Momentum intuition, had me thinking last night

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  • #1
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Main Question or Discussion Point

I came up with last night and I couldn't figure it out.

Can you actually predict what happens in a collision?

For instance, suppose a truck and a toy duck is coming from opposite ends, they both have a velocity, now suppose they crash, it should be obvious that the collison isn't going to be elastic. But how do we know this? Is it because of the significantly large momentum of the truck?

Suppose I give you another situation

Situation #2

A truck and a car coming towards each other, with the same speed.

And they clashed. What happens?

1) The truck has a greater momentum and hence will inelastically collide with the car and become one
2) Even though the truck has a greater momentum, they will collide and elastically bounce off each other

How do we predict this? Please only use linear momentum. Use these

M = mass of truck
m = mass of car
vtruck
vcar
vtruck + car
vtruck'
vcar'

Situation # 3 (other)

Suppose I throw a snowball at an 500-year-old oak tree and the ball sticks to the tree and eventually falls to the ground when it melts

So that

Mtree >> msnowball

Now we have

Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'


But

0 + msnowballvsnowball = (Mtree + msnowball)v'

0 + msnowballvsnowball = (Mtree)v'

For Mtree >> msnowball

And that the tree doesn't move (maybe the roots sustained the net force impaled?)



msnowballvsnowball = 0

But this doesn't make sense seeing neither are 0

Does this have to do with Newton's third law?
 

Answers and Replies

  • #2
Doc Al
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I came up with last night and I couldn't figure it out.

Can you actually predict what happens in a collision?

For instance, suppose a truck and a toy duck is coming from opposite ends, they both have a velocity, now suppose they crash, it should be obvious that the collison isn't going to be elastic. But how do we know this? Is it because of the significantly large momentum of the truck?
No. You can only predict that the collision will be inelastic due to your experience with such materials. Momentum is always conserved.

Suppose I give you another situation

Situation #2

A truck and a car coming towards each other, with the same speed.

And they clashed. What happens?

1) The truck has a greater momentum and hence will inelastically collide with the car and become one
2) Even though the truck has a greater momentum, they will collide and elastically bounce off each other

How do we predict this? Please only use linear momentum. Use these
Since momentum is always conserved, that won't be enough to predict whether the collision is elastic.

Situation # 3 (other)

Suppose I throw a snowball at an 500-year-old oak tree and the ball sticks to the tree and eventually falls to the ground when it melts

So that

Mtree >> msnowball

Now we have

Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'


But

0 + msnowballvsnowball = (Mtree + msnowball)v'

0 + msnowballvsnowball = (Mtree)v'

For Mtree >> msnowball

And that the tree doesn't move (maybe the roots sustained the net force impaled?)



msnowballvsnowball = 0

But this doesn't make sense seeing neither are 0

Does this have to do with Newton's third law?
Since the tree is attached to the earth, other forces are involved besides that from the snowball. For practical purposes, the speed of the tree remains 0. (Of course, you know that the momentum of the earth+tree+snowball system must remain constant. )
 
  • #3
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Situation # 3 (other)[/b]

Suppose I throw a snowball at an 500-year-old oak tree and the ball sticks to the tree and eventually falls to the ground when it melts

So that

Mtree >> msnowball

Now we have

Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'


But

0 + msnowballvsnowball = (Mtree + msnowball)v'

0 + msnowballvsnowball = (Mtree)v'

For Mtree >> msnowball

And that the tree doesn't move (maybe the roots sustained the net force impaled?)



msnowballvsnowball = 0

But this doesn't make sense seeing neither are 0

Does this have to do with Newton's third law?
You are right in your 'maybe...'. There is another external force (besides that of the snowball) acting on the tree and so linear momentum of (snowball + tree) is not conserved.
 
Last edited:
  • #4
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Doc Al said:
No. You can only predict that the collision will be inelastic due to your experience with such materials. Momentum is always conserved.
So physics can't tell us why trucks can run over toy ducks?

Doc Al said:
Since momentum is always conserved, that won't be enough to predict whether the collision is elastic.
Or inelastic either? Is what you are implying in the first quote?

Doc Al said:
Since the tree is attached to the earth, other forces are involved besides that from the snowball. For practical purposes, the speed of the tree remains 0. (Of course, you know that the momentum of the earth+tree+snowball system must remain constant. )
But what's wrong with my math though? How do I include the earth into my equation? Or is this entirely up to energy?
 
  • #5
Doc Al
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So physics can't tell us why trucks can run over toy ducks?
There's a lot more to physics than simple conservation of momentum.
Or inelastic either? Is what you are implying in the first quote?
Right.
But what's wrong with my math though?
Now we have

Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'
Your equation would be true if the tree were not attached to the earth.
 
  • #6
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1
Doc AL said:
Your equation would be true if the tree were not attached to the earth.


But I could use that equation again and arrive at

msnowballvsnowball = 0

Doc Al said:
Right.
Can't we still tell what would happen in a collison based on the material of the bodies slamming?

In my duck and truck example, it isn['t just experience

It is

1) Truck mass/density >> toy duck mass/density
2) Truck speed >> duck speed (0)
3) Duck made of rubber? I don't know and truck made of metal
 
  • #7
Doc Al
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But I could use that equation again and arrive at

msnowballvsnowball = 0
Only by treating the mass of the tree as 'infinitely' larger than the mass of the snowball. A reasonable approximation. Just depends on how accurate you want to be.
Can't we still tell what would happen in a collison based on the material of the bodies slamming?
Of course, given knowledge of the physics of materials. That is not always trivial.
 
  • #8
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Doc Al said:
Only by treating the mass of the tree as 'infinitely' larger than the mass of the snowball. A reasonable approximation. Just depends on how accurate you want to be.
But that's it, m and v aren't 0, how could I have gotten 0 from their product?
 
  • #9
Doc Al
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But that's it, m and v aren't 0, how could I have gotten 0 from their product?
I assumed you meant the speed after the collision. Of course, before the collision the speed of the snowball isn't zero.

The momentum before equals the momentum after. (Assuming the tree is not attached to the earth and is free to slide, as if on ice.) Of course, since the mass of the tree >> the mass of the snowball, the speed of both after the collision is approximately zero. So what? Exactly zero? No. Total mass X final speed = mass of snowball X initial speed.
 
  • #10
Mtree >> msnowball

Now we have

Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'


But

0 + msnowballvsnowball = (Mtree + msnowball)v'

0 + msnowballvsnowball = (Mtree)v'

For Mtree >> msnowball
So isn't the answer that msnowballvsnowball (Mtree)v' ? or is = correct?
 
  • #11
sophiecentaur
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I don't understand how anyone would think that elastic or inelastic collision is something to do with the relative masses or speeds. It's entirely to do with the construction of the objects. A ball bearing hitting the steel plate on the front of an Express train is a pretty well 100% elastic collision. Two equal masses of rice pudding coming together are an inelastic collision. A (real) duck and a truck is inelastic. Two ball bearings colliding is elastic. Need I go further?

Also, it is pointless trying to find errors in the Momentum Conservation law. The law refers to the total system involved and always applies; you just have to take everything relevant into account if you want to apply the Law. When a 1kg duck hits a massive truck, which is freewheeling over the ground, it is only the masses of duck and truck that need to be taken into consideration for that collision. If the truck is exerting significant force on the Earth's surface (e.g. driving), the mass (or moment of inertia) of the Earth becomes, strictly speaking, part of the calculation. There is a tiny change in the Earth's rotation - as there was when the truck was initially accelerating - when anything happens to the truck. At the end of the process, the Earth will still be rotating at the same rate.
The fact is that in collisions, any change in the Earths motion will be insignificant. If there is an elastic collision then the small colliding object will have its velocity of approach reversed.

When I learned about 'Dynamics' we were taught about the Coefficient of Restitution. This was the ratio of speed of separation after to the speed of approach before collision. This can be used to calculate what will happen in partly elastic collisions (very common events). You could set up equations for Momentum Conservation and for velocities before and after and find out what would happen as a result of a collision. This Coefficient is not easy to measure and it will often depend upon the actual speeds and masses involved (most objects do not behave linearly) but it provided hours of 'fun' for us.
 

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