# Confused by Newtons Laws

1. Feb 9, 2004

### adamg

Could anyone please help me with the following question? A rocket rises steadily upwards at a constant speed of 500 metres per second. Its initial mass is 5 x 10^4 kg; after 10 mins, this has decreased to 4.4 x 10^4 kg. Calculate the average force acting on the rocket during this time. This force does work on the rocket. Explain how this changes the rocket's energy.

I am confused because a changing momentum means an unbalanced force is acting down, yet the rocket has constant velocity so surely the forces acting on the rocket are balanced??? Hope someone can help, thank you. *adam*

2. Feb 9, 2004

### ZapperZ

Staff Emeritus
It would be helpful if you give a bit of a background on the level you are at (and this applies to everyone who posts school questions on here). For example, do you know a bit of calculus?

If you do, then assuming that this is a standard intro physics course, you should have known that a force is generally defined as the rate of change of momentum (which you have mentioned), but if you carry that through, you would have noticed an extra component to it, i.e.

F = dp/dt = d/dt (mv) = m dv/dt + v dm/dt

Now, in many instances, m is a constant in time, so dm/dt=0 and you only have

F = m dv/dt = ma

which is the familiar form of Newton's 2nd law. However, in YOUR case, m isn't a constant with time, only v is.... So dv/dt=0 and you have to rederive the force expression, i.e.

F = v dm/dt.

You are told how much the mass has changed in a certain time interval. Assume that this rate of change is uniform, you then know what dm/dt is, i.e. time rate of change of the mass. You then have enough information to find the force acting on the rocket.

Zz.

3. Feb 9, 2004

### pmb_phy

There are two different forces at work here. There is the force of gravity and there is the force on the rocket due to the exhaust. The total force on the rocket is zero since the rocket is not accelerating. However the gravitational force is doing work on the rocket and that is equal to force times distance integrated over the distance traveled. The gravitational force is equal to the weight of the rocket and that weight is decreasing as a function of time.

That should get you started

4. Feb 9, 2004

### HallsofIvy

Staff Emeritus
Another important point is that force is not "mass times accelration" in this problem. More generally, force= "rate of change of momentum". Since momentum= mass*velocity, if mass is a constant, then this is the same as "mass*rate of change of velocity" = "mass times acceleration". However, here, it is velocity that is constant, not mass. In this problem force= "rate of change of mass"*velocity".

5. Feb 10, 2004

### adamg

Ok, so i work this through and the force acting on the rocket comes out as -5000N. But my confusion is whether this is an unbalanced force of 5000N down, or is it balanced by a force of 5000N up? Because it must be unbalanced to cause the change in momentum(?) yet it must also be balanced because the velocity is constant(?). (In reply to ZZ, i am currently studying A2 phyiscs and maths, so yes i have met calculus. Thanks for the help)

6. Feb 10, 2004

### HallsofIvy

Staff Emeritus
The rocket was initially 50000kg and moving at 500 m/s so its initial momentum was 25000000 kgm/s. 10 minutes later(= 600 seconds later), its mass was 44000kg and it was still moving at 500 m/s so its final momentum was 22000000 kgm/s, a reduction of 3000000 kgm/s. Since it did that in 600 seconds, that was an average force of 5000 Newtons (downward) just as you say.
That is a net force: yes it unbalanced "Because it must be unbalanced to cause the change in momentum(?) " just as you say.
However, (once again!) "yet it must also be balanced because the velocity is constant(?)" is NOT true! That's only true when mass is a constant so that momentum is proportional to velocity. Force changes momentum (mass times velocity), not just velocity.

7. Feb 10, 2004

### adamg

Thats great, thanks for the help. I suspected that was the case. I just thought that it sort of contradicted Newtons first law (when there is no mention of mass needing to be constant) and I didnt feel in a position to do that. So Newtons first law assumes mass to be constant, right.

8. Feb 11, 2004

### HallsofIvy

Staff Emeritus
Newton's first law is "Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it." Which doesn't say anything about mass because it doesn't have to. It doesn't say anything about what happens if there is a force on the object.

Newton's second law is often written as "f= ma". Stated that way it is incorrect. What Newton actually said was that "force is the rate of change of momentum"- that is, f= dp/dt which is, in fact, a perfectly good definition of force.

9. Feb 12, 2004

### adamg

Ok, so imagine an object of mass 50kg moving through space at 20ms-1, there is no force acting on it.
Over time, it disintegrates, and in 100 seconds, its mass has fallen to 45kg.

Because the mass has decreased, does that mean a force must be acting? What will happen to the object - will its velocity increase so that mv remains constant, or will it remain at the same speed, and therefore mv is reduced?

10. Feb 13, 2004

### adamg

Thinking about this I have a suggestion. Every time a particle is emitted from the object, it exerts on equal and opposite force on the object. The cumulative effect of all these forces gives the total force. The size of this force will correspond to the change in the objects momentum. This change, however, could be anything because it depends on aspects such as the direction the particles are emitted in and the velocity they have. A force must be acting if there is a loss in mass, because if no force was acting, all the particles would surely just stay together and not move apart, hence maintaining the mass.

Is this anywhere near the actual answer? Please correct me if i'm wrong.

11. Feb 14, 2004

### HallsofIvy

Staff Emeritus
No, since there is no external force, the momentum must stay the same. The velocity increases as m decreases so that mv remains constant.
With no external force, d(mv)/dt= mdv/dt+ vdm/dt= 0.
The body has lost mass (actually, we ought to account for the momentum of every part that it loses but lets just assume it just disappears) of 5 kg in 100 seconds. Assuming that is a uniform loss, m(t)= 50- .05 t where m is measured in kg and t in seconds.
Now, momentum mv= (50- 0.05t)v(t) so d(mv)/dt= (50- 0.05t)dv/dt- 0.05v= 0. We can solve that differential equation for v:
It's separable. (50- 0.05t)dv/dt= 0.05v => dv/v= (0.05/(50-0.05t))dt. Integrating dv/v given ln(v). To integrate 0.05dt/(50-0.05t)dt, let u= 50- 0.05t so du= -0.05dt. The integral becomes -du/u which is -ln(u). We must have ln(v)= -ln(50-0.05t)+ C
or v= C/(50-0.05t). If the original velocity at t= 0 was v0 then v(t)= 50v0/(50-0.05t). That increases as t increases. In fact, there is a vertical asymptote at t= 50/(0.05)= 1000 sec. That is, of course, because at that point all the mass would have disappeared. (All of this is classical- it gets a lot harder if we take into account the increase of mass with velocity!)

12. Feb 16, 2004

### adamg

So in Newtons first law, does uniform motion refer to constant momentum? I was taught it meant constant velocity in a straight line, but im guessing this is a bit basic. In the example I used above, there was no external force, yet velocity was changing. That means uniform motion must refer to momentum and not just velocity doesnt it?

13. Feb 16, 2004

### HallsofIvy

Staff Emeritus
Yes!

14. Feb 16, 2004

### adamg

(phew) thanks

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