Explaining Newton's First Law & the Dropped Weight Trolley

[pkc111]

I read the following question in a Yr11 High School Physics Test.

A trolley is traveling along at unifrom velocity when a weight is dropped onto it which stays there.

The velocity of the trolley will...

A) Decrease
B) Increase
C) Stay the same

?

The answer says "decrease" because of law of conservation of momentum.

But this doesn't seem right to me because it seems to disobey Newtons first Law which states that an object won't change its motion unless it is acted upon by a net force. Where is the net force??

Can anyone help explain?

Thanks

 

[Hootenanny]

Do you know the mathematical equation which represents Newton's first law?

~H

 

[pkc111]

Do you know the mathematical equation which represents Newton's first law?

~H

Well sorta...

F(net)=ma

describes both Newton's 1st and 2nd laws. Right? How does this answer the question?

 

[Hootenanny]

Well sorta...

F(net)=ma

describes both Newton's 1st and 2nd laws. Right? How does this answer the question?

You are correct, m represents mass in Newton's second law. What happens to the mass of the object (in this case a trolley) if a weight is dropped into it.

~H

 

[pkc111]

You are correct, m represents mass. What happens to the mass of the object (in this case a trolley) if a weight is dropped into it.

~H

m increases,
but a=0 (because a=F/m, F=0).

How does this show that the trolley decelerates exactly?

 

[Hootenanny]

As your in year 11, I'll assume you have done anything on friction (F = \mu mg) for example. Think of it this way, if your pushing a trolley around tesco, when the trolley is full, you need to push harder than if it was empty to keep it going at a constane speed.

~H

 

[pkc111]

As your in year 11, I'll assume you have done anything on friction (F = \mu mg) for example. Think of it this way, if your pushing a trolley around tesco, when the trolley is full, you need to push harder than if it was empty to keep it going at a constane speed.

~H

So conservation of momentum doesn't have anything to do with it?

What if it was on a frictionless surface and there was no force from friction, would conservation of momentum be disobeyed?

 

[Hootenanny]

So conservation of momentum doesn't have anything to do with it?

What if it was on a frictionless surface and there was no force from friction, would conservation of momentum be disobeyed?

Yes, conservation of momentum has everything to do with it; I apologise, its been a while since High school and I can't remember whether momentum eas covered at GCSE. (And thinking about Newton's laws through me a bit ). Can you write an equation for momentum representing this collision?

~H

 

[pkc111]

Law of Conservation of Momentum:
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

or,

Sum(m1v1) = Sum(m2v2)


But the system in the question not an isolated system. Trolleys run on surfaces, and surfaces apply reaction forces.

 

[Hootenanny]

But the system in the question not an isolated system. Trolleys run on surfaces, and surfaces apply reaction forces.

Yes but going back to the question, "A trolley is traveling along at unifrom velocity", this means there are no net external forces and therefore, momentum is conserved.

~H

 

[pkc111]

Ok so if conservation is suppose to apply then how is momentum conserved if the trolley slows down?

When you add m1v1's (of trolley and dropping mass) you get a vector sum which is in a diagonal direction. However the final monetum in the trolley example is horizontal (appearing to disobey the conservation law).

 

[Hootenanny]

Just consider the mometum in the x direction (horizontal plane). Also, initally before it is dropped, the weight is at rest, and hence has no momentum.

~H

 

[pkc111]

Just consider the mometum in the x direction (horizontal plane).

~H

This comes back to the earlier question...

If the experiment were done on a frictionless surface, where would the net force come from to create the deceleration of the trolley ?

This is assuming that deceleration does occur in accordance with your explanation using the law of conservation of momentum in the x-direction?

 

[arildno]

Remember that the trolley accelerates the falling mass up to its own horizontal velocity so that they become co-moving in the horizontal direction. Hence, by Newton's third law, the falling mass exerts a retarding force on the trolley of equal magnitude.
Thus, the resultant common velocity is lower than the trolley's original one.

This, of course, is in accordance with conservation of momentum.

 

[Hootenanny]

Conservation of momentum is implied in by Newton's first law. Newton's second law as F = ma only applies for constant mass. When mass is not constant you need to apply Newton's law in its differential form;

F = \frac{dp}{dt}

As, momentum is conserved dp = 0 and hence F = 0, thus no net external forces act.

~H

 

[arildno]

This comes back to the earlier question...

If the experiment were done on a frictionless surface, where would the net force come from to create the deceleration of the trolley ?

This is assuming that deceleration does occur in accordance with your explanation using the law of conservation of momentum in the x-direction?

If the surface of the trolley were frictionless, then the falling mass wouldn't attain the trolley's velocity by any means, i.e, would eventually slide off it. No change would be observed in the trolly's velocity.

 

[arildno]

Conservation of momentum is implied in by Newton's first law. Newton's second law as F = ma only applies for constant mass. When mass is not constant you need to apply Newton's law in its differential form;

F = \frac{dp}{dt}

As, momentum is conserved dp = 0 and hence F = 0, thus no net external forces act.

~H

There doesn't exist a single example in classical mechanics where a material system's mass change over time. Conservation of mass holds absolutely.

For geometric systems (which does not contain the same material particles over time) , the correct version of Newton's 2.law is:
F=dp/dt+M, where dp/dt is the rate of change of momentum inside the system and M is the momentum flux out of the system.
F is the net external force acting upon the coinciding material system.

 

[pkc111]

Newton's second law as F = ma only applies for constant mass.

I haven't heard of this before, is there a reference?

 

[pkc111]

If the surface of the trolley were frictionless, then the falling mass wouldn't attain the trolley's velocity by any means, i.e, would eventually slide off it. No change would be observed in the trolly's velocity.

Thank you very much arildno, that explains a lot.

 

[Hootenanny]

For geometric systems (which does not contain the same material particles over time) , the correct version of Newton's 2.law is:
F=dp/dt+M, where dp/dt is the rate of change of momentum inside the system and M is the momentum flux out of the system.
F is the net external force acting upon the coinciding material system.

Thank's for that aldrino, haven't seen that before. I'm goin to look it up now.

~H

 

[Hootenanny]

Thank you very much arildno, that explains a lot.

See here
http://en.wikipedia.org/wiki/Newton's_laws_of_motion

under the heading "Newton's second law: law of motion".

~H

 

[arildno]

Thank's for that aldrino, haven't seen that before. I'm goin to look it up now.

~H

I've made a tutorial on it here:
https://www.physicsforums.com/showthread.php?t=72176

 

[pkc111]

Thanks guys, you are champions! xxx :!)

 

[pkc111]

Another question on the topic of conservation of momentum if I may...

If a tennis ball hits a brick wall does conservation of momentum apply?
Similarly if a tennis ball is thrown which is made of plasticine and sticks to the wall does the conservation law apply?

Im guessing no, because its not a free system?

 

[Hootenanny]

I've made a tutorial on it here:
https://www.physicsforums.com/showthread.php?t=72176

Well well, arildno. I've just finished (briefly) reading your tutorial and I am very impressed, my compliments.

~H

 

[Hootenanny]

Another question on the topic of conservation of momentum if I may...

If a tennis ball hits a brick wall does conservation of momentum apply?
Similarly if a tennis ball is thrown which is made of plasticine and sticks to the wall does the conservation law apply?

Im guessing no, because its not a free system?

In reality, yes you are correct, because the tennis ball is being acted on by unbalanced external forces (gravity and air resistance for example), momentum would not be conserved, however, for simplicty when modelling such collisions it is assumed that the system is not acted on by any external forces.

~H

 

[pkc111]

In reality, yes you are correct, because the tennis ball is being acted on by unbalanced external forces (gravity and air resistance for example), momentum would not be conserved, however, for simplicty when modelling such collisions it is assumed that the system is not acted on by any external forces.

~H

I was thinking more of the wall ? Isnt it being held by external forces??

 

[arildno]

Momentum is definitely NOT conserved, neither for the tennis ball system, or the tennis ball+brick wall system.

In both cases, non-negligible external forces acts upon the system:
1. In the tennis ball system, there is a huge external force acting upon it from the brick wall, effectively REVERSING, not conserving, the ball's momentum.

2. In the wall+ball system, the force from the ground keeps the wall at rest.
This force must balance the ball's huge collision force upon the wall, and is therefore not negligible to the magnitude of the internal forces in the collision.
Hence, momentum conservation does not apply here either.

ENERGY, however, might well be conserved; in that case, we have an elastic collision.

 

[pkc111]

You don't need to consider the external forces, only the unbalanced / net external forces. If any object experiences external forces, but they are balanced, for example an objects weight and the normal reaction force when the object is on a table, there is no net external force, so it is as if no external forces act.

~H

OK, so wouldn't a loosely held wall (exaggerated example - on ice) react differently to a tightly held wall ?

 

[arildno]

OK, so wouldn't a loosely held wall (exaggerated example - on ice) react differently to a tightly held wall ?

Then, you WOULD have momentum conservation.

 

[Hootenanny]

OK, so wouldn't a loosely held wall (exaggerated example - on ice) react differently to a tightly held wall ?

My explanation isn't phrased correctly, arildino's is phrased much better, I'm going to have to work on my explanations.

~H

 

[pkc111]

Momentum is definitely NOT conserved, neither for the tennis ball system, or the tennis ball+brick wall system.

In both cases, non-negligible external forces acts upon the system:
1. In the tennis ball system, there is a huge external force acting upon it from the brick wall, effectively REVERSING, not conserving, the ball's momentum.

2. In the wall+ball system, the force from the ground keeps the wall at rest.
This force must balance the ball's huge collision force upon the wall, and is therefore not negligible to the magnitude of the internal forces in the collision.
Hence, momentum conservation does not apply here either.

ENERGY, however, might well be conserved; in that case, we have an elastic collision.

Thanks arildno, that makes sense.

I guess if you think of a fixed wall as an extension of the Earth then momentum may be conserved in the tennis ball + wall/earth system but the resulting velocity of the wall/earth would only be undetectably small for conservation to occur ?

 

[arildno]

In the ball+EARTH system, momentum is, indeed, conserved (neglecting any external forces from stars and such).

 

[Hootenanny]

Thanks arildno, that makes sense.

I guess if you think of a fixed wall as an extension of the Earth then momentum may be conserved in the tennis ball + wall/earth system but the resulting velocity of the wall/earth would only be undetectably small for conservation to occur ?

Yes, if you consider the Earth as part of the system, then you are correct, but as you stated only the tenis ball and the wall is part of the 'system', hence momentum for that system is not conserved.

~H

Edit: arildno has got there before me

 

[pkc111]

Thanks again guys, nite nite.

 

[arildno]

As a note, in the elastic collision between an object A (approaching B with velocity V) and B (at rest initially) where external forces are negligible, the final velocities are:
For Object A: (m-M)/(M+m)V, For B: 2mV/(M+m)
where m is the mass of A and M the mass of B.

Thus, if M>>m, Object A will effectively reverse its velocity, whereas B remains practically at rest.