Calculating Heights of Swinging Balls Colliding

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In summary, the conversation discusses the scenario of two balls suspended side by side and colliding with each other. The first ball, with a mass of m, is released from a height h and collides with the second ball, with a mass of 2m, at the lowest point. After the collision, both balls swing down and collide once more at the lowest point. The height each ball will swing after the second collision is being calculated using the equations for conservation of momentum and energy. The maximum potential energy is equal to the maximum kinetic energy, and the initial and final velocities of the first and second balls are determined to find the height the second ball reaches after the impact. A question is raised about a possible mistake in the equations.
  • #1
rpullo
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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.
 
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  • #2
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.
 
  • #3
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
 
  • #4
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 dosen't make sense =\
Can someone please point out where I went wrong?
Thanks
 
  • #5
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.
 
  • #6
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?
 
  • #7
Yes, it should be 0.5m(u2)^2.
 
  • #8
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?
 

What is the phenomenon of "two balls hanging"?

The phenomenon of "two balls hanging" refers to a common physics demonstration where two balls of equal mass are suspended from strings of equal length, side by side. When one ball is pulled back and released, it will swing forward, colliding with the other ball and causing it to swing in the opposite direction. This is also known as a Newton's Cradle.

What is the purpose of the "two balls hanging" demonstration?

The purpose of this demonstration is to illustrate the principles of conservation of momentum and energy. When the first ball is released, it transfers its energy and momentum to the second ball, which then transfers it to the third, and so on. This shows how energy and momentum are conserved in a closed system.

What factors affect the motion of the "two balls hanging" demonstration?

The motion of the balls in this demonstration is affected by several factors, including the mass of the balls, the length of the strings, and the angle at which the first ball is released. The elasticity of the balls and strings can also impact the motion.

What real-world applications does the "two balls hanging" demonstration have?

The principles demonstrated by "two balls hanging" are used in various real-world applications, such as in engineering and physics experiments. It can also be used to explain the concept of conservation of energy and momentum in sports, such as billiards and bowling.

Are there any limitations to the "two balls hanging" demonstration?

The "two balls hanging" demonstration is a simplified model of the conservation of energy and momentum, and therefore, it has limitations. It does not take into account external forces, such as air resistance, which can affect the motion of the balls. Additionally, it assumes perfectly elastic collisions, which may not always be the case in real-world situations.

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