Unlocking the Mystery of Kinetic Energy

[digalumps]

Homework Statement

Question about kinetic energy

Relevant Equations

KE=1/2 mv squared

I have a problem regarding Kinetic Energy which as we know is 1/2 m v squared.

Say I have a 1kg mass moving at 10 meters/second. I have a 1 Newton rocket which I attach to the back and it burns for 1 second accelerating the mass by 1 m/sec/sec to 11 m/sec. The KE originally was 50 joules and it has now increased to 60.5 joules.

Now say I have the same mass moving at 1000 m/sec and I do the same thing. It accelerates to 1001 m/sec. The KE goes from originally 1000 squared to 1001 squared ie an increase of 2001 joules - a much bigger increase.

It is the same rocket with the same amount of gunpowder so where has the 'extra' energy come from?

I know in the second case the distance moved in the 1 second of rocket burning is larger, and work=force x distance moved so equation-wise I can see but not intuitively as it is the same rocket with the same amount of input energy.

 

[jbriggs444]

Have you considered the entire closed system?

 

[digalumps]

I think so - the work done by the 1 Newton rocket is used to increase the kinetic energy.

The work done is the same in both scenarios (the rocket burns for 1 second and accelerates the mass) but the KE increases are not the same.

 

[robphy]

Possibly useful:
[itex] K=\frac{1}{2}mv^2 [/itex] implies
[itex] \Delta K=\frac{1}{2}m (v_f^2 -v_i^2)=\frac{1}{2}m(v_f+v_i)(v_f-v_i)[/itex]
One way to factor this is
[itex]\Delta K=\displaystyle\frac{m(v_f+v_i)}{2} \Delta v \approx p_{avg}\Delta v [/itex] (*assuming a constant force during the interval)
but this might be more intuitive
[itex]\Delta K=\displaystyle\frac{(v_f+v_i)}{2} \Delta (mv) \approx v_{avg}\Delta p [/itex] (*)
since [itex] dK= v\ dp [/itex] is a more natural definition of the increment of kinetic energy.

Can you explicitly calculate the work done by [itex]F_{net}[/itex] as [itex] \int F_{net}\ dx [/itex]?
(It may help to think about the area under the [itex] F_{net} [/itex] vs [itex] x [/itex] graph.)

 

[digalumps]

Thank you for your help and I understand the equations above.

I think what I don't understand in the intuitive bit - the thought experiment if you will.

I accelerate the same mass by the same amount for the same time (1 sec) and get different KE increase depending on the initial velocity.

The SAME rocket with the SAME weight of gunpowder in it (enough to burn for 1 second and produce a 1 Newton force) attached to the SAME mass produces the same acceleration but different increases in KE. How?

The rocket burns. In case 1 v goes from 10 to 11 m/sec and in case 2 v goes from 1000 to 1001 m/sec.

Where does the 'extra' KE come from?

 

[robphy]

Thank you for your help and I understand the equations above.

I think what I don't understand in the intuitive bit - the thought experiment if you will.

I accelerate the same mass by the same amount for the same time (1 sec) and get different KE increase depending on the initial velocity.

Yes, that's what [itex] \Delta K =v_{avg}\Delta p [/itex] suggests.

Throughout let's assume that the gunpowder mass is negligible compared to the rocket mass.

The SAME rocket with the SAME weight of gunpowder in it (enough to burn for 1 second and produce a 1 Newton force) attached to the SAME mass produces the same acceleration but different increases in KE. How?

A constant force of 1 N applied for 1 sec results in an impluse [itex] \Delta p= F\Delta t [/itex].
So, you applied the same impulse (so, [itex] (\Delta p)_A=(\Delta p)_B=(1 N\cdot s) [/itex])
when [itex] v_A=10\rm\ m/s [/itex] and when [itex] v_B=1000\rm\ m/s [/itex].

The change in kinetic energy is equal to the
net work done [itex]\Delta K= F_{net} dx= \displaystyle\frac{d p}{d t} dx = v\ dp [/itex]

The rocket burns. In case 1 v goes from 10 to 11 m/sec and in case 2 v goes from 1000 to 1001 m/sec.

Where does the 'extra' KE come from?

As you said in your first post,

I know in the second case the distance moved in the 1 second of rocket burning is larger, and work=force x distance moved so equation-wise I can see but not intuitively as it is the same rocket with the same amount of input energy.

This is correct because the displacement [itex] dx_A [/itex] traveled during the first impulse
is less than the displacement [itex] dx_B [/itex] traveled during the second (equal to the first) impulse.

Here's a desmos visualization to see how the work done increases for equal impulses
because the displacement increases for the same elapsed 1-sec interval.

https://www.desmos.com/calculator/w9z3m57udi
This is a net-force vs displacement plot.
The net-force is constant.
The dots mark the displacement at t=0s, 1s, 2s, etc.. (same [itex] \Delta t=1 s [/itex]).
The work done during each interval (= the change in the kinetic energy)
is the area under the graph.
Note how successive impulses are accompanied by increasing work-done due to the increasing-displacements.

So, the extra KE comes from the increasing [itex] \Delta x [/itex] needed to obtain the same impulse [itex] \Delta p [/itex].

[I should have encoded [itex] F_{net}=ma [/itex] (by using the a-slider for F) in the desmos visualization.]

 

[jack action]

It is the same rocket with the same amount of gunpowder so where has the 'extra' energy come from?

[...] as it is the same rocket with the same amount of input energy.

But is it?

Say your force is produced by combustion of a fuel. The fuel has a constant heat of combustion with respect to the mass of fuel ##m_f## burned (heating value, ##HV##). Assuming 100% efficiency, the energy released by the heat is equal to the work done, so:
$$F\Delta x = m_f HV$$
Or:
$$m_f = \frac{F}{HV}\Delta x$$
So the quantity of fuel burned is proportional to the distance traveled, not the time interval it burned.

To visualize the amount of energy it takes at higher velocities, the concept of power may also be helpful:
$$P = F\frac{\Delta x}{\Delta t} = Fv$$
So you need more power to produce the same force during the same period of time while traveling a longer distance, i.e. at a larger velocity.

Producing 1 N at 10 m/s requires an engine that produces 10 W of power. At 1000 m/s, the engine needs to produce 1000 W of power.

 

[TSny]

According to @jbriggs444, we should look at the entire system. If something exerts a force on the rocket, then the rocket exerts a force back on the something. We need to look at the whole system.

So, consider a general situation where two objects interact.

On the left, we are in an inertial reference frame where the two masses are initially at rest. The red spot between the masses indicates some mechanism such as a spring or explosive device that can propel the masses apart. The final speeds of the masses in this frame are ##v_1## and ##v_2##. Assume that the system of two objects is isolated.

Now look at the same thing, but viewed in a different inertial frame of reference such that initially, the masses are moving to the right with speed ##u##. Fill in the question marks with the final speeds of each mass in terms of ##u##, ##v_1##, and ##v_2##, where ##v_1## and ##v_2## still denote the speeds in the first reference frame. The picture assumes that ##u## is large enough that ##m_1## is still moving to the right after the interaction. But, that's not essential.

Compare the change in total KE of the system for the two reference frames.

 

[digalumps]

Hi everyone thank you for taking the time and trouble to reply. I still do not really understand but perhaps it is because I am just starting out in Physics maybe one day I will.

My rocket produces F and thus accelerates my mass by a=F/m (F=ma)

From 10->11 m/sec or 1000->1001 m/sec

If I were trying to explain this to a 12 yr old without equations I would say the KE has increased by different amounts for the same input energy ie the same rocket burn for 1 second.

 

[PeroK]

Hi everyone thank you for taking the time and trouble to reply. I still do not really understand but perhaps it is because I am just starting out in Physics maybe one day I will.

My rocket produces F and thus accelerates my mass by a=F/m (F=ma)

From 10->11 m/sec or 1000->1001 m/sec

If I were trying to explain this to a 12 yr old without equations I would say the KE has increased by different amounts for the same input energy ie the same rocket burn for 1 second.

Did you look at post #8? The rocket must have some propulsion mechanism. An object, by Newton's third law, cannot provide an accelerating force on itself. In the case of a rocket, almost the only mechanism available is to "throw stuff out the back" - and, hence, rely on conservation of momentum to provide the acceleration for what remains of the rocket (minus the expellant).

When you consider conservation of momentum for the rocket and expellant you find that the change in total KE (of the system) is the same in all reference frames. It's a good exercise to work this out.

 

[archaic]

The work done cares about the distance traveled. In both scenarios, the net force on the block is ##F=1\mathrm N##.
First scenario:
We apply the force for a duration of one second. We have ##v(t)=v_0+at##.
$$\Delta x = \int_0^1v(t)\,dt=\int_0^1 v_0\,dt+\int_0^1at\,dt=v_0+\frac a2=10.5\mathrm\,m$$$$\begin{align*}
W&=\int_0^{10.5}Fdx(=10.5\mathrm\,J)\text{, let $x=x(t)$, then $dx=v(t)dt$}\\
&=\int_0^1Fv(t)dt=10.5\mathrm\,J
\end{align*}$$
Second scenario:
We want to find the time it takes the block to move ##10.5\mathrm\,m##.
Frankly, I can just redo the ##\int_0^{10.5}dx## integration, since work depends on distance not time, but I'll transform it to time for the sake of argument.
$$\Delta x = \int_0^\tau v(t)\,dt=\int_0^\tau v_0\,dt+\int_0^\tau at\,dt=v_0\tau+\frac a2\tau^2 = 10.5$$$$
\tau=-v_0+\sqrt{v_0^2+21}$$$$\begin{align*}W&=\int_0^\tau Fv(t)\,dt\\
&=v_0\tau+\frac{\tau^2}{2}\\
&=-v_0^2+v_0\sqrt{v_0^2+21}+\frac{2v_0^2+21-2v_0\sqrt{v_0^2+21}}{2}\\
&=\frac{21}{2}=10.5\mathrm\,J
\end{align*}$$

Same change in KE in both scenarios, you just need to notice that the work is originally defined using integration over distance, not time.

If you are curious, you can find the distance the block moves in one second in the second scenario, then plug it into the work formula for the first scenario.

 

[jbriggs444]

The work done cares about the distance traveled. In both scenarios, the net force on the block is ##F=1\mathrm N##.
First scenario:
We apply the force for a duration of one second. We have ##v(t)=v_0+at##.
$$\Delta x = \int_0^1v(t)\,dt=\int_0^1 v_0\,dt+\int_0^1at\,dt=v_0+\frac a2=10.5\mathrm\,m$$$$\begin{align*}
W&=\int_0^{10.5}Fdx(=10.5\mathrm\,J)\text{, let $x=x(t)$, then $dx=v(t)dt$}\\
&=\int_0^1Fv(t)dt=10.5\mathrm\,J
\end{align*}$$
Second scenario:
We want to find the time it takes the block to move ##10.5\mathrm\,m##.
Frankly, I can just redo the ##\int_0^{10.5}dx## integration, since work depends on distance not time, but I'll transform it to time for the sake of argument.
$$\Delta x = \int_0^\tau v(t)\,dt=\int_0^\tau v_0\,dt+\int_0^\tau at\,dt=v_0\tau+\frac a2\tau^2 = 10.5$$$$
\tau=-v_0+\sqrt{v_0^2+21}$$$$\begin{align*}W&=\int_0^\tau Fv(t)\,dt\\
&=v_0\tau+\frac{\tau^2}{2}\\
&=-v_0^2+v_0\sqrt{v_0^2+21}+\frac{2v_0^2+21-2v_0\sqrt{v_0^2+21}}{2}\\
&=\frac{21}{2}=10.5\mathrm\,J
\end{align*}$$

Same change in KE in both scenarios, you just need to notice that the work is originally defined using integration over distance, not time.

While true, this does not address the conundrum posed by the original poster. He knows quite well that there is more energy gained by the rocket in the one case than the other. But the energy source is the same in both cases.

You have one tank of gas. One source of energy. You can gain a little kinetic energy with it or a lot. On the face of it, that would be a problem for conservation of energy.

The answer, as @TSny, @PeroK and myself have pointed out is that you need to consider the kinetic energy in the exhaust plume that you are leaving behind. The proper way to convince yourself of that is to actually do the exercise that @TSny set in #8.

 

[jack action]

Knowing that the initial statement is:

Say I have a 1kg mass moving at 10 meters/second. I have a 1 Newton rocket which I attach to the back and it burns for 1 second

[...]

Now say I have the same mass moving at 1000 m/sec and I do the same thing.

Am I the only one who thinks all the followings are false:

It is the same rocket with the same amount of gunpowder

The work done is the same in both scenarios

The SAME rocket with the SAME weight of gunpowder in it

I would say the KE has increased by different amounts for the same input energy ie the same rocket burn for 1 second.

i.e. you cannot burn the same amount of gunpowder and obtain the same force AND the same burning time in both cases?

Or am I wrong and misunderstand something?

 

[digalumps]

Hi - I did not say the work was the same as work=force x distance and the distance is more in the second case.

I just could not understand intuitively how if you attach the same rocket to the same mass and burn it for eg 1 second in the two cases and get an acceleration the same (f=ma - nothing about initial velocity) then the KE energy increase is different.

I am thinking about what the other posters say about the rocket expellent but surely that is also the same in both cases?

I am obviously missing something very fundamental here.

In space exploration (ignore gravity, friction etc), you launch a rocket from Earth and the speed increases and fuel is used and the rocket KE increases. However, the fuel energy required is not thousands times less for the rocket to go from 101 ->101 km/hr as it takes of compared with eg 10000 -> 10001 km/hr later in the flight. The rocket KE energy increase can only come from the fuel burn, nowhere else.

 

[PeroK]

I am obviously missing something very fundamental here.

1) Conservation of momentum.

2) The importance of doing calculations. Calculations always trump speculation.

 

[gleem]

The masses receives an impulse of 1kg⋅m/sec. Impulse = Δp = 1 no matter their initial momentum each masses momentum is increased by 1 kg⋅m/sec.

I = F⋅t = F⋅s/v --> work/v --> work = I⋅v where v is the average velocity during the acceleration.

If the mass is at rest the rocket would only gain an energy of 0.5J. If you apply 1 Nt to a 1 kg mass for 1 second the acceleration is 1m/sec² moving the mass 0.5 m.

Perhaps a flaw in the logic in the OP is that a rocket can produce the same Δp with the same amount of fuel in both situations if you believe in the conservation of energy. Does anybody see a problem?

 

[jbriggs444]

I = F⋅t = F⋅s/v --> work/v --> work = I⋅v where v is the average velocity during the acceleration.

That's one part of the work done by the expanding gunpower. Now write down the equation for the other part.

If you have a closed system, there has to be another part.
If you have an open system, energy does not have to be conserved.

 

[Lnewqban]

The masses receives an impulse of 1kg⋅m/sec. Impulse = Δp = 1 no matter their initial momentum each masses momentum is increased by 1 kg⋅m/sec.
...
Perhaps a flaw in the logic in the OP is that a rocket can produce the same Δp with the same amount of fuel in both situations if you believe in the conservation of energy. Does anybody see a problem?

...
Now say I have the same mass moving at 1000 m/sec and I do the same thing. It accelerates to 1001 m/sec. The KE goes from originally 1000 squared to 1001 squared ie an increase of 2001 joules - a much bigger increase.
...

##{\Delta{v^2}\neq{v^2}}##

 

[etotheipi]

And once the OP has puzzled this one out, it might be interesting to consider a little corollary...

Suppose a race car is at rest at the start line of a straight race track. Someone watching in the stands sees the race car accelerate from ##0 \text{ms}^{-1}## to ##10 \text{ms}^{-1}## in a time ##t##. Someone in a helicopter moving at ##10\text{ms}^{-1}## w.r.t. the Earth's surface, in the opposite direction to the car, sees the car accelerate from ##10\text{ms}^{-1}## to ##20\text{ms}^{-1}## in the same time ##t##.

They must both agree on how much petrol has been consumed. How do you resolve the difference in gain of kinetic energies? As you might guess, the answer is to do with the Earth...

 

[digalumps]

First of all a big thank you for everyone who posts on here it is incredibly kind of you to give up your time this way. I mean that.

Also, although there have been many 'complicated' answers including calculus, advice to run the equations etc this is not pure mathematics or string theory where asking 'what is really going on' is a bit meaningless - this is classical mechanics.

There must be a paragraph answer to the question that could be understood by a 12 yr old.

If I stand on a car running on a frictionless rail at 10 km/sec and throw a bowling ball off the end it accelerates to 11 m/sec and the KE increases correspondingly.

If i throw the bowling ball off the end when the car is traveling at 1000 km/sec if accelerates to 1001 m/sec and the KE increases by a lot more. Why?

 

[Lnewqban]

...
If I stand on a car running on a frictionless rail at 10 km/sec and throw a bowling ball off the end it accelerates to 11 m/sec and the KE increases correspondingly.

If i throw the bowling ball off the end when the car is traveling at 1000 km/sec if accelerates to 1001 m/sec and the KE increases by a lot more. Why?

It doesn't.
Your arm only gave the ball enough impulse to accelerate ##1 m/s^2##

 

[digalumps]

Why doesn't it? The same force is applied and therefore the same acceleration? a=f/m.

 

[PeroK]

There must be a paragraph answer to the question that could be understood by a 12 yr old.

Why must there be? I didn't understand any of this when I was 12. At some point in your studies you will have to move past what any 12-year-old can understand.

 

[kuruman]

If I understand OP's question correctly, it can be recast as follows: If a fixed amount of energy ##E## is added to mass ##m## at a constant rate (as exemplified by a fixed amount of fuel being consumed over a fixed time interval ##\Delta t##), why does the change in kinetic energy apparently depend on the initial velocity? After all, if we add the same amount of energy the kinetic energy must change by the same amount according to the work-energy theorem.
OP's reasoning in post #1 is faulty in that the acceleration is not he same if the initial velocities are different. Let's do some calculations assuming that the acceleration of the mass is constant. The displacement of the mass is $$ \Delta x = v_0t+\frac{1}{2}at^2. $$The work done by an external (net) force ##F_{net}## is $$W=F_{net}\Delta x=ma\Delta x=mv_0at+\frac{1}{2}ma^2t^2. $$The change in kinetic energy is $$\Delta K=\frac{1}{2}m(v_0+at)^2-\frac{1}{2}mv_0^2=mv_0at+\frac{1}{2}ma^2t^2.$$ Note that both the kinetic energy change and the work done by the net force depend on the acceleration, but the validity of the W-E theorem does not because the two are equal for any acceleration.

Now let's see what happens when we add energy ##E## to an already moving object and change its kinetic energy by that amount.$$\Delta K=E=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2 ~\Rightarrow v=\sqrt{v_0^2+\frac{2E}{m}}.$$This gives an acceleration$$a=\frac{v-v_0}{\Delta t}=\frac{\sqrt{v_0^2+2E/m}-v_0}{\Delta t}.$$For objects with different initial velocities, one may not fix ##E## and ##\Delta t## and assert that the acceleration is the same.

 

[Lnewqban]

Why doesn't it? The same force is applied and therefore the same acceleration? a=f/m.

Please, see:
https://en.wikipedia.org/wiki/Impulse_(physics)

The ball had huge kinetic energy before the muscles of your hand transferred some more to it.

 

[digalumps]

I will have one last try then shut up as I don't want to waste people's time.

I come home one morning after shopping. "Hey son, I got a 2 for 1 deal on rockets this morning.."

Great Dad let's go down to the disused airfield and try them on my bike.

(At airfield) OK son, attach both rockets to your bike and pedal at 10 km/hr and we'll light the first rocket

(10 km/hr rocket lights) Look dad I'm traveling at 11 km/hr - cool!

Well done son, now pedal at 20 km/hr and light the rocket.

OK Dad. Hey look Dad I'm traveling at 21 km/hr (F=MA)

Blimey son, you are the first person ever to beat the law of conservation of energy. Now, about that perpetual motion machine I was designing...

 

[TSny]

If I stand on a car running on a frictionless rail at 10 km/sec and throw a bowling ball off the end it accelerates to 11 m/sec and the KE increases correspondingly.

If i throw the bowling ball off the end when the car is traveling at 1000 km/sec if accelerates to 1001 m/sec and the KE increases by a lot more. Why?

Let a spring mechanism perform the throwing. Initially, you have a car and a bowling ball moving with the same velocity and you have a spring that has potential energy.

After the spring throws the ball, the car has gained KE. The gain in KE of the car does not equal the potential energy stored in the spring. The gain in KE of the car equals the potential energy stored in the spring plus the loss of KE of the ball. The faster the car is moving initially, the greater the loss of KE of the ball. The larger loss of KE of the ball accounts for the larger increase in KE of the car.

 

[kuruman]

I will have one last try then shut up as I don't want to waste people's time.

I come home one morning after shopping. "Hey son, I got a 2 for 1 deal on rockets this morning.."

Great Dad let's go down to the disused airfield and try them on my bike.

(At airfield) OK son, attach both rockets to your bike and pedal at 10 km/hr and we'll light the first rocket

(10 km/hr rocket lights) Look dad I'm traveling at 11 km/hr - cool!

Well done son, now pedal at 20 km/hr and light the rocket.

OK Dad. Hey look Dad I'm traveling at 21 km/hr (F=MA)

Blimey son, you are the first person ever to beat the law of conservation of energy. Now, about that perpetual motion machine I was designing...

Blimey father and son, why don't you stop guessing and calculate somethin'? If the rocket adds constant energy ##E## to whatever the kinetic energy of the bicycle is, then in terms of the change of speed ##\Delta v##, $$E=\frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2=mv\Delta v+\frac{1}{2}m(\Delta v)^2.$$ In the two scenarios you present, ##v_1=10;~ \Delta v_1=1;~v_2=20;~\Delta v_2=?## (All quantities are in some units.) Let's find ##\Delta v_2## without guessing. We assume energy conservation in the sense that the rocket releases the same amount of energy regardless of how fast the bicycle is moving. Then, $$\frac{E}{m}=v_1\Delta v_1+\frac{1}{2}(\Delta v_1)^2=v_2\Delta v_2+\frac{1}{2}(\Delta v_2)^2$$ In terms of numbers, this gives the quadratic equation $$10\times 1+\frac{1}{2}(1)^2=20\Delta v_2+\frac{1}{2}(\Delta v_2)^2$$which has only one positive root, ##\Delta v_2=0.53##. Your argument that ##\Delta v_1## is equal to ##\Delta v_2## because ##F=MA## violates energy conservation.

Thus, dear father and son, you end up "showing" that energy conservation is violated having assumed beforehand that it actually is. This is a logical fallacy called "begging the question."

 

[jbriggs444]

I will have one last try then shut up as I don't want to waste people's time.

I come home one morning after shopping. "Hey son, I got a 2 for 1 deal on rockets this morning.."

Great Dad let's go down to the disused airfield and try them on my bike.

(At airfield) OK son, attach both rockets to your bike and pedal at 10 km/hr and we'll light the first rocket

(10 km/hr rocket lights) Look dad I'm traveling at 11 km/hr - cool!

Well done son, now pedal at 20 km/hr and light the rocket.

OK Dad. Hey look Dad I'm traveling at 21 km/hr (F=MA)

Blimey son, you are the first person ever to beat the law of conservation of energy. Now, about that perpetual motion machine I was designing...

To repeat the explanation for the conundrum yet another time...

You are failing to look at all of the energy. The gunpowder in those toy rockets attached to your bike did not just supply energy to your bike. It supplied energy to the exhaust stream. That cloud of smoke rapidly going out the back is carrying energy. A whole lot of energy. In fact, most of the energy in the gunpower was expended making those move rearward very rapidly. Only a small fraction went into making your bike go faster.

What happens when your bike starts out moving at 20 km/hr instead of 10 km/hr? Those gasses whooshing rearward very rapidly are whooshing rearward at 10 km/hr less than before. That may not sound like much, but if you put pencil to paper and calculate it, that's enough to eat up all of the excess energy you seem to have gained -- exactly.

You've been urged to calculate. You have refused to do so. Your loss. Education comes with a price. Ignorance is free.

 

[digalumps]

Or it could just be that I have not reached the place in my studies where I am ABLE to calculate what you are asking. If I could then I would. Everyone has to start somewhere.

This is a basic homework thread, not a masterclass in differential calculus.

I did do calculations to the limit of my current ability and worked out the increase in KE, and tried to correlate it to the input energy which I thought using F=ma would be the same and tried to visualise it using a bike, rocket or whatever. I could not see where I was going wrong.

However, your explanation above is the first reply that actually explains what is going on in an understandable format so thank you for that.

 

[jbriggs444]

Or it could just be that I have not reached the place in my studies where I am ABLE to calculate what you are asking. If I could then I would. Everyone has to start somewhere.

This is a basic homework thread, not a masterclass in differential calculus.

No calculus is involved. Just addition, subtraction, multiplication and division. No fancy formulas that you do not already know. The one useful insight is the one that @TSny gave in post #8.

Rather than considering a continuous exhaust stream produced from the starting speed to the ending speed (which does require calculus and leads to the Tsiolkovsky rocket equation) go back to post #8 and review.

Let me reproduce that idea here.

We will work with symbols. One could, instead, substitute in fixed values, e.g. "A 10 kg rifle shoots a 10 g bullet rearward at 500 meters per second relative to the recoiling rifle . The rifle begins moving at either 1 meter per second or 10 meters per second. Perform an energy analysis". In my opinion, the algebra is more clear with symbols rather than numbers. But some people feel more comfortable with numbers.

You have a large mass M, a small mass m and a massless spring between them. They are lined up on a horizontal axis. You are free to imagine carts with frictionless wheels, pucks on ice, boats on a lake, ships in space or a rifle and a bullet.

You have a massless cord tied around all three holding them together. The spring is compressed and contains a certain amount of potential energy. That energy is sufficient to result in a change in velocity ##\Delta V## in the large mass M. Of course, this will result in a change of ##\Delta v## in the velocity of the small mass m.

The entire assembly is initially moving together in a straight line with the M in front the the m behind. The initial velocity is ##V_i = v_i##. You cut the cord and examine the results.

Determine the final velocities ##V_f## of ##M## and ##v_f## of ##m##.

Determine the initial and final kinetic energy of just ##M##. What is the difference between the two? Does it depend on ##v_i##? [This is the part that you've done previously].

Determine the initial and final kinetic energy of just ##m##. What is the difference between the two? Does it depend on ##v_i##?

Determine the initial total kinetic energy. Determine the final total kinetic energy. Compare the two. What is the difference between the two? Does that difference depend on ##v_i##?

Can you perform the calculations to determine these results? How far can you get? If you show your work, we can help when you get stuck.

Edit: I've edited the above a few times to make the algebra simpler. I hope you did not begin work on one of the tougher-to-get-started-on versions.

 

[digalumps]

Thank you very much I will work on this now.

 

[digalumps]

Ok my first working out. An infinitely light person is sitting on top of a 'stationary' 10kg rocket in space

KE = 0 (v=0)

Then person throws a 1kg bowling ball of the back at 1 m/sec

Conservation of momentum. m1v1=m2v2

Thus the rocket moves off in opposite direction at 0.1 m/sec

KE=1/2 mv squared.

Bowling ball: 0.5 x 1 x 1 x 1 = 0.5 Joules

Rocket: 0.5 x 10 x 0.1 x0.1 = 0.05 joules

Why are the two KEs not the same (but opposite directions) given the conservation of energy?

 

[PeroK]

Why are the two KEs not the same (but opposite directions) given the conservation of energy?

If there were conservation of energy then nothing could change. You have zero KE before and you would need zero KE after. The energy comes from the propulsion mechanism: in this case energy stored in the muscles.

There is equal and opposite momentum, but KE is not a vector, so there is no concept of equal and opposite KE. The KE's are not of equal magnitude unless the masses are the same.

 

[digalumps]

Thanks that makes sense - basically I have added 0.5 + 0.05 =0.55 joules from my muscles. Thank you.