the length of the string.Using the fact that the string is inextensible you will have to rule out one component right?(More precisely the fact that string is inextensible and the velocity component acts precisely to increase its length.)Do the new calculations give you the correct answer?(Apply energy conservation after that)

regards
Yukoel

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