Finding the momentum after a collision

In summary, the conversation is discussing the calculation of the momentum before and after a collision between two balls attached to a rod. The conservation of angular momentum is used to solve for the momentum after the collision, with consideration for the angular velocity of the rod and the velocity of the balls. The final expression for the momentum includes the mass and velocity of the two balls and the mass, length, and angular velocity of the rod.
  • #1
PhyIsOhSoHard
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0

Homework Statement


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A ball is attached to one end of a rod. A second ball hits the ball with velocity u. The collision is inelastic and both balls after the collision travel together with a new velocity v.
The rod is attached to the ceiling at a point with no friction. The moment of inertia for the rod is known and the mass of the rod is known and its length is L. The mass of the two balls are each known.

Find an expression for the momentum before and after the collision.


Homework Equations


[itex]p=mv[/itex]
[itex]v_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex]


The Attempt at a Solution


I know how to find the momentum for the balls by simply multiplying their speed with their mass.
My problem is with the rod. I'm not sure how to implement the rod so that I can use this in the equation for momentum. I know that I'm supposed to use the velocity at the center of mass but how do I calculate the velocity of center of mass?

[itex]v_{cm}=\frac{m_{ball1}v_{ball1}+m_{ball2}v_{ball2}+m_{rod}\cdot?}{m_{ball1}+m_{ball2}+m_{rod}}[/itex]

You see, I'm not sure what velocity to use for the rod when calculating the velocity of the center of mass.
 
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  • #2
This asks for the momentum before and after the collision. So what is (and what kind is) the momentum before the collision?
 
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  • #3
Sentin3l said:
This asks for the momentum before and after the collision. So what is (and what kind is) the momentum before the collision?


The momentum before the collision would be the ball with the velocity u:
p=mu

Where m is the known mass of the ball.
 
  • #4
What is going to be conserved during the collision (that you can make use of)? hint:
it is not linear momentum.
 
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  • #5
Linear momentum is conserved when net force acting is zero, here there is a force acting, guess what's that.
And you may get close to what haruspex is asking!
 
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  • #6
haruspex said:
What is going to be conserved during the collision (that you can make use of)? hint:
it is not linear momentum.


Im pretty sure it is the angular momentum about the rods axis, right?
 
  • #7
phoenixXL said:
Linear momentum is conserved when net force acting is zero, here there is a force acting, guess what's that.
And you may get close to what haruspex is asking!

Is it the angular momentum?
 
  • #8
PhyIsOhSoHard said:
Is it the angular momentum?

Yes, if you take moments about the right point. When the collision occurs, there is also an impulse at the pivot point of the rod. (Otherwise it would move.) In order to keep that unknown out of the equation, what point will you take moments about?
 
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  • #9
haruspex said:
Yes, if you take moments about the right point. When the collision occurs, there is also an impulse at the pivot point of the rod. (Otherwise it would move.) In order to keep that unknown out of the equation, what point will you take moments about?

I would take moment about the rods axis, the point where it is attached to the ceiling, right?
 
  • #10
PhyIsOhSoHard said:
I would take moment about the rods axis, the point where it is attached to the ceiling, right?

Yes.
 
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  • #11
haruspex said:
Yes.

But I'm still confused as to how I find the momentum after the collision.
I know that the equation for momentum is:
[itex]p=mv[/itex]

I know that for the balls it is simply their mass multiplied by their velocity which are all known variables. But I also have to take the rod's momentum into consideration. And while I know the mass of the rod, I have a hard time figuring out the velocity of the rod to insert into the momentum equation. I just can't see how I'm supposed to use the conservation of angular momentum in this situation.
 
  • #12
PhyIsOhSoHard said:
But I'm still confused as to how I find the momentum after the collision.
I know that the equation for momentum is:
[itex]p=mv[/itex]

I know that for the balls it is simply their mass multiplied by their velocity which are all known variables. But I also have to take the rod's momentum into consideration. And while I know the mass of the rod, I have a hard time figuring out the velocity of the rod to insert into the momentum equation. I just can't see how I'm supposed to use the conservation of angular momentum in this situation.
First, what is the angular momentum of the incoming ball about the rod's axis before collision?
After collision, the balls are moving, initially, at speed v. What is the angular momentum of the two balls about the rod's axis now?
For the rod itself, you know its length and how fast the lower end is moving, so what is its angular velocity?
 
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  • #13
haruspex said:
First, what is the angular momentum of the incoming ball about the rod's axis before collision?
After collision, the balls are moving, initially, at speed v. What is the angular momentum of the two balls about the rod's axis now?
For the rod itself, you know its length and how fast the lower end is moving, so what is its angular velocity?

The incoming ball's angular momentum about the rod's axis before collision is [itex]L=Mul[/itex]

The two ball's angular momentum after collision is [itex]L=2Mvl[/itex]

The rod moves at velocity v and its angular velocity is [itex]\omega=\frac{v}{l}[/itex]

Therefore the momentum before the collision is just the incoming ball:
[itex]p=Mul[/itex]

And the momentum after the collision is both balls together and the rod:
[itex]p=2Mvl+m\frac{v}{l}[/itex]

Would that be correct?
 
  • #14
PhyIsOhSoHard said:
And the momentum after the collision is both balls together and the rod:
[itex]p=2Mvl+m\frac{v}{l}[/itex]

Would that be correct?
Not quite. You don't multiply an angular velocity by a mass to get angular momentum. What equation do you need instead?
 
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  • #15
haruspex said:
Not quite. You don't multiply an angular velocity by a mass to get angular momentum. What equation do you need instead?

Is it [itex]L=I\omega[/itex]?

If my angular momentum is conserved, that means the angular momentum before and after are equal each other.

The angular momentum before the collision was [itex]L=Mul[/itex] so that has to be equal to the angular momentum after the collision:
[itex]Mul=I\frac{v}{l}[/itex]

Then I can find an expression for v. But isn't this the same velocity as the two balls? Don't the balls and rod share the same velocity so I can write it as:
[itex]p=2Mv+mv[/itex]?
 
  • #16
PhyIsOhSoHard said:
Is it [itex]L=I\omega[/itex]?
Yes. What is the moment of inertia of a uniform rod of given mass and length about one end?
 
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  • #17
haruspex said:
Yes. What is the moment of inertia of a uniform rod of given mass and length about one end?

That is [itex]I=\frac{1}{3}ml^2[/itex]

So then I have [itex]L=\frac{1}{3}ml^2\frac{v}{l}[/itex]

But how do I go from the angular momentum to the momentum? I need to incorporate it into [itex]p=mv[/itex] somehow but I haven't been able to figure it out.
 
  • #18
PhyIsOhSoHard said:
That is [itex]I=\frac{1}{3}ml^2[/itex]

So then I have [itex]L=\frac{1}{3}ml^2\frac{v}{l}[/itex]

But how do I go from the angular momentum to the momentum? I need to incorporate it into [itex]p=mv[/itex] somehow but I haven't been able to figure it out.

Using angular momentum you can find v. Once you have v you can find the linear momentum. If the balls are moving at speed v, how fast is the mass centre of the rod moving?
 
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  • #19
haruspex said:
Using angular momentum you can find v. Once you have v you can find the linear momentum. If the balls are moving at speed v, how fast is the mass centre of the rod moving?

I found my v to:
[itex]v=\frac{u}{2+\frac{m}{3M}}[/itex]

The CM velocity would be:
[itex]v_{CM}=\frac{2mv}{2m}=v[/itex]

So the velocity of CM would be equal to the speed of the balls?
 
  • #20
PhyIsOhSoHard said:
The CM velocity would be:
[itex]v_{CM}=\frac{2mv}{2m}=v[/itex]

So the velocity of CM would be equal to the speed of the balls?

No. This is just geometry. One end of the rod is moving at speed v, the other is stationary, so how fast is its centre moving?
 
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  • #21
haruspex said:
No. This is just geometry. One end of the rod is moving at speed v, the other is stationary, so how fast is its centre moving?

It's half. So that would make the linear momentum after the collision:
[itex]p=2Mv+m\frac{v}{2}[/itex]?
 
  • #22
phoenixXL said:
Linear momentum is conserved when net force acting is zero, here there is a force acting, guess what's that.
And you may get close to what haruspex is asking!
The force acting is the Tension in the string. As discussed earlier, angular momentum is conserved here.

What would be the initial and final angular momentum, when taken about the pivot point ?
 
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  • #23
phoenixXL said:
The force acting is the Tension in the string. As discussed earlier, angular momentum is conserved here.

What would be the initial and final angular momentum, when taken about the pivot point ?

According to my calculations:
Before:
[itex]L=Mul[/itex]

After:
[itex]L=(2Ml^2+1/3ml^2)v/l[/itex]
 
  • #24
Right, equating both you get the final velocity. The velocity you found is correct.
Now when finding
[tex]v_{cm}\ =\ ?[/tex]we divide total momentum by total mass.
Doesn't the rod even have momentum ?

And you get it as
PhyIsOhSoHard said:
It's half. So that would make the linear momentum after the collision:
p=2Mv+mv2?
So now what would be [tex]v_{cm}\ ?[/tex]
 
Last edited:
  • #25
phoenixXL said:
So now what would be [tex]v_{cm}\ ?[/tex]
The question only asks for the momentum.
 

Related to Finding the momentum after a collision

1. What is momentum?

Momentum is a physical quantity that measures the motion of an object, taking into account its mass and velocity. In other words, it is the product of an object's mass and its velocity.

2. Why is momentum important in collisions?

Momentum is important in collisions because it helps us understand the changes in motion of objects during a collision. By calculating the momentum before and after a collision, we can determine the forces involved and predict the resulting motion of the objects involved.

3. How is momentum conserved in a collision?

In a closed system where there are no external forces acting on the objects, momentum is conserved. This means that the total momentum before the collision will be equal to the total momentum after the collision.

4. What is the equation for finding momentum after a collision?

The equation for finding momentum after a collision is: p = m x v, where p is momentum, m is mass, and v is velocity.

5. How do you calculate the total momentum in a collision involving multiple objects?

To calculate the total momentum in a collision involving multiple objects, you need to calculate the individual momentum of each object and then add them together. This can be done using the equation p = m x v for each object and then adding the resulting momenta together.

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