Collision Between Two Balls on Strings

In summary, the heavier sphere swings to a maximum angle of 75 degrees, while the lighter sphere swings to a maximum angle of 18.9 degrees.
  • #1
Megzzy
22
0

Homework Statement


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?

Homework Equations


mgh=1/2mv^2
m1v1 +m2v2=m1v1



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?
 
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  • #2
Megzzy said:

The Attempt at a Solution


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?

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
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.
 
  • #5
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]

I think I went about this a more difficult and long way if its even correct.
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.
 
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  • #6
It is really very complicated. To get h, see attached picture.

ehild
 

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  • #7
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:
  • #8
It looks correct.

ehild
 

1. How do you calculate the velocity of the balls after the collision?

The velocity of the balls after the collision can be calculated using the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. This can be expressed mathematically as m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the balls and v1, v2, v1', and v2' are their velocities before and after the collision, respectively. By solving for v1' and v2', you can determine their values after the collision.

2. What factors affect the outcome of the collision between two balls on strings?

The outcome of the collision between two balls on strings is affected by several factors, including the mass and velocity of the balls, the length and elasticity of the strings, and the angle at which the balls collide. These factors can influence the amount of energy transferred between the balls and the direction in which they move after the collision.

3. Can the collision between two balls on strings be elastic?

Yes, the collision between two balls on strings can be elastic if there is no energy lost during the collision. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. However, this is only possible if the strings and balls are perfectly elastic, which is not always the case in real-world situations.

4. How does the angle of collision affect the outcome of the collision between two balls on strings?

The angle at which the balls collide can greatly impact the outcome of the collision. If the balls collide head-on (at a 180-degree angle), they will transfer the maximum amount of energy to each other and move in opposite directions after the collision. However, if the angle of collision is less than 180 degrees, the balls will transfer less energy and may move in different directions than expected.

5. Can the collision between two balls on strings be used to demonstrate conservation of momentum?

Yes, the collision between two balls on strings is a classic example of conservation of momentum. Due to the strings' flexibility, the total momentum of the system (balls and strings) remains constant before and after the collision. This can be observed by measuring the velocities of the balls before and after the collision and comparing them to the calculated values using the conservation of momentum equation.

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