Conservation of Momentum - Inelastic Collisions

[SkyrimKhajiit]

Hello, I recently posted a few conceptual questions regarding conservation of momentum. I'm processing the information more easily now, but I've hit yet another bump in the road. I've learned that in an inelastic collision in which "velocity is the same before and after the collision" the velocity can be determined by dividing the initial velocity by the ratio of m_new:m_old. In other words, mass and velocity are inversely proportional:

http://www.physicsclassroom.com/class/momentum/Lesson-2/Using-Equations-as-a-Guide-to-Thinking

(This is the animation): http://www.physicsclassroom.com/mmedia/momentum/fca.cfm

But then, how is velocity the same before and after if it's obviously changing? And when would this situation of being able to find the new velocity by a simple ratio not work?

Thanks, and I know this is going to be a very simple answer pointing out something I missed, but I'm just stumped :D.

 

[Doc Al]

Two objects undergoing an inelastic collision end up with the same final velocity. That velocity is different from their initial velocities.

 

[mfb]

I've learned that in an inelastic collision in which "velocity is the same before and after the collision"

That is not true, independent of the question which velocities you mean.

the velocity can be determined by dividing the initial velocity by the ratio of m_new:m_old

What are new and old? What exactly do you mean with "dividing [...] velocity by the ratio"? I think I know what you mean, this is a special case - a perfectly inelastic collision where the relative velocity between the collision partners after the collision is zero.

 

[haruspex]

There appears to be a mistake in the video. In the last frame it says the combined speed is 5km/hr, but if you note what it says in the preceding frame you can see that the last frame should say 4km/hr.

 

[HalfLostAndSearching]

Hello,

Thank you for your question and for sharing your thought process as you continue to learn about the conservation of momentum in inelastic collisions. You are correct in your understanding that in an inelastic collision, the velocity is the same before and after the collision. This may seem counterintuitive, as we often think of collisions as involving a change in velocity.

However, in an inelastic collision, the objects involved stick together after the collision and move together as one object. This means that the velocity of the combined object is the same as the velocity of the individual objects before the collision.

To understand why the ratio method works, let's look at the equation for conservation of momentum in an inelastic collision: m1v1 + m2v2 = (m1 + m2)v. If we rearrange this equation, we get v = (m1v1 + m2v2) / (m1 + m2). This is the same equation you referenced in your post.

Now, let's look at what happens when we divide the initial velocity by the ratio of mnew:mold. This can be rewritten as (mold/mnew) * v1. If we plug this into the equation for conservation of momentum, we get v = ((mold/mnew) * v1) + v2. This shows that the new velocity is a combination of the initial velocities of the two objects, with a weight of (mold/mnew) given to the initial velocity of the first object.

So to answer your question, the velocity is the same before and after the collision because we are considering the combined velocity of the objects after the collision. The ratio method works because it takes into account the relative masses of the objects and how they contribute to the combined velocity.

As for when this method may not work, it will not work in situations where the objects involved do not stick together after the collision. In this case, the objects will have different velocities before and after the collision, and the ratio method will not accurately predict the new velocity.

I hope this helps clarify your understanding of conservation of momentum in inelastic collisions. Keep asking questions and exploring the concepts, and you will continue to deepen your understanding. Best of luck in your studies.