Calculating Heights of Swinging Balls Colliding

  • Thread starter Thread starter rpullo
  • Start date Start date
  • Tags Tags
    Balls
Click For Summary

Homework Help Overview

The problem involves two small balls suspended from strings, where one ball is released from a height and collides with the other at the lowest point. The inquiry focuses on determining how high each ball swings after a second collision, assuming elastic collisions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and energy principles to analyze the collisions. There are attempts to derive expressions for velocities before and after collisions, with some participants questioning the initial conditions and energy transformations involved.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between kinetic and potential energy. Some guidance has been offered regarding the calculations needed to find the velocities, but there remains uncertainty about specific expressions and initial conditions.

Contextual Notes

There are mentions of potential confusion regarding the initial velocities of the balls and the application of energy conservation principles. Participants are also clarifying the correct forms of energy equations in the context of elastic collisions.

rpullo
Messages
14
Reaction score
0

Homework Statement



Two small balls are suspended side by side from two strings of length L so that they just touch when in their lowest positions. Their masses are m and 2m, repectively. If the left ball (of mass m) is pulled aside and released from a height h, it will swing down and collide with the right ball (of mass 2m) at the lowest point. Both balls again swing down, and they collide once more at the lowest point. How high will each swing after this second collision? Assume both collisions are elastic.

Homework Equations



So, for now, I'm just using:

mv1 +mv2 = mv1' + mv2'
Ek = 0.5mv^2
Ep = mgh

The Attempt at a Solution



So I know v2 = 0, but I'm not sure if v1 would also be 0.
For ball#1 (mass m), Ek = Ep => 0.5v1'^2 = mgh
Sort of stuck as with what to do next...
Any help is appreciated.
 
Physics news on Phys.org
For the first collision:

Your mass m1 will reach its maximum velocity at the point of impact (why?) so yes, that also means that its kinetic energy will be a maximum there and its potential energy will be zero (again, why?). From these conditions you can calculate the initial velocity of the first mass m. We also know that, since it's an elastic collision energy will be conserved.

That should be enough to get you started.
 
So I basically tried what you were saying...but I can't seem to find an expression for v1 to find h. I was wondering if you could help me get started with finding it. THanks
 
Ok from:

Conservation of momentum:
mv1 + mv2 = mv1' + mv2'

and Conservation of energy:
0.5mv1^2 + 0.5mv1'^2 = 0.5mv2^2 + 0.5mv2'^2

...I got:

(1)... m(v1+v1') = 2m(v2'-v2)
(2)... 0.5m(v1+v1')(v1-v1') = m(v2'+v2)(v2'-v2)

Since v1' and v2 are 0 and m's can cancel:

(3)... v1 = 2(v2')
(4)... 0.5(v1)^2 = (v2')^2

However, substituting (3) into (4) results into an expression that makes no sense:

0.5[4(v2')^2] = (v2')^2 --> 2(v2')^2 = (v2')^2 ?

It just doesn't make sense =\
Can someone please point out where I went wrong?
Thanks
 
I'm sorry, I made a typo here :redface:

It should be:

phyzmatix said:
From these conditions you can calculate the final velocity of the first mass m

The maximum energy is the maximum potential energy (where the first mass is at its highest point) so

[tex]E_{max} = E_p = m_1gh[/tex]

at the bottom of its swing, this potential energy will be zero, but its kinetic energy will be a maximum and from conservation of energy we have

[tex]E_p = E_k[/tex]
[tex]m_1gh = \frac{1}{2}m_1v_1^2[/tex]
[tex]v_1=\sqrt{2gh}[/tex]

Now, we have the initial conditions (I use u for initial velocity and v for final, all those primes just confuse me)

[tex]u_1 = 0[/tex] (the initial velocity of the first mass before it's released)

[tex]v_1 = \sqrt{2gh}[/tex] (the final velocity of the first mass at impact)

[tex]v_2 = 0[/tex] (the final velocity of the second mass when it reaches the top of its swing)

[tex]u_2 = ?[/tex] (what we're looking for)

From conservation of momentum

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
[tex]m_1u_1 + 2m_1u_2 = m_1v_1 + 2m_1v_2[/tex]
[tex]u_1 + 2u_2 = v_1 + 2v_2[/tex]
[tex]0 + 2u_2 = \sqrt{2gh} + 0[/tex]
[tex]u_2 =\frac{1}{2} \sqrt{2gh}[/tex]

The energy of the second mass is at a max shortly after the collision and we can calculate its kinetic energy

[tex]E_k = \frac{1}{2}u_2^2[/tex]

This should get you going towards finding the height the second mass reaches after the impact.

I've given you quite a lot here and I hope it helps. I'm not going to be around for the next couple of days, so please contact one of the homework helpers if you're still stuck (have a look under the "Staff" tab at the top of the PF page).

Good luck.
 
Alright...one question.
I'm not sure if this is a mistake on your part of if I'm missing something but..
Why does Ek = 0.5(u2)^2?...Shouldn't it be Ek = 0.5m(u2)^2?
 
Yes, it should be 0.5m(u2)^2.
 
phyzmatix said:
[tex]u_1 = 0[/tex] (the initial velocity of the first mass before it's released)

[tex]v_1 = \sqrt{2gh}[/tex] (the final velocity of the first mass at impact)

[tex]v_2 = 0[/tex] (the final velocity of the second mass when it reaches the top of its swing)

[tex]u_2 = ?[/tex] (what we're looking for)

I'm confused. Shouldn't u2 be 0, since the 2nd ball is initially hanging?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
21
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K