How much kinetic energy do I need to impart?

[fewme]

Homework Statement

I'm actually making up my own problem here to try to resolve some confusion I'm running into on understanding kinetic energy and reference frames...

Let's launch a potato of mass m (just call it 1) horizontally with a potato gun on a fast vehicle moving 50 m/s, while I'm watching and measuring from a stationary position. I want to see a potato flying by at 100m/s. Assume that whatever potential energy is "charged" in the gun is fully transferred into kinetic energy of the potato. How much energy do I need to precharge the gun with before putting it on the vehicle and having it fired?

Homework Equations

KE=1/2 mv^2

The Attempt at a Solution

I assume that both the moving vehicle and my stationary position are valid inertial reference frames. From a reference frame of my position on the ground, then it would appear that the KE needed to add to the potato to go from 50m/s (vehicle speed) to 100m/s (post-launch speed) is
KE = 1/2 * 1 * 100^2 - 1/2 * 1 * 50*2 = 3750J

From the reference frame of the vehicle, it would appear that the potato only would need to go from the "apparent standstill" of 0m/s up to 50m/s, which would require 1/2 * 1 * 50^2 = 1250J

Of course, these reference frames are moving 50m/s relative to each other, which changes the KE of the potato measured relative to observers in each frame.

But here's where I get confused in my example... From my stationary point of view, I need the gun to accelerate a potato from 50 to 100 m/s, and so would need to add 3750J. But from a vehicle occupant's point of view, it would only need to accelerate from 0 to 50m/s, and so need 1250J added. Both inertial reference frames are equally valid. The change in velocity is the same in each case (0 to 50 or 50 to 100), while only the absolute velocity range being looked at is different. But the end result from the stationary view is simply a potato moving at 100m/s.

So to see the result I want, when I hand off the potato gun into the vehicle, do I pre-charge the potato gun with 1250J or with 3750J?

Or put another way, from the stationary reference frame, it seems the vehicle's motion contains the 1250J (doing the initial accelerating from 0 to 50 within the vehicle) and the gun needs to add the 3750J, while from the vehicle's reference frame, the moving vehicle seems to give the potato the 3750J based on the stationary viewer and the gun only needs to add the 1250J. In any case, the potato should end up with KE = 5000J and going 100m/s relative to the stationary observer.

 

[Ll Judd]

I think the missing link you have is the V² I havn't studied your question closely, but you will need have 3750j in the gun and the remaining 1250j comes from the car.

take 100² = 10,000 and 50 ² = 2,500 (this is only looking at the v² section of EK formula).
so the car will supply 25% of total energy and the gun the remaining 75%

On second thoughts;
I am going to challenge your assumption that the moving vehicle is a valid reference frame. The potato has Ek already from the moving vehicle, so you can't use that as a static reference frame.
Velocity is directional, and the gun is increasing velocity of an already moving object. Say as an example you shot the potato 'backwards' with 1250j the velocity would be 0. Because it is already moving you need to increase its velocity.

The underlying principles in this scenario is the difference between velocity and acceleration.

 

[haruspex]

You need to consider conservation of momentum. The vehicle cannot keep moving at the same speed, no matter how massive, so it is not an inertial frame. If you take into account the change in the vehicle's velocity and use inertial frames (e.g., an observer on the ground and another moving steadily at the initial velocity of the vehicle) you will find there is no paradox.

 

[Cutter Ketch]

To make the math easy say you have two potatoes launched through the air at 50m/s. At some time a charge between them goes off accelerating one potato to 100m/s and stopping the other one dead. In the ground reference frame you had 2 kg (heavy potatoes!) going 50 m/s with 2500 J of energy become one potato going 100 m/s with 5000J of energy. The energy added by the charge is 2500 J. In the center of mass coords traveling with the two potatoes you start with 0 and end up with 2 potatoes traveling +/- 50 m/s with net 2500 J.

 

[TomHart]

So to see the result I want, when I hand off the potato gun into the vehicle, do I pre-charge the potato gun with 1250J or with 3750J?

Have you ever walked down the aisle of a commercial airliner when it is flying at a cruising speed of approximately 600 mph? Did you notice if it takes a lot more effort to go from stopped to a normal walking speed? You might want to run some numbers on that situation like you did with the potato.

 

[Cutter Ketch]

Have you ever walked down the aisle of a commercial airliner when it is flying at a cruising speed of approximately 600 mph? Did you notice if it takes a lot more effort to go from stopped to a normal walking speed? You might want to run some numbers on that situation like you did with the potato.

Oh wow! That's fun. 0 to 2 m/s vs 278 to 280 m/s. Done the same way as the original calculation without conserving momentum it would take 280 X more effort to walk on the jet. We'd all be pinned to our seats!

 

[haruspex]

Have you ever walked down the aisle of a commercial airliner when it is flying at a cruising speed of approximately 600 mph? Did you notice if it takes a lot more effort to go from stopped to a normal walking speed? You might want to run some numbers on that situation like you did with the potato.

I think fewme understands that one of the calculations must be wrong. What is sought is an explanation of which is wrong and why.

 

[haruspex]

The potato has Ek already from the moving vehicle, so you can't use that as a static reference frame.

The reference frame does not need to be static, it only needs to be inertial. As I posted above, if momentum is considered, we see that the vehicle is not an inertial frame.

 

[fewme]

Thanks for the answers so far, and I have a little more clarity now. What I gather is this...

The moving vehicle has momentum relative to the stationary observer, along with the potato contained within. When the potato gun is fired from the vehicle, the vehicle absorbs the recoil of the gun as an action/reaction, and loses a tiny amount of velocity and momentum relative to the stationary observer. Therefore, the vehicle is not a consistent and independent inertial frame relative to the stationary observer, because its state changes with the action being observed.

Because the vehicle loses some momentum, which the potato gains, the full amount of energy in the potato gun is divided between the fired potato and the vehicle.

Unfortunately, I'm still a little confused about the correct answer to my original question... should the gun be charged with 1250J or 3750J? Or some other value?

If I put the full 3750J in the gun, would the potato overshoot its target speed? Or would it hit its target speed, with the extra energy (difference between 1250J and 3750J) going into reducing the momentum of the vehicle? Or is that distribution of KE not correct? (Which situation happens: Would the potato and vehicle both receive the same absolute value of KE from the firing (in opposite directions), but with their respective momentums changing by unequal amounts relative to the stationary observer? Or would both the potato and vehicle receive equal momentum from the firing, by conservation of momentum, but unequal kinetic energies?)

Based on the airplane example, it seems like the gun should be charged with 1250J (but this adds the 50m/s to the potato only relative to the vehicle occupants). But it also seems that the vehicle absorbs some of the energy via recoil of the gun (slowing its momentum), and so it would seem that perhaps I do need a higher amount than 1250J in the gun in order to get a true 100m/s measured from the stationary point of view.

So, if I put 1250J in the gun, and firing it slows the vehicle to say 49.8m/s, then would the final speed of the potato be 99.8m/s? And then how much "extra" energy must I charge the gun with above the 1250J to get the true 100m/s for the stationary observer? Would it work out to an "extra" 2500J (the difference between the 1250? and the 3750J), thereby requiring an initial 3750J in the potato gun?

 

[fewme]

Also, here is another variant I just thought of...

Let's move this whole experiment into deep space, and let's build a self-contained rocket-propelled potato, so there is no vehicle involved to absorb momentum.

When the rocket is fired, there is conservation of momentum between the potato and the individual particles ejected in the rocket thrust. Would the thrust particles take the same place as the vehicle in my initial example, and act as the absorber of momentum?

If this were measured in space with a rocket potato and no vehicle, would the same math as above apply here when judging the KE required to accelerate the potato from 0m/s to 50m/s relative to itself, versus from 50m/s to 100m/s relative to a "stationary" observer (moving initially 50m/s "slower" than the potato)?

 

[TomHart]

I'm going to try another silly one. Maybe it is better to think of this as being in a vacuum. If you are riding in the back seat of a car traveling at 100 mph, does it take any more effort to toss, say, a nerf ball into the front seat than it would if the car was sitting at a stoplight? The answer is obviously, no, it doesn't take any more effort. Does the tossing of the nerf ball from the back seat into the front seat affect the speed of the car? Well, yes, negligibly, while the ball is in the air. However, once the ball lands in the front seat, the car will return to its original speed. Now if the nerf ball was tossed out the window (instead of tossed into the front seat), then the reduction in the speed of the car (although negligible) would be permanent. So could you add the car's initial speed to the speed of the nerf ball relative to the car to get the final speed of the ball relative to a fixed position outside of the car? Almost, but it would be negligibly less than that. So how much additional energy would be required to negligibly increase the nerf ball speed? A negligible amount.

Rocket powered potato? I love it.

 

[Cutter Ketch]

Based on the airplane example, it seems like the gun should be charged with 1250J (but this adds the 50m/s to the potato only relative to the vehicle occupants). But it also seems that the vehicle absorbs some of the energy via recoil of the gun (slowing its momentum), and so it would seem that perhaps I do need a higher amount than 1250J in the gun in order to get a true 100m/s measured from the stationary point of view.

Yes, more than 1250J. I think the thing you are missing here is that from the stationary observer's point of view that apparent extra energy came FROM the car. You didn't specify a mass for the car, so in the two potato version I gave the car a mass of 1 the same as the potato to make the math easy. However we can be more realistic. Say the potato has a mass of 1 kg and the car has a mass of 1000 kg. In a reference frame traveling with the center of mass in order to fire the potato at 50 m/s the car must recoil at 0.05 m/s to conserve momentum. The energy required is 1250J + 1.25 J = 1251.25 J. From the stationary observer's POV the potato accelerated from 50 to 100 m/s increasing by 3750J, but the car slowed from 50 to 49.95 m/s losing 2498.75 J and once again the energy expended is 1251.75 J.

 

[Cutter Ketch]

And now algebraically.

Initially the ball and car are moving together in some reference frame with an initial velocity v_R. From conservation of momentum

m₁ Δv₁ = - m₂ Δv₂

⇒Δv₂ = -(m₁/m₂)Δv₁

The change in energy is
m₁ (v_1f² - v_1i²)/2 + m₂ (v_2f² - v_2i²)/2

Oops. I see in copying and pasting while editing I deleted a step here, but you get it= m₁ (Δv₁²+2v_RΔv₁)/2 + m₂ (Δv₂²+2v_RΔv₂)/2

= m₁ Δv₁²+m₁v_RΔv₁ + m₂ Δv₂²+m₂v_RΔv₂Substituting for Δv₂ in the last term from the momentum conservation above

= m₁ Δv₁²/2+m₁ v_RΔv₁ + m₂ Δv₂²/2 - m₁ v_RΔv₁

= m₁ Δv₁²/2+ m₂ Δv₂²/2

So the energy is the same as in the center of mass system regardless of choice of reference frame.