How can I find work, horizontal force, and time contact?

[athena810]

Homework Statement

For a physics project, i need to answer a bunch of questions after conducting an experiment with projectile motion. I have answered most of them. The questions with the work shown are here: http://i.imgur.com/JmYPMC9.png

I cannot answer number 10 and 11, because I don't have the values for Force or acceleration. However, I'm pretty sure there's a way to do it because this is a project that people have done before.
10. What is the horizontal force you exerted on the ball?
11. What is the amount of time you were in contact with the ball when throwing it?

I'm also not sure if question 9 is correct about Work. I figured W = KE, but I'm not sure if that's correct.

Homework Equations

W = Fd = mgh = change in KE
Impulse (J) = p' - p = mv' - mv = Ft

The Attempt at a Solution

W in the horizontal direction = F* d = F (15.5m) = (J/t)(15.5m) = (19.705 Nm)/t)(15.5m)...
W in vertical direction = mgh = (.055kg)(9.81 m/s^2)(1.226 m) = .66 J

Horizontal Force = m * a = (.055kg)(0 m/s^2)... except this does not seem right. So i think I have to find this value from impulse which is Ft. I do not have time.

Amount of time in contact with the ball... I can only think of getting this from impulse. But i do not have the force.

 

[micole]

Can you clarify the experiment? It is called Projectile but you indicate the ball only moved horizontally? In order to complete the problem correctly, one needs to know if this was a ball sliding on the ground or a parabolic path in the air

 

[athena810]

micole,

It was parabolic in the air.

 

[BvU]

What a nice exercise. physics and physical at the same time.
If you don't know, you have to find out, or make an educated guess.
For example, when you throw you start next to your shoulder and let go with a stretched arm. That would be a certain distance, for example 0.8 m. Make an approximation that the force you exert is constant over that 0.8 m and then you have a value for F from your first eqn and for t from the second !

I also have a few questions:

In your report the last line is often a rather coarse rounding off; is there a good reason for that?

Under 2 there is mention of a $\theta$. What is $\theta$ ?

Under 3, is $d_{y_{max}}$ really 0 ? 0 with respect to what ? And does it continue down for another 1.2 m ? Or is your shoulder height 1.2 m ?

Under 4, where does the launch angle appear in the expressions? I only see a $\theta$ mentioned at the very end, as a sort of comment.

Under 6, I suddenly see a 13.1 m/s pop up. What is that and where does it come from ?

Under 7, 31 and 5 m/s are squared and added up, the result is a meagre 30 m/s ?
Also under 7: you can not compute the inverse tangent of a speed. Only of e.g. a ratio of speeds.

Under 8,9: rounding off to 1 digit is too coarse

 

[micole]

Parabolic projectile

Athena - you indicate you threw a ball at a parabolic path. I recommend you draw a parabolic path on your paper from left to right. The starting angle theta θ, is not zero since you indicate the ball arcs.

How can you find θ? Once you solve for initial vx and vy, you will know the adjacent and opposite sides of the right triangle with an angle θ, so what arc-trig function can you use to find theta?

Once you release the ball, are there any forces acting In the x direction? We neglect air resistance ...if there are no forces acting, then vx initial = vx final, right? From your table you correctly identified the vx is range/time elapsed.

Now you need vy. Here is a calculation trick: call the time elapsed from start to finish 2t (instead of t). I believe you had .5 seconds elapse, so that's 2t, so t is half that. Since parabolas are symmetrical, that time t (.5/2) is the time it takes for your ball to reach its peak. At that moment right when the ball is about to come back down, what is vy? One moment vy is positive and decreasing, next moment it's negative, so for one moment at the peak, vy =?
Now you want vy initial, you have vy at peak which I'll call vy final, and you have acceleration -9.8m/s^2 and time t (.5/2). So what equation can you use to get vy initial?

Now you have vx and vy from the moment you release the ball. You correctly use Pythagorean to get the magnitude initial vector v, well done on that method. Just need to update the numbers.

How about the height reached at time t=.5/2 seconds? You know vy initial now, and vy final is that speed right before the ball turns around, so you know that...you know time elapses .5/2 and you know acceleration is -g. You have the right equation on your he but wrong numbers. Try again now

#4 is not asking vx it's asking v and angle. You should now know v (like in your #7, but going up vs down) and θ we addressed with trig above

Now that you have v right, #5 should follow. You have the right equation

For 6, did we decide there are forces acting in the x plane?if not, vx stays the same. Vy...well luckily parabola are symmetrical so now the ball heads down instead of up but you can see by your diagram the angle, and the magnitude by symmetry is the same as when you let the ball go

7 covered above

8 - right equation. Update now that you have v right

9 yes the work you put in is the initial and final KE of the ball...one caveat is, here I am assuming you have no hEight. You would have a more complex problem if you incorporate your height..all the same ideas and equations but you go beyond the symmetry and allow gravity to keep acting...giving where you got into trouble I think you are probably to the level of assuming symmetry.

10 and 11. the horizontal force you exerted f= Am and m is the mass of the ball, a is from zero to vx (you already calculated) in however many seconds till you released the ball. For example if the ball weighs .5kg and vx is 5m/s and were in contact with the ball throwing it for 3 sec, then am is .5kgx(5m/s-0m/s)/3 seconds =2.5/3 N. I think you need a sense of how long you were in contact to answer

I hope that helps!

 

[athena810]

For example, when you throw you start next to your shoulder and let go with a stretched arm. That would be a certain distance, for example 0.8 m. Make an approximation that the force you exert is constant over that 0.8 m and then you have a value for F from your first eqn and for t from the second !

I don't understand, do i just time the amount of time I swing my arm and the length of my arm?

In your report the last line is often a rather coarse rounding off; is there a good reason for that?

significant figures, the time i got was .5 seconds which is 1 significant figure.

Under 2 there is mention of a $\theta$. What is $\theta$ ?

Direction. Vectors require both magnitude and direction. It is 0 because it is relative to the ground.

Under 3, is $d_{y_{max}}$ really 0 ? 0 with respect to what ? And does it continue down for another 1.2 m ? Or is your shoulder height 1.2 m ?

With respect to the height of launch. I threw it horizontally as in my arm is parallel to the ground and i accelerate the ball by swinging my arm. It would be 0 because it never gained height i think.

Under 4, where does the launch angle appear in the expressions? I only see a $\theta$ mentioned at the very end, as a sort of comment.

I tried to measure theta with respect to the ground but I guess there would be no theta since it changes as the ball moves. The question asks for the launch angle though so since my arm is parallel to the ground, launch is 0 degrees.

Under 6, I suddenly see a 13.1 m/s pop up. What is that and where does it come from ?

Oh, that is a typo. it should be 31 m/s. Good catch, thanks.

Under 7, 31 and 5 m/s are squared and added up, the result is a meagre 30 m/s ?

Yeah, it's 31.4 m/s.

Also under 7: you can not compute the inverse tangent of a speed. Only of e.g. a ratio of speeds.

The inverse tangent was to find theta? Because theta changes so I need to find the angle the ball makes with the floor.

Under 8,9: rounding off to 1 digit is too coarse

Don't i have to because 30 m/s is 1 digit and the answer must be the same significant digit of the least precise.

Thanks

 

[athena810]

Athena - you indicate you threw a ball at a parabolic path. I recommend you draw a parabolic path on your paper from left to right. The starting angle theta θ, is not zero since you indicate the ball arcs.

I don't quite understand...is there no theta then? And I did not really throw it in the parabolic path. I thought it would be easier to throw it horizontally like this: http://awwapp.com/s/e5/a7/af.png

How can you find θ? Once you solve for initial vx and vy, you will know the adjacent and opposite sides of the right triangle with an angle θ, so what arc-trig function can you use to find theta?

But the initial Velocity in y direction is 0 m/s...or at least I think that's what I was taught. it is not?

Now you need vy. Here is a calculation trick: call the time elapsed from start to finish 2t (instead of t). I believe you had .5 seconds elapse, so that's 2t, so t is half that. Since parabolas are symmetrical, that time t (.5/2) is the time it takes for your ball to reach its peak. At that moment right when the ball is about to come back down, what is vy? One moment vy is positive and decreasing, next moment it's negative, so for one moment at the peak, vy =?

0 m/s?

Now you want vy initial, you have vy at peak which I'll call vy final, and you have acceleration -9.8m/s^2 and time t (.5/2). So what equation can you use to get vy initial?

a = (Vf - Vi)/t -> t * a - Vf = - Vi -> Vi = Vf - at?
So Vi= at = 9.81 m/s^2 * .5/2 = 2.4525 m/s?

10 and 11. the horizontal force you exerted f= Am and m is the mass of the ball, a is from zero to vx (you already calculated) in however many seconds till you released the ball. For example if the ball weighs .5kg and vx is 5m/s and were in contact with the ball throwing it for 3 sec, then am is .5kgx(5m/s-0m/s)/3 seconds =2.5/3 N. I think you need a sense of how long you were in contact to answer

Ok so this is something that i need to measure? I cannot just find this from what I know now? And contact with the ball as in how long i hold the ball before i throw it or how long it takes for me to throw the ball?

Thanks.

 

[BvU]

Hello again. What a pile of stuff! I go one by one until I really, really have to go to bed.

For example, when you throw you start next to your shoulder and let go with a stretched arm. That would be a certain distance, for example 0.8 m. Make an approximation that the force you exert is constant over that 0.8 m and then you have a value for F from your first eqn and for t from the second !

I don't understand, do i just time the amount of time I swing my arm and the length of my arm?

No for the first one, yes for the second: I meant you measure the length of your arm. You have an initial horizontal velocity (under 1.: 31 m/s) so you have a kinetic energy change. Using the first equation

2. Homework Equations
W = Fd = change in KE (here not mgh because we are talking horizontal here) v is horizontal (v_{i y} = 0), F is horizontal (I hope).
Impulse (J) = p' - p = mv' - mv = Ft​

you find F. With the second you find t

 

[BvU]

In your report the last line is often a rather coarse rounding off; is there a good reason for that?

significant figures, the time i got was .5 seconds which is 1 significant figure.

Actually, you did five measurements that average out to 0.50 +/- 0.03 seconds (the standard deviation is 0.07 s). Good reason to keep at least two digits.

Generally we keep significant digits (or significant digits + 1 if the first is a 1). And we make sure that calculations don't make it worse (or make it look unjustifiably more accurate).

In your case, the range is a bit more worrying: 15.5 +/- 2 would be my best attempt.

I am puzzled why you change from v_{i x} to d_{y max} for the same thing...

2.: I would stick to 31 m/s.
3.: Ok, I understand the 0 m. for d I would report -1.23 m (first digit is a 1).
4.: Is a repetition of 2 that makes me suspect that your instructions did ask you to somehow do something with the launch angle.
5.: 1.71 Nm Good to have when we are at 11.
6. Looks like a typo: 31 m/s can not change into 13.1 m/s. Vertically 4.9 m/s.
As for $\theta$: if $\tan \theta = {-v_{f, y} \over v_{f, x} }= - {4.9\over 31} = \tan (-9^\circ)$, I would write $\theta = 351^\circ$; how do the 270 come about ?

 

[athena810]

Hi,

You have an initial horizontal velocity (under 1.: 31 m/s) so you have a kinetic energy change. Using the first equation

2. Homework Equations
W = Fd = change in KE (here not mgh because we are talking horizontal here) v is horizontal (v_{i y} = 0), F is horizontal (I hope).
Impulse (J) = p' - p = mv' - mv = Ft​

you find F. With the second you find t

So how does measuring my arm get the Kinetic Energy? Sorry, I'm really confused.

Actually, you did five measurements that average out to 0.50 +/- 0.03 seconds (the standard deviation is 0.07 s). Good reason to keep at least two digits.

Generally we keep significant digits (or significant digits + 1 if the first is a 1). And we make sure that calculations don't make it worse (or make it look unjustifiably more accurate).

In your case, the range is a bit more worrying: 15.5 +/- 2 would be my best attempt.

So I should put the measurements at 2 significant digits?

As for θ: if tanθ=−vf,yvf,x=−4.931=tan(−9deg), I would write θ=351deg; how do the 270 come about

Oh, it asked for horizontal and vertical and not together, i should have put theta = 0 for Vx and theta = 270 for Vy.

 

[BvU]

So how does measuring my arm get the Kinetic Energy? Sorry, I'm really confused.

Kinetic energy = mass times (initial horizontal initial speed) squared. This has been brought about by you exercising a force over a distance, approximately equal to the length of your arm. So
KE = Force x length of arm Divide KE by length of arm and you have an estimate for force.

Then divide initial horizontal momentum by force and you have time

 

[BvU]

So I should put the measurements at 2 significant digits? Yes. 3 if the first is a 1.

I am puzzled why you change from vi x to dy max for the same thing...

For which question? The question 1. where you report vi x and then continue with dy max . It makes no sense.

 

[BvU]

Bedtime for me. 2:30 AM and I have to get to work at 9.

 

[athena810]

Bedtime for me. 2:30 AM and I have to get to work at 9.

Alright, thanks!

 

[micole]

replies to projectile follow ups

Please see below:

I don't quite understand...is there no theta then? And I did not really throw it in the parabolic path. I thought it would be easier to throw it horizontally like this: http://awwapp.com/s/e5/a7/af.png
ah okay then its not parabolic. All the same ideas apply, but you're starting with vy = 0 as you indicate below. You can think of it as if your hand was the top of the parabola, and now the ball has to fall down the back side of the path to the ground.

But the initial Velocity in y direction is 0 m/s...or at least I think that's what I was taught. it is not?

0 m/s?
yes based on the image you sent

a = (Vf - Vi)/t -> t * a - Vf = - Vi -> Vi = Vf - at?
So Vi= at = 9.81 m/s^2 * .5/2 = 2.4525 m/s?
on the y direction:
vi = 0 as you say.
correct formula above but not used exactly right:
Vf = Vi + at, where you know:
-vi is zero
-a is -9.8m/s^2 (negative, not positive, right? because its toward the ground and presumably you are treating up as positive, down as negative)
- you have t in your table as .5 seconds
so Vf =...you can do this part...just substitute


Ok so this is something that i need to measure? I cannot just find this from what I know now? And contact with the ball as in how long i hold the ball before i throw it or how long it takes for me to throw the ball?
My belief is you cannot measure this, because you impart momentum mv to the ball by applying force F for time Δt. You can only know what F*Δt is, cannot know one without the other. And since you didn't wear a force measuring device (: Δt is easier.

Hope that helps

Thanks.