# Homework Help: Collision Between Two Balls on Strings

1. Aug 14, 2011

### Megzzy

1. The problem statement, all variables and given/known data
Two marble spheres of masses 30 and 20 grams, respectively, are suspended from the ceiling by massless strings. The lighter sphere is pulled aside, as shown in the diagram, through an angle of 75° and let go. It swings and collides elastically with the other sphere at the bottom of the swing.
a. To what maximum angle will the heavier sphere swing?
b. To what maximum angle will the lighter sphere swing?

2. Relevant equations
mgh=1/2mv^2
m1v1 +m2v2=m1v1

3. The attempt at a solution
I'm actually really not sure on this one. Would you find the velocity of the smaller ball by using mgh=1/2mv^2? Then could I use that velocity in a momentum equation?

2. Aug 15, 2011

### ehild

Yes, use conservation of energy first, to get the velocity of the lighter ball at the impact, then use conservation of momentum, and also conservation of energy, as the collision is elastic.

ehild

3. Sep 4, 2011

### Megzzy

ahhh I'm sorry I've been and completely forgot I even asked this question.
I'm still somewhat stuck but this is what I have so far.

1/2mv^2=mgh
=mg(L sin theta)
so v therefore is v=sqrt(2g L sin theta)

Also 1/2 m v^2 + 0 = m g L sin(theta)

and then the two equations
mv=mv'+MV'
1/2mv^2=1/2MV^2

Then I am lost again because of the unknowns.

4. Sep 4, 2011

5. Sep 4, 2011

### Megzzy

The massless string is 95 cm by the way. I didn't realize the problem didn't state that.
Here is the diagram.
http://img.photobucket.com/albums/v330/JrJeter2/Screenshot2011-09-04at90602PM.png [Broken]

Because it is an isosceles triangle the 2 remaining angles had to be 52.5 degrees. So I then found the missing side of the triangle which turned out to be 115.6cm. Then I made a right triangle with one side being 95 and one 115.6 while keeping that 75 degree angle and found the side to be 30.8. I then subtracted this from 95 to get a height of 64.2. This all sounds very confusing I'm aware (sorry for the horrible description).

I then used the formula 1/2mv^2=mgh.
1/2(20)v^2=(20)(9.8)(0.642)
and found v to be equal to 3.55 m/s.

Then I used the equations mentioned in my other post and found the v'=2.31m/s and V'=2.2m/s.
Then finally I plugged these into the formula 1/2mv^2=mg 95sintheta n found the angle for the first sphere to be 20.8 degrees and the second to be 18.9 degrees.

I'm probably way off with this aren't I? Sorry for the trouble.

Last edited by a moderator: May 5, 2017
6. Sep 4, 2011

### ehild

It is really very complicated. To get h, see attached picture.

ehild

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7. Sep 5, 2011

### Megzzy

Thanks so much! I think I have it now.

Would you mind checking the work of another problem I did? I don't really want to start a new thread just to get it checked over.

The system at the right is in static equilibrium, and the string in the middle is exactly horizontal. Find
a. tension T1. b. tension T2. c. tension T3. d. angle θ.
This is the diagram provided.
http://img.photobucket.com/albums/v330/JrJeter2/Screenshot2011-09-05at125358AM.png [Broken]

I converted to the weights to Newtons then assumed the inner angle for T1 would be 30 degrees to make the 90 degree angle.

a)T1=29.4N/cos30
T1=33.9N

b)T2=T1sin30
=33.9sin30
=16.95N

c)Tx=T2
Tx=16.95
Ty=19.6(2kg)

T3=sqrt(tx^2+ty^2)
T3=25.9N

d)tan-1(16.95/19.6)
=40.9degrees

Thanks so much for your help!

Last edited by a moderator: May 5, 2017
8. Sep 5, 2011

### ehild

It looks correct.

ehild