AP Physics C Momentum Problem: A dart launched horizonally by a spring gun

[Kush]

Homework Statement

A boy launches a 20 g dart horizontally by a spring gun from a balcony 45 m above the ground. The dart lands 15 m away from the balcony. If the length of the gun’s barrel is 10 cm, what is the average horizontal force applied by the spring?(A) 1.0N (B) 2.0 N (C) 2.5 N (D) 5. N (E) 7.5 N

Homework Equations

Law of Conservation of Energy, Law of Conservation of Momentum, Kinematics Equations

The Attempt at a Solution

I tried to use kinematics, but was unable to solve it.

 

[haruspex]

I tried to use kinematics, but was unable to solve it.

Nevertheless, post your work as far as you got.
Did you find the launch velocity?

By the way, none of the given answers is correct.

 

[Kush]

I calculated these using the kinematics equations.
Initial Horizontal Velocity (V0x) 5
Final Velocity (Vf) 30.41381 ----
Final Horizontal Velocity (Vfx) 5 ----
Final Vertical Velocity (Vfy) -30
Flight duration (t) : 3 sec
I was unsure what equation to use at this point.

 

[haruspex]

I calculated these using the kinematics equations.
Initial Horizontal Velocity (V0x) 5
Final Velocity (Vf) 30.41381 ----
Final Horizontal Velocity (Vfx) 5 ----
Final Vertical Velocity (Vfy) -30
Flight duration (t) : 3 sec
I was unsure what equation to use at this point.

Ok, so you found the launch velocity as 5m/s. (Always specify the units!)

Next, consider the motion of the spring. Have you studied SHM?

 

[Kush]

I have studied work done by a spring force which is W = -1/2 k x^2 but have not studied simple harmonic motion yet. Maybe force of a spring F=Kx applies?

 

[haruspex]

I have studied work done by a spring force which is W = -1/2 k x^2 but have not studied simple harmonic motion yet. Maybe force of a spring F=Kx applies?

Did you notice my edit to post #2?
You know what work is done by a constant force that advances a certain distance, right?
What the question setter intends you to do is to assume that this provides a valid way to calculate the average of a force that does a known amount of work over a known distance. I can tell that because doing it that way does match one of the offered answers. Sadly, that is invalid.

The correct definition of average force is mass x average acceleration, which can also be written as change in momentum / elapsed time, Δp/Δt. You already know Δp, so it becomes a matter of finding how long the spring took to expand to its relaxed position. That involves knowing a bit about SHM.

 

[kuruman]

By the way, none of the given answers is correct.

Is this your edit to post #2? I was able to get one of the given answers by assuming that g = 10 m/s².

 

[haruspex]

Is this your edit to post #2? I was able to get one of the given answers by assuming that g = 10 m/s².

Yes, so can I, but only by using the invalid method I describe in post #6. Analysis of the spring behaviour and using Δp/Δt yields 4/π times as much.

 

[kuruman]

I have a different interpretation of the question that is more general and does not tie the average force to modeling the mechanism that produces it. I think the spirit of the problem is "given that the projectile follows the given trajectory, what is the average force acting on it within a tube of given length?" In that case the work-energy theorem has to come into play.

 

[Kush]

How could this be solved using work-energy theorem?

 

[haruspex]

I have a different interpretation of the question that is more general and does not tie the average force to modeling the mechanism that produces it. I think the spirit of the problem is "given that the projectile follows the given trajectory, what is the average force acting on it within a tube of given length?" In that case the work-energy theorem has to come into play.

You seem to be missing my point. In order to solve it using the work-energy theorem you have to assume that acceleration in the tube is constant. This is because the definition of average force is Δp/Δt, not ΔE/Δs. (It has to be that way to be consistent with average acceleration = Δv/Δt.)
If we do not assume constant acceleration then in general we can find Δp but not Δt. However, here we are told it is a spring, so in this case we can use SHM to find Δt.
The resulting (correct) value of average force is $\frac{\Delta p}{\Delta t}=\frac 4{\pi}\frac{\Delta E}{\Delta s}$.
Note that the virtue of defining average force this way is even clearer in vectors. We can write $\vec F_{avg}=\frac{\Delta \vec p}{\Delta t}$. There's no equivalent using energy.

 

[Delta2]

There are many different ways to define average, average over distance, average over time, e.t.c, which do not necessarily coincide and in this case they do not (because acceleration is not constant but I think they don't coincide even if acceleration is constant). @haruspex the way this question is stated it is like "it is begging" to calculate the average over distance, not the average over time.

 

[haruspex]

There are many different ways to define average, average over distance, average over time,

Yes, you can specify different bases for the averaging when you intend something other than the standard definition. You can specify average over distance for an acceleration or velocity, for example. But if you write "average acceleration", with no such qualification, it is well defined to mean average over time.
The same applies to average force, for the reasons I gave.

I think they don't coincide even if acceleration is constant

They do.

 

[Delta2]

Well @haruspex, as I said it already, the question might not explicitly state it, but the way the question is stated, giving the length of the barrel of the spring gun as a given data, it is BEGGING to consider the average over distance.

 

[haruspex]

giving the length of the barrel of the spring gun as a given data, it is BEGGING to consider the average over distance.

a. If it had given the distance without mentioning the means of propulsion that would be a valid argument, but the spring + the length is exactly what you need to solve it correctly. Thus, basing it on what info is provided fits better with the correct definition.
However,
b. From the answers offered, it is clear the questioner is ignoring the spring aspect and just using work-energy and distance. So I quite agree with you that this is the questioner's intent.
But,
c. My complaint is that the answer is wrong because the questioner is using the wrong definition. This is a widespread blunder leading to cohorts of students being left with the belief that this is a valid way to find an average force in general. It is not. So I make it one of my missions to point this out to those students unfortunate enough to have these shoddy teachers/textboooks.

A quick scan of the net:

http://www.softschools.com/formulas/physics/average_force_formula/46/:
"average Force is equal to the mass of the body multiplied by the average velocity over the defined time."
http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html:
"The product of average force and the time it is exerted is called the impulse of force."
http://www.phys.ufl.edu/courses/phy2053/fall09/lecture14.pdf:
"The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force."
https://books.google.com.au/books?i...KEwjLmOPSgMfgAhUOAXIKHV1qA_s4ChDoATALegQIARAB:
" impulse = momentum transferred during contact in a collision and it equals average force x contact time."
https://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection: (numerous examples)

Now, there are a few out there which make the same blunder as in this thread, but they are fairly amateurish sites.

 

[jbriggs444]

As I see it, one is forced to make assumptions to idealize the problem and make it unambiguously solvable. Unfortunately, the choice of assumption will change the resulting answer. This is, accordingly, a poorly posed problem.

Assumption 1: We are after a time average force. The ideal spring in an unloaded gun is relaxed => @haruspex answer
Assumption 2a: We are after a distance average force. The spring is irrelevant.
Assumption 2b: We are after a time average force. The tension in the spring is negligibly different from constant during the firing action, so a distance average will amount to the same thing.
Assumption 3: The instructor does not know any better. Use the distance average because that is the instructor's intent. [Usually safe, but eventually you'll come across instructors with brains].

 

[SammyS]

As I see it, one is forced to make assumptions to idealize the problem and make it unambiguously solvable. Unfortunately, the choice of assumption will change the resulting answer. This is, accordingly, a poorly posed problem.

Assumption 1: We are after a time average force. The ideal spring in an unloaded gun is relaxed => @haruspex answer
Assumption 2a: We are after a distance average force. The spring is irrelevant.
Assumption 2b: We are after a time average force. The tension in the spring is negligibly different from constant during the firing action, so a distance average will amount to the same thing.
Assumption 3: The instructor does not know any better. Use the distance average because that is the instructor's intent. [Usually safe, but eventually you'll come across instructors with brains].

It seems to me that Assumption 2b is quite reasonable.

So now the argument can be boiled down to the following:

2b or NOT 2b? That's the question.

(Apologies to Shakespeare or whoever actually wrote the literary version.)

 

[haruspex]

The tension in the spring is negligibly different from constant during the firing action

That would imply it thuds into a retaining ring at the end of the tube, without which it would have projected well beyond. Possible, but unlikely?

But in the present problem we do not need to make any assumption. Since it is multiple choice we can tell we are expected to use ΔE/Δs. (I do not refer to this as being expected to use average over distance as that would suggest the question setter understood the error.)

 

[jbriggs444]

That would imply it thuds into a retaining ring at the end of the tube, without which it would have projected well beyond. Possible, but unlikely?

The toy dart guns that I have played with do feature a retaining mechanism of some sort and remain tensioned after firing, yes. A dart gun with a spring that was relaxed after firing, I would consider "worn out".

Sadly, I have no such device immediately available for destructive testing.

 

[haruspex]

The toy dart guns that I have played with do feature a retaining mechanism of some sort and remain tensioned after firing, yes. A dart gun with a spring that was relaxed after firing, I would consider "worn out".

Sadly, I have no such device immediately available for destructive testing.

It's the "well beyond" that I was suggesting as unlikely.

 

[kuruman]

You seem to be missing my point. In order to solve it using the work-energy theorem you have to assume that acceleration in the tube is constant.

I understood your point and my reply would have been something like post #16 but @jbriggs444 preempted me. However, I will attempt to provide the following set of beguiling statements to show how easy it is to deceive oneself.
1. Asking for an average force implies replacing the actual force with a constant force in the calculations.
2. If the force is constant, it does not matter whether it is constant over a time interval Δt or over a displacement Δx; constant is constant.
3. Constant force implies constant acceleration therefore the standard SUVAT equations are applicable.
4. We are given the length over which the object accelerates and we can infer the muzzle speed from the trajectory, so we use v²=2aL to find the constant (average) acceleration hence the force.

 

[haruspex]

Asking for an average force implies replacing the actual force with a constant force in the calculations.

Yes, but, specifically, average force means that constant force which would have produced the same momentum change in the same time. That is different from the constant force which would have produced the same energy change over the same distance.
If (unqualified) average force is to mean anything then all valid definitions must lead to the same answer.

 

[kuruman]

Yes, but, specifically, average force means that constant force which would have produced the same momentum change in the same time. That is different from the constant force which would have produced the same energy change over the same distance.
If (unqualified) average force is to mean anything then all valid definitions must lead to the same answer.

And that's exactly the pitfall in this reasoning.