Momentum: why don't the two carts become one?

[paulb203]

Homework Statement

A 1kg cart moving rightward at 5m/s collides with a second cart moving rightward at 2m/s.
After the collision the 1kg cart is moving rightward at 0.5m/s and the second cart is moving rightward at 2.3m/s.
What is the mass of the second cart?
[Consider rightward the positive direction. Assume friction and air resistance are negligible. Round your answer to two significant digits.]

Relevant Equations

Pi=Pf
(m1)(u1)+(m2)(u2)=(m1+m2)v

So far I’ve dealt with scenarios in which the vehicle colliding with the rear of the other vehicle ‘joins’ the other vehicle, and the ‘two become one’, so there is one final velocity.

Why have these two vehicles not ‘become one’?

I’ll have a go at the maths first then have a go at answering the above.

Pi=Pf

Pi=(1x5)+(2x)

Pi=5+2x

Pf=(1x0.5)+(2.3x)

Pf=0.5+2.3x

5+2x=0.5+2.3x

5=0.5+0.3x

4.5=0.3x

x=4.5/0.3

x=15kg

So, if my maths is correct, the second cart has a mass of 15kg

So a 1kg cart at 5m/s goes into the back of a 15kg cart at 2m/s...

...and the 1kg slows to 1/10 of its speed...

...and the 15kg speeds up by 3/20 of its speed...

...and the second cart is 15 times the mass of the first...

I'm guessing there is some complex maths involving ratio/direct/inverse proportion, that kind of thing, to explain this (?).

Q If the first cart doesn’t ‘join’ with the second but instead continues at a new velocity, one different from the second cart, what happens to it on collision?

Does it initially get pushed backwards?

Does it push the second cart causing itself to stop momentarily?

I’m trying to imagine driving down the motorway in a car and going into the back of the lorry, with me moving at 2 and half times the speed of the lorry, and 1/15 the mass...

If the lorry was parked I'm guessing the two of us would become one, although I'm also wondering if the masses and or my speed was different I might 'bounce' back off it (I suppose that could depend on how elastic our bumpers were?).

 

[PeroK]

Why have these two vehicles not ‘become one’?

Such a collision could be totally inelastic, meaning that the two carts would be stuck together. That would be a different problem. In this case, the collision has some elasticity (rebound of one cart of the other). The mass can be determined from conservation of momentum.

You could also, if you want, calculate how much energy (if any) was lost in the collision.

 

[PeroK]

PS note that there is nothing fundamentally different about a collision where both objects are initialy moving in the same direction. All you have to do is change reference frames to have the slower moving object at rest. Or, change to the centre of momentum frame - which might be quite instructive in this case.

 

[kuruman]

Why have these two vehicles not ‘become one’?

Because you are told that they haven't. Vehicles that "become one" when they collide move "as one" after the collision. Here you are told they move with different velocities. It happens often. The game of billiards, for example, would be quite different if the balls "became one" after colliding.

 

[DaveC426913]

Why have these two vehicles not ‘become one’?

Newton's Cradle is a good demonstration of why:

 

[paulb203]

Such a collision could be totally inelastic, meaning that the two carts would be stuck together. That would be a different problem. In this case, the collision has some elasticity (rebound of one cart of the other). The mass can be determined from conservation of momentum.

You could also, if you want, calculate how much energy (if any) was lost in the collision.

Thanks.
When you said this collison has some elasticity does that mean it is, simply put, an elastic collision?
When you said 'rebound of one cart of the other' does that mean the first cart, the one with the smaller mass, is pushed backwards, momentarily, before moving forwards again (or, in this case, it is pushed leftwards, before moving rightwards again?
Is my maths correct (mass of second vehicle=15kg)?

 

[jbriggs444]

the one with the smaller mass, is pushed backwards, momentarily, before moving forwards again (or, in this case, it is pushed leftwards, before moving rightwards again?

A leftward acceleration can coexist with a rightward velocity.

 

[paulb203]

Because you are told that they haven't. Vehicles that "become one" when they collide move "as one" after the collision. Here you are told they move with different velocities. It happens often. The game of billiards, for example, would be quite different if the balls "became one" after colliding.

Thanks.
In some vehicle collision scenarios the two do become one, when, for example, a more massive vehicle goes into the back of a less massive one, when the two are moving at the same velocity; what would happen if a more massive billiard ball 'went into the back' of a less massive one, when the two were moving at the same velocity?

 

[paulb203]

Newton's Cradle is a good demonstration of why:
View attachment 358664

Thanks.
If there were just two balls in the cradle and the one on the left was more massive and was moved up and to the left then let go; would the two 'become one'?

 

[paulb203]

A leftward acceleration can coexist with a rightward velocity.

Thanks.
I think I get that. A leftward acceleration in this case would be a case of slowing down, of deceleration, I think; but if the first cart 'rebounds' does that mean it actually MOVES LEFTWARDS, initially, before resuming it's rightwards path? Or does it simply decelerate (accelerate leftwards, accelerate in the negative direction)?

 

[haruspex]

if the first cart 'rebounds' does that mean it actually MOVES LEFTWARDS,

That's what rebounding means. In general, it may rebound or continue at a lower velocity. It depends on the masses.

initially, before resuming it's rightwards path?

What force would make it switch to moving right again?

 

[nasu]

Thanks.
If there were just two balls in the cradle and the one on the left was more massive and was moved up and to the left then let go; would the two 'become one'?

No. The nature of collision (stick together or move apart) depends on the elastic properties of the objects and not on the ratio of the masses.

 

[jbriggs444]

Thanks.
I think I get that. A leftward acceleration in this case would be a case of slowing down, of deceleration, I think; but if the first cart 'rebounds' does that mean it actually MOVES LEFTWARDS, initially, before resuming it's rightwards path? Or does it simply decelerate (accelerate leftwards, accelerate in the negative direction)?

It simply decelerates.

If one adopts the center-of-momentum frame as @PeroK suggests, one can see that this deceleration is also a true rebound.

 

[DaveC426913]

Thanks.
I think I get that. A leftward acceleration in this case would be a case of slowing down, of deceleration, I think; but if the first cart 'rebounds' does that mean it actually MOVES LEFTWARDS, initially, before resuming it's rightwards path? Or does it simply decelerate (accelerate leftwards, accelerate in the negative direction)?

The second is correct. A negative acceleration can still result in a positive (though smaller) velocity.

Read this diagram as if time is a horizontal line, sweeping downwards (blue arrows):

 

[paulb203]

That's what rebounding means. In general, it may rebound or continue at a lower velocity. It depends on the masses.

What force would make it switch to moving right again?

Thanks.

So rebounding can sometimes mean, NOT bouncing back in the opposite direction, it can sometimes simply mean continuing at a lower velocity?

In this case has the first cart simply slowed down (its always been moving rightwards)?

 

[paulb203]

It simply decelerates.

If one adopts the center-of-momentum frame as @PeroK suggests, one can see that this deceleration is also a true rebound.

Thanks.
So is rebound relative? Observer in cart 1 sees cart 1 move rightwards, then continue to move rightwards but slower. Observer in truck 2 sees cart 1 coming up the rear at 5m/s but then rebounding off cart 2 (it appears like this to the observer in truck 2 because after the collision truck 2 speeds up rightwards, but truck 1 slows down)?

 

[paulb203]

“All you have to do is change reference frames to have the slower moving object at rest.”

If I changed cart 2 from 2m/s to 0m/s I would be subtracting 2m/s from its velocity. Do I then subtract 2m/s from cart 1? And then have cart 1 moving at 3m/s?

 

[jbriggs444]

So is rebound relative?

It depends on what you mean by "rebound" and what you mean by "relative". Those are just words. Words that are not even used in the problem description.

I consider "rebound" to accurately characterize an interaction where two objects get closer, have a repulsive interaction of some sort and then separate.

I consider "relative" to mean dependent on a choice of reference frame. By contrast, "invariant" would mean frame independent.

By those definitions, rebound is invariant, not relative.

 

[jbriggs444]

“All you have to do is change reference frames to have the slower moving object at rest.”

If I changed cart 2 from 2m/s to 0m/s I would be subtracting 2m/s from its velocity. Do I then subtract 2m/s from cart 1? And then have cart 1 moving at 3m/s?

Yes.

Instead of a 5 m/s cart running into the back of a 2 m/s cart, you would have a 3 m/s cart running into a stationary (0 m/s) cart.

The closing rate is 3 m/s either way.

 

[Lnewqban]

In this case has the first cart simply slowed down (its always been moving rightwards)?

If we consider both moving bodies as an isolated system having a common center of mass (which also moves at an average velocity), the total momentum of both bodies, pre and post-collision, should remain constant unless an external force is applied, according to the law of conservation of momentum.
Is it conserved in this case?

 

[haruspex]

I consider "rebound" to accurately characterize an interaction where two objects get closer, have a repulsive interaction of some sort and then separate.

Doesn’t that reduce its meaning to "bounce"?
If the word is to be useful, it would have to be in the context of a frame. In everyday meaning, object A strikes stationary object B and reverses direction in consequence.

 

[jbriggs444]

Doesn’t that reduce its meaning to "bounce"?

Sure, I consider "rebound" and "bounce" to be essentially synonymous.

Also, when discussing physics, I dislike being tied to a particular rest frame.

 

[paulb203]

No. The nature of collision (stick together or move apart) depends on the elastic properties of the objects and not on the ratio of the masses.

Thanks.
Why, in the various 'carts' scenarios do the two carts sometimes stick together, sometimes don't, if its down to the elastic properties of the carts and not the ratio of the masses (and/or the velocities)? I'm assuming we are to imagine the properties are the same in all of the various carts in all of the various examples.

 

[paulb203]

Yes.

Instead of a 5 m/s cart running into the back of a 2 m/s cart, you would have a 3 m/s cart running into a stationary (0 m/s) cart.

The closing rate is 3 m/s either way.

Thanks.
So now we have a 1kg cart at 3m/s rightwards running into a 15kg cart that is stationary.
Pi=Pf
Pi=3+zero
Pi=3

So, Pf=3

But how do we infer the velocities after the event?
In the orginal example we are told cart 1 is at 0.5m/s rightwards after the event, and cart 2 is at 2.3m/s rightwards after the event.
Do we subtract 2m/s from both of these?
That would give -1.5m/s and 0.3m/s

So, Pf=(1kg)(-1.5m/s)+(15kg)(0.3m/s)
=3kg.m/s

Which adds up.

 

[paulb203]

If we consider both moving bodies as an isolated system having a common center of mass (which also moves at an average velocity), the total momentum of both bodies, pre and post-collision, should remain constant unless an external force is applied, according to the law of conservation of momentum.
Is it conserved in this case?

Thanks.
By, 'having a common center of mass', do you mean we add the two masses, to give 16kg? We think of them as one vehicle?

 

[jbriggs444]

Thanks.
So now we have a 1kg cart at 3m/s rightwards running into a 15kg cart that is stationary.
Pi=Pf
Pi=3+zero
Pi=3

So, Pf=3

Yes. In the frame where the target cart is stationary, the total momentum is 3 kg m/s and remains so.

But how do we infer the velocities after the event?

In the orginal example we are told cart 1 is at 0.5m/s rightwards after the event, and cart 2 is at 2.3m/s rightwards after the event.
Do we subtract 2m/s from both of these?

Yes.

We were given velocities relative to the road's rest frame. The new frame is moving rightward at 2 m/s relative to this. So we subtract 2 m/s from all velocities to get them relative to the new frame instead of relative to the old.

That would give -1.5m/s and 0.3m/s

So, Pf=(1kg)(-1.5m/s)+(15kg)(0.3m/s)
=3kg.m/s

Which adds up.

Indeed, yes.

 

[nasu]

Thanks.
Why, in the various 'carts' scenarios do the two carts sometimes stick together, sometimes don't, if its down to the elastic properties of the carts and not the ratio of the masses (and/or the velocities)? I'm assuming we are to imagine the properties are the same in all of the various carts in all of the various examples.

The same two carts will always have the same type of collision (elastic, inelastic or totally inelastic) as long as the velocities are not so high as to break them to pieces or deform them over the elastic limit (in case of an elastic collision). But these are not situations you encounter in introductory physics examples. You are usually told what type of outcome to expect (move together or move separately) for a particular problem. Do you actually have an example where the nature of collision changes due to changes in initial velocities, without changing the nature if the colliding objects?

 

[PeroK]

Thanks.
Why, in the various 'carts' scenarios do the two carts sometimes stick together, sometimes don't, if its down to the elastic properties of the carts and not the ratio of the masses (and/or the velocities)? I'm assuming we are to imagine the properties are the same in all of the various carts in all of the various examples.

I must confess, I don't understand the thought processes that would lead to such a question.

 

[Steve4Physics]

@paulb203,I'd like to add to what @nasu said in Post #27.

Why, in the various 'carts' scenarios do the two carts sometimes stick together, sometimes don't, if its down to the elastic properties of the carts and not the ratio of the masses (and/or the velocities)?

For the simple examples used in teaching, sticking or not-sticking only depends on the carts' construction/materials. It does not depend on the carts' masses or velocities.

I'm assuming we are to imagine the properties are the same in all of the various carts in all of the various examples.

No. Some examples will involve carts that stick together; some examples will involve carts that separate after a collision (because they are made of different materials from the carts that stick together).

Minor edit.

 

[Lnewqban]

Thanks.
By, 'having a common center of mass', do you mean we add the two masses, to give 16kg? We think of them as one vehicle?

I just trying to use a different approach that could help you.

Not as one vehicle, but as a dynamic system formed by two masses, each moving at a different velocity.
The center of mass of that system has its own velocity.

As the masses do not change prior or after the impact, if the velocity of the center of mass of our system remains the same prior to anf after the impact, momentum of the system has been conserved.

Otherwise, the impact has taken away some kinetic energy from both masses, reducing the momentum of the system.

 

[haruspex]

I'm assuming we are to imagine the properties are the same in all of the various carts in all of the various examples.

Then don’t.
In the real (macroscale) world, nothing is perfectly elastic or perfectly inelastic. Even putty is a tiny bit elastic, but it is also sticky, so does tend to coalesce.
In physics problems, try not to assume anything about elasticity, but sometimes you are not given that information and are expected to make a reasonable assumption given the materials involved, like putty or billiard balls.

 

[paulb203]

The same two carts will always have the same type of collision (elastic, inelastic or totally inelastic) as long as the velocities are not so high as to break them to pieces or deform them over the elastic limit (in case of an elastic collision). But these are not situations you encounter in introductory physics examples. You are usually told what type of outcome to expect (move together or move separately) for a particular problem. Do you actually have an example where the nature of collision changes due to changes in initial velocities, without changing the nature if the colliding objects?

Thanks.
No, I don't have an example. I was incorrectly under the impression, from the outset, that all the carts were made of the same stuff, and that the sticking together or not was down to mass and/or velocities. I'm getting the hang now of doing the maths, of getting the correct answer, but will have to work on understanding what's actually going on :). Elastic and inelastic collisions haven't came up, as such, yet, in the course; I look forward to it.

 

[paulb203]

Thanks, guys. Really helfpul, as ever :)