Elastic Collisions of steel balls

[kristen151027]

Here's the problem:

"Two steel balls are suspended on (massless) wires so that their centers align. One ball, with mass 2.30 kg, is pulled up and to the side so that it is 0.0110 m above its original position. Then it is released and strikes the other ball in an elastic collision. If the second ball has a mass of 3.10 kg, to what height does it rise above its original position?"

I started by assuming that the change in upward momentum would equal zero, but I came out with the wrong answer. Do I need to incorporate potential and kinetic energy into this problem? If so, I'm unsure exactly how to do so.

 

[Andrew Mason]

Here's the problem:

"Two steel balls are suspended on (massless) wires so that their centers align. One ball, with mass 2.30 kg, is pulled up and to the side so that it is 0.0110 m above its original position. Then it is released and strikes the other ball in an elastic collision. If the second ball has a mass of 3.10 kg, to what height does it rise above its original position?"

I started by assuming that the change in upward momentum would equal zero, but I came out with the wrong answer. Do I need to incorporate potential and kinetic energy into this problem? If so, I'm unsure exactly how to do so.

This question is all about energy.

[Note: Because the masses are suspended, momentum is not conserved generally. Momentum and energy is conserved, however, at the moment of collision since there is no vertical momentum and no constraint on horizontal motion (no horizontal force from the wires). ]

Use conservation of momentum to determine the speed of the first and second ball immediately after collision and then use energy conservation to determine how high the second ball rises.

AM

 

[kristen151027]

Okay, but I was told to use the conservation of energy first (the kinetic energy of the ball as soon as it is released is equal to its initial potential energy...so you can find it's initial velocity). I set its initial kinetic energy equal to the final kinetic energy of the two balls combined. Then, I solved for the final velocity of the first ball.

Then, using the conservation of momentum, I solved the the final velocity of the first ball again. I set the equations equal to each other, but I ended up with a quadratic and the square root of a negative number.

Any ideas?

 

[Saketh]

Okay, but I was told to use the conservation of energy first [...]

It doesn't really matter which one you do "first", as long as you have both before you start doing algebra.

Here are the two things that you've already established:

Kinetic energy is conserved, as is momentum at the collision:
[tex]\frac{m_1v_1^2}2+\frac{m_2v_2^2}2=\frac{m_1v_1'^2}2+\frac{m_2v_2'^2}2[/tex]
[tex]m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'[/tex]

You already know how to solve for [tex]v_1[/tex], and you know that [itex]v_2 = 0[/itex]. Solve for [itex]v_{1}'[/itex] and [itex]v_{2}'[/itex]. Make sure you keep your algebra straight, because otherwise you will become extremely frustrated with wrong answers.

From there, you know that energy is conserved in the second ball. So, after solving for and obtaining the value of [itex]v_{2}'[/itex], you can use conservation of energy to find the final height of the second ball.

 

[kristen151027]

...So, after solving for and obtaining the value of [itex]v_{2}'[/itex], you can use conservation of energy to find the final height of the second ball.

What you're saying is that you would set its KE equal to its PE?

Did you actually work out the problem? Because I couldn't get an answer by doing the algebra. I did it several times.

 

[Andrew Mason]

What you're saying is that you would set its KE equal to its PE?

Did you actually work out the problem? Because I couldn't get an answer by doing the algebra. I did it several times.

The total energy of the system is the potential energy of the first ball at the moment of its release. This total energy never changes. So:

[tex]E_{sys} = m_1gh_1[/tex]

Immediately prior to the collision:

[tex]E_{sys} = \frac{1}{2}m_1v_{1i}^2[/tex]

After collision by cons. of momentum:

[tex]m_1v_{1f} + m_2v_{2f} = m_1v_{1i} = m_1\sqrt{2gh_1}[/tex]

(1) [tex]v_{1f} = \sqrt{2gh_1} - m_2v_{2f}/m_1[/tex]

and by cons.of energy:

[tex]\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 = E_{sys} = m_1gh_1[/tex]

[tex]\frac{1}{2}m_1v_{1f}^2 = m_1gh_1 - \frac{1}{2}m_2v_{2f}^2 [/tex]

(2) [tex]v_{1f} = \sqrt{2(m_1gh_1 - \frac{1}{2}m_2v_{2f}^2)/m_1}[/tex]

Subtracting 1 from 2 will allow you to solve for v_{2f}:

[tex]\sqrt{2gh_1} - m_2v_{2f}/m_1 = \sqrt{2(m_1gh_1 - \frac{1}{2}m_2v_{2f}^2)/m_1}[/tex]

[tex]2gh_1 + m_2^2v_{2f}^2/m_1^2 - 2\sqrt{2gh_1}m_2v_{2f}/m_1 = 2gh_1 - m_2v_{2f}^2/m_1[/tex][tex]\left(\frac{m_2^2}{m_1^2} + \frac{m_2}{m_1}\right)v_{2f} - \left(\frac{2\sqrt{2gh_1}m_2}{m_1}\right) = 0[/tex]

[tex]v_{2f} = \left(\frac{2\sqrt{2gh_1}m_2}{m_1}\right) / \left(\frac{m_2^2}{m_1^2} + \frac{m_2}{m_1}\right)[/tex]

AM

 

[Saketh]

What you're saying is that you would set its KE equal to its PE?

Did you actually work out the problem? Because I couldn't get an answer by doing the algebra. I did it several times.

After the collision, the second ball has a certain kinetic energy. This energy has to be converted into potential energy. When all of the KE has been converted to PE, then the ball reaches its maximum height. You can solve for the height by setting KE = PE.

After you find [itex]v_{2f}[/itex], use the following expression (KE = PE) to find the height of the second ball.
[tex]\frac{1}{2}m_2v_{2f}^2 = m_2gh[/tex], where h is the height.