Jump to content

[SOLVED] Can someone help me with this problem?


ARS

Recommended Posts

Okay guys, I need a help to solve this problem. I'm given this in my school:

"A projectile with a mass of 100 gram is being fired at horizontal trajectory, 20 centimeters above the ground with a speed of 900m/s. As the projectile travels 1 meter from firing point, a force of 0.298 N is being applied to the projectile to accelerate it. Assuming there's no wind resistance, calculate:

1. The speed of the projectile after it's being accelerated

2. The range at which the projectile hit the ground (like an actual bullet, it's affected by gravity at long range), compare it after being accelerated with as if it isn't accelerated

3. If the projectile is being changed with the one with a mass of 30 gram, what's the speed after it's being accelerated and it's maximum range?"

I've been spending 3 days to think this through and tried to solve it and I cannot find an answer for this problem T T. So if anyone have a solution/ formula/ answer for this, I'm really appreciate it

And thanks for helping

Edited by ARS
Link to comment
Share on other sites

Generally speaking, we don't do homework here.

But, I'll give you a nudge.

This is a problem of horizontal gun shot. It being horizontal makes it somewhat simpler.

The projectile starts 20 cm above ground, and at t=0 has zero vertical velocity. Time of flight is purely a function of gravitational acceleration and height. If this was a school assignment, you should have covered that and be able to calculate how long it will take for the stationary projectile to fall down to ground from 20 cm.

Sub-question 1 doesn't make much sense, since it doesn't specify for how long the force is being applied. One could probably assume that it is applied until the projectile hits the ground.

To calculate the speed after acceleration, you need to know how long the projectile travels, from that you subtract the time it takes to travel the first 1 meter. Now you know the time during which the force is being applied. From force and mass, calculate the acceleration. Add the acceleration due to gravity (don't just add them up, keep in mind that they are vectors). Now that you have the total acceleration apply it for the amount of time the flight takes (don't forget the starting velocity). That should give you the answer to 1.

2 is just a sum of distance of flight of 1 meter at constant speed and the rest under acceleration, vs all under constant speed. (hint: acceleration due to gravity doesn't affect the distance, other than limiting the flight time).

3 is the same as 1, only the mass is different, which means the acceleration due to force is greater, so it will go further. Math is the same.

 

v = v0 + a * t

d = v0 * t + 0,5 * a * t2

Edited by Shpaget
Link to comment
Share on other sites

No - as far as we know, the projectile is unpowered, so the rocket equation isn't required.

We need to make four assumptions.

1.  This scenario takes place on Earth and the projectile is falling under normal gravity.
2.  Collisions with the ground are perfectly inelastic - i.e. the projectile hits the ground and stops.
3.  The secondary acceleration applied, after 1m of travel, is applied horizontally.
4.  The secondary acceleration is instantaneous and not applied over time.

Equations of motion

1.  F = ma     - where F = force applied, m = mass and a = acceleration
2.  v = u + at     - where v = final velocity, u = initial velocity, a = acceleration and t= time. Turns out we don't need this :) 
3.  s = ut + 1/2at^2  -  where s = displacement, a and t are as previously defined and t^2 = time squared.

To solve this problem we resolve the projectile motion horizontally and vertically. In other words we can treat horizontal motion and vertical motion of the projectile separately. We calculate time of flight of the projectile by considering it's vertical motion only and then range of the projectile by considering its horizontal motion only.

For question 1,

Equation 1 allows you to calculate horizontal acceleration and equation 2 then allows you to calculate horizontal velocity of the projectile after the force is applied.

For question 2:

Equation 3 lets you calculate the time of flight of your projectile by considering the vertical motion of the projectile only. Here: u=0 (since the projectile is fired horizontally and initial vertical speed is zero) so equation 3 simplifies to s = 1/2at^2 where a = acceleration due to gravity and s = 0.2m. Plug the numbers in and you can calculate t(total), i.e the time it takes the projectile to hit the ground. 

To calculate the range we need to calculate time t1, that is the time at which the secondary force is applied

Using equation 3 again and considering horizontal motion only.  a = 0 (since no horizontal acceleration is applied between time t=0 and time t1), so the equation simplifies to s = ut1. We know s (1m) and we know u (900m/s) so we can calculate time t1.

We can now break the horizontal motion of the projectile into two stages:

In the first stage, the projectile travels at horizontal speed 900m/s for a time t1. In the second stage, the projectile travels at a new horizontal velocity (which we calculated in question 1) for an unknown time t2. We can calculate time t2 as t2 = t(total)-t1, where t(total) = time of flight (which we calculated above) and t1 is the time at which the force was applied (which we calculated above)

Using equation 3 we can calculate the distance travelled in each stage and sum them to give the total range. EDIT. Because we assume that the force is applied instantaneously, the projectile is not accelerating horizontally during each stage. This allows us to simplify equation 3.

For the case where no secondary force is applied, we simply use the same simplified equation 3 to calculate the distance travelled in time t(total).

For question 3:

As per question 2 but with different numbers!

 

Hope this helps!

 

 

 

 

 

Edited by KSK
Link to comment
Share on other sites

32 minutes ago, Shpaget said:

Sub-question 1 doesn't make much sense, since it doesn't specify for how long the force is being applied. One could probably assume that it is applied until the projectile hits the ground.

Enough information has been supplied to calculate the time. We have the initial velocity, the force and the distance at which the force ceases to accelerate the bullet. Use @KSK 's 1. equation, solve for a, then use 3. equation and solve for t.

30 minutes ago, KSK said:

4.  The secondary acceleration is instantaneous and not applied over time.

Wrong assumptions. The force is a force, given in Newtons, it has to act over time in order to produce any acceleration. Impulses can be approximated to occur instantaneously, but that results in infinite force being applied.

@ARS, in general when asking for help in school assignments you should post not only the assignment but also what you have already tried to solve it. First it shows that you are not trying to slack off but are actually putting effort into learning and genuinely stuck, second we can much better help you when we see how close you are. You have been given some very good advice here, but do double check everything. One last tip, convert all numbers into SI base units before calculating, that way your units will just fall into place properly. Or if they don't you know there is a mistake to be found and fixed. For this assignment the units would be seconds, meters and kilograms. Trivia: Kilogram is the only SI base unit that includes a prefix.

Link to comment
Share on other sites

20 minutes ago, monophonic said:

Enough information has been supplied to calculate the time. We have the initial velocity, the force and the distance at which the force ceases to accelerate the bullet. Use @KSK 's 1. equation, solve for a, then use 3. equation and solve for t.

Hm, I interpreted the problem as the force starting to act after d= 1 m and not acting from t=0 until d = 1.

Reading it again, I suppose I could be wrong, but I still find the wording to be weird.

Link to comment
Share on other sites

2 minutes ago, Shpaget said:

Hm, I interpreted the problem as the force starting to act after d= 1 m and not acting from t=0 until d = 1.

Reading it again, I suppose I could be wrong, but I still find the wording to be weird.

Now I'm not sure anymore either. Ambiguous wording is the worst kind you can have in an assignment. No way around that but to write out your interpretation and hope for the best. I had some teachers who would not penalize for solving the wrong thing if the assignment was ambiguous, but others still would.

Link to comment
Share on other sites

47 minutes ago, monophonic said:

Enough information has been supplied to calculate the time. We have the initial velocity, the force and the distance at which the force ceases to accelerate the bullet. Use @KSK 's 1. equation, solve for a, then use 3. equation and solve for t.

Wrong assumptions. The force is a force, given in Newtons, it has to act over time in order to produce any acceleration. Impulses can be approximated to occur instantaneously, but that results in infinite force being applied.

D'oh. Good point - thanks. @ARS - definitely double check my working then, although hopefully the general approach to solving the problem should still help.

Link to comment
Share on other sites

2 hours ago, KSK said:

We need to make four assumptions.

1.  This scenario takes place on Earth and the projectile is falling under normal gravity.
2.  Collisions with the ground are perfectly inelastic - i.e. the projectile hits the ground and stops.
3.  The secondary acceleration applied, after 1m of travel, is applied horizontally.
4.  The secondary acceleration is instantaneous and not applied over time.

About these assumptions, all of the above is correct

2 hours ago, KSK said:

Hope this helps!

Thanks mate! You really help me in this case! I've been trying to solve this problem for 3 days, and now you saved my life!

1 hour ago, monophonic said:

@ARS, in general when asking for help in school assignments you should post not only the assignment but also what you have already tried to solve it. First it shows that you are not trying to slack off but are actually putting effort into learning and genuinely stuck, second we can much better help you when we see how close you are.

I'm sorry, this is my first time asking something like this on the forum. I'm working on this assignment for 3 days and I got stuck. It's easy for me to figure out the secondary question of point B and C (maximum range and alternated projectile mass without being accelerated) but I got stuck when I had to figure out about the projectile being accelerated mid-flight since all I know is how to calculate the projectile from motionless state to being accelerated once (I had a problem to figuring out how to find the velocity if it's accelerated while it's already moving) anyway, thanks for your advice! :) (also, nice trivia, I only realized it now)

Edited by ARS
Link to comment
Share on other sites

3 hours ago, ARS said:

About these assumptions, all of the above is correct

Thanks mate! You really help me in this case! I've been trying to solve this problem for 3 days, and now you saved my life!

You're welcome. :) 

Just check that last assumption OK? The other folks on this thread are correct - you do need to know how long that force is applied for.

Link to comment
Share on other sites

20 minutes ago, KSK said:

You're welcome. :) 

Just check that last assumption OK? The other folks on this thread are correct - you do need to know how long that force is applied for.

Sure. Thanks :)

Link to comment
Share on other sites

@ARS Should all work great! The first bit with the acceleration after the barrel can also be a little easier when calculated with energy rather than momentum. In this case, we can see the projectile as having a kinetic energy (speed) which gets added to by the force as it travels over that distance. The energy that gets added is conveniently force*distance no matter how fast the projectile is moving! 

Initial KE  + Energy Added = Final KE

1/2* m * v12  + F * d  = 1/2* m * v22

 

How do you know when to use energy (W=f*d), and when to use momentum (f=m*a)? It's tricky, sometimes you can use one, the other, both... sometimes you need both! In general, if something is happening over a certain distance it'll be easiest with energy, while if something is happening over a certain time it's easiest with momentum. Rocket stuff (like engines) are almost always firing for a certain time, so momentum reigns king around here.

Finally, how is it accelerating after it leaves the barrel? Especially at 30g, it could be a rail gun :D . Railguns have a hard time getting the projectile started, so one concept has been to start the projectile moving with another means, and then finish accelerating it with the rail gun (in this case a snubby little 1m one). Of course the practical cons heavily outweigh this in general, but it's a fun rationalization!

Where is the right place for homework questions, @Shpaget?
Edit: If there happens to not really be a place, would you mind if I made an offshoot thread for them?

Edited by Cunjo Carl
Link to comment
Share on other sites

2 hours ago, Cunjo Carl said:

Where is the right place for homework questions, @Shpaget?
Edit: If there happens to not really be a place, would you mind if I made an offshoot thread for them?

My point was that we don't write other peoples homework, but we will give guidance. The student should be the one doing the calculations. Furthermore, when first coming with the question, he should show his previous attempts and efforts.

If someone came here with "Guys, can somebody solve this for me, plox? I need full step by step solution." it would not be beneficial at all for the student if this community just provides the solution. The idea is that the student does the work while we guide and correct errors. kind of like this thread.

 

2 hours ago, Cunjo Carl said:

Finally, how is it accelerating after it leaves the barrel? Especially at 30g, it could be a rail gun

It doesn't really matter. It could be a railgun, it could be a perfect propellantless rocket, a strong wind or just magic, as it is usually the case with school problems.

Link to comment
Share on other sites

Thoughts: Sounds to me like the wording is meant to imply a bullet with a quick initial speed, then a further increase in speed from escaping gasses until it goes 1 meter, then no more acceleration. Seems odd, but the only way it works. It's the sort of bad wording I'd check, but when writing the answer I'd make sure to point out how I read it.

Secondly, to cover all my bases for the first question, I'd make sure to point out the horizontal speed, vertical speed, and overall speed. Pedantic, but that's how I learned to answer questions, as you don't want to assume which of those three the teacher really wants.

 

Link to comment
Share on other sites

5 hours ago, Cunjo Carl said:

Finally, how is it accelerating after it leaves the barrel? Especially at 30g, it could be a rail gun :D . Railguns have a hard time getting the projectile started, so one concept has been to start the projectile moving with another means, and then finish accelerating it with the rail gun (in this case a snubby little 1m one). Of course the practical cons heavily outweigh this in general, but it's a fun rationalization!

 

@Cunjo Carl, you have a really cunning thought to guess it! It is indeed a Rail gun. I am designing a Rail gun for my thesis, and my teacher said I had to calculate the parameters of it's bullet's trajectory and comparing it with non-accelerated bullet (my railgun uses 2-stage acceleration, first by using propellant, then using lorentz force from rails inside the barrel, with the bullet acting as a switch). When my teacher said I had to calculate about that, I thought: "Hmm... okay, let's try this". Then I'm stuck when I had to figure out about the projectile being accelerated mid-flight since all I know is how to calculate the projectile from motionless state to being accelerated once (I had a problem to figuring out how to find the velocity if it's accelerated while it's already moving). Anyway, thanks for all of you! Now I can finally continue my work! :D

Here's my design:

oupjbRb.jpg

Link to comment
Share on other sites

And with one fell stroke, all our calculations are ruined. Ruined I tell you. :) 

Or at least mine were. Zeroth assumption - this was a projectile in free flight, not one being shot down a barrel.

Cool looking design though!

Link to comment
Share on other sites

1. If we assume that the Earth is flat, the fall time is constant. Any object 20 cm above the ground under 9.8 m/s^2 will impact at t+0.20 seconds.

2. The initial velocity of the Projectile is 900 m/s. The mass is 100 grams, and the net force excluding gravity is 298 mN forward. This translates to 2.98 m/s^2, for a final velocity of 900.6 m/s.

3. This means the mean velocity was 900.3 m/s. So we are on the ground by 180 meters. (Actually 182, I rounded the time to 2 sig figs).

4. Practically speaking, changing the acceleration isn't that important. An average speed of 900, 900.6, or 902 m/s will make less than a meter of difference in the result, which is well below the threshold of 2 significant figures given by the initial height.

Edited by Pds314
Link to comment
Share on other sites

I'm still not sure of how the force is actually applied, but back of the hand calculation over energy conservation, assuming no mass loss ("magic" extra force) :

- Work is distance * force. So the extra energy is only 0.298 J.

- Initial KE (which sums up the whole mechanical energy available) : 0.5*0.1*(900)^2 = 40,500 J

The extra energy is quite insignificant. I'm not sure how it affect the length though, given the rather shallow trajectory. You could compare the unpowered trajectory between a projectile with initial energy of 40,500 J (900 m/s) and 40,500.298 J (900.0033... m/s) if you want, then put up the conclusion.

Edited by YNM
Link to comment
Share on other sites

@ARS I'm a connoisseur of science minutia, so it's nice to know I can still call 'em! That is the coolest looking iso for your design mockup, did you draft that? I had to look up the name- cute reference :D

Railguns are notoriously resistant to simple theoretical analysis. It's frustrating, because the elements that go into them are all pleasantly fundamental and politely linear, but when you get them all together it turns into a jumbly mess unless you make some.... not-entirely-justifiable assumptions. Especially in regards to heating.

But

They are double happy funtime for finite element analysis spreadsheets! We can Googlesheets a model together if it sounds like fun? For fair warning from the start, a system like the one you're proposing won't do much for boosting velocity, but it will be fun :) . It'd require about 20 hours of work, most of which would of course be up to you. I'd be happy to help get you started, though. Not trying to butt in, just an offer :)

In any case, best of luck on your thesis!

 

Link to comment
Share on other sites

On 5/24/2017 at 4:35 AM, Shpaget said:

Sub-question 1 doesn't make much sense, since it doesn't specify for how long the force is being applied. One could probably assume that it is applied until the projectile hits the ground.

It gives you the force and distance of the acceleration event (0.298N though 1m). From this the work done can be derived, and that work turns into additional kinetic energy. At least that's my reading of the problem.

Link to comment
Share on other sites

4 hours ago, Red Iron Crown said:

It gives you the force and distance of the acceleration event (0.298N though 1m). From this the work done can be derived, and that work turns into additional kinetic energy. At least that's my reading of the problem.

Could be. I fully accept the possibility that I misunderstood the problem, as I stated previously.

On 5/24/2017 at 10:20 AM, Shpaget said:

Hm, I interpreted the problem as the force starting to act after d= 1 m and not acting from t=0 until d = 1.

Reading it again, I suppose I could be wrong, but I still find the wording to be weird.

 

Link to comment
Share on other sites

  • 6 years later...
On 5/24/2017 at 11:51 AM, ARS said:

Okay guys, I need a help to solve this problem. I'm given this in my school:

"A projectile with a mass of 100 gram is being fired at horizontal trajectory, 20 centimeters above the ground with a speed of 900m/s. As the projectile travels 1 meter from firing point, a force of 0.298 N is being applied to the projectile to accelerate it. Assuming there's no wind resistance, calculate:

1. The speed of the projectile after it's being accelerated

2. The range at which the projectile hit the ground (like an actual bullet, it's affected by gravity at long range), compare it after being accelerated with as if it isn't accelerated

3. If the projectile is being changed with the one with a mass of 30 gram, what's the speed after it's being accelerated and it's maximum range?"

I've been spending 3 days to think this through and tried to solve it and I cannot find an answer for this problem T T. So if anyone have a solution/ formula/ answer for this, I'm really appreciate it

And thanks for helping

Hi, can i see your design

It looks interesting thought

 

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...