Is the Force Acting on this Rocket Balanced or Unbalanced?

In summary: If mass is changing, the velocity will change even if there isn't any external force. If the external force is zero, the velocity will remain constant.
  • #1
adamg
48
0
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*
 
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  • #2
Originally posted by 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*

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
Originally posted by 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*

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
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
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
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
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
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
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
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
Originally posted by 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?

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 let's 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
So in Newtons first law, does uniform motion refer to constant momentum? I was taught it meant constant velocity in a straight line, but I am 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 doesn't it?
 
  • #13
Yes!
 
  • #14
(phew) thanks
 

1. What are Newton's Laws of Motion?

Newton's Laws of Motion are three fundamental principles that explain the behavior of objects in motion. They were developed by Sir Isaac Newton in the 17th century and are considered the basis of classical mechanics.

2. What is the First Law of Motion?

The First Law of Motion, also known as the Law of Inertia, states that an object at rest will remain at rest and an object in motion will continue in motion at a constant velocity unless acted upon by an external force.

3. What is the Second Law of Motion?

The Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In simpler terms, the greater the force applied to an object, the greater its acceleration will be, and the more massive an object is, the less it will accelerate.

4. What is the Third Law of Motion?

The Third Law of Motion, also known as the Law of Action and Reaction, states that for every action, there is an equal and opposite reaction. This means that when a force is applied to an object, the object will exert an equal force in the opposite direction.

5. How do Newton's Laws apply to everyday life?

Newton's Laws of Motion can be seen in many everyday situations, such as riding a bike, driving a car, or playing sports. They also play a crucial role in engineering, space travel, and other scientific fields. Understanding these laws allows us to predict and control the motion of objects in our daily lives.

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