# Homework Help: Finding initial velocity of a projectile. Help with force and acceleration

1. Aug 12, 2012

### Sony95

Hello everyone,
I was doing some experiment using a spring compressed gun, that shoots up ball beraings. I made it using a desoldering pump and a PVC pipe,its quite small. ok, so I set it in 45° angles for maximum range. A desoldering pump has a spring inside, when you push down on top the spring compresses and stays compressed until you click on the button which releases it. This force of the spring getting released exerts a force on the ball bearing causing it to launch into the air. so, i used different bearing balls with same diameter but different weights. The force exerted by the gun is the same so I thought the initial velocity is constant. But later i found Newton's 2nd law stating F=ma
so if F is constant, and mass is increased, 'a' will change resulting in a change in Initial velcoity (Vi) beacuse, a=(Vf-Vi)/t
so later i found hooke's law F=kx
F=ma
a=(vf-vi)/(tf-ti)
putting them all together i got an equation to find the initial velocity of a projectile
-vi=(kχx/m)(t)-Vf
where, k is the spring constant(which i found) , x is the displacement of spring due to compression(which i can see) and time(i used a stopwatch) and vf is zero as it is the maximum point of proejctile. ok so i worked it out for each mass of ball bearing and got Vi for four ball bearings which were all the same (6m/s).
Can some one check if this equation and the way I am doing this is right beacuse when u use the same intial velcoity for launching a ball aren't all the ball bearings suppposed to land in the same position and have the same flight time.
But, when I did the actual investigation, the time were different and the distance were different too . I am totally going paranoid with this problem . some one please help me! its due in 3 days and I am not even half way through.

Thank u so much for your help!!!!!!!!!!!!!

2. Aug 12, 2012

### Simon Bridge

You did your calculations in the absence of air resistance :)
On top of that, small jolts in the setup will send the balls to different places... the recoil of the gun (conservation of momentum) will be different for different masses.

3. Aug 12, 2012

### Sony95

Hi Simon Bridge, :)
Oh true. I did it in the absence of air resisitance , but I have no idea how to do it in the presence of air resistance. So, the recoil of the gun (conservation of momentum) will be differnt for different masses. Does that mean I will not be able to have the same Initial velcoity. And can you please also check if the equation that I did is the right way to find the Vi. Sorry, if I am asking too much because my brain is dead. Can you please explain it in a little bit more detail.
Thank you very much for the quick reply,
Sony

4. Aug 12, 2012

### Sony95

ok . I checked this website <http://www.bsharp.org/physics/recoil> [Broken]
and understood what gun recoil was. Ok, so every action has an equal and opposite reaction. So when you launch something, the force taken for launch will cause a recoil in the spring loaded gun. Causing it to move back. But in this case, the gun is fixed to a table and the forces used for launching are all below 20N. Would this force cause the gun to have some serious recoil?

Last edited by a moderator: May 6, 2017
5. Aug 12, 2012

### Sony95

the answer that I got from that equation says the ball bearing's initial velcoity is 6m/s . well, I thought about that now and that doesn'y make any sense beacuse that is pretty fast 600cm/s . I am sure this spring loaded gun will not go more than 1.5m/s.
I have been working on this for the whole week and I have no idea what to do next. Everything is going wrong!!!!!!!!!!!!!!!!!!!!!!!11

6. Aug 12, 2012

### Simon Bridge

I bet the gun barrel vibrates when you fire it.
This is due to the recoil. Loose components too. The vibration introduces a random uncertainty to your initial velocity ... if you arrange for the balls to land in something that records where they hit (sand tray, on a bit of carbon-paper, something) and launch the one ball lots of times, you'll see a pattern.

The ball will also roll a bit in the tube on the way out, try firing a cylindrical slug.

If you want to see the air resistance effect - fire the gun horizontally and compare that time of flight with just dropping the ball from the same height.

It does not sound like it's "going wrong" as such, it's just that the real world is happening to you. Welcome to experimental science :) experiments never go the way you expect.

The way to find the expected muzzle-speed is to use conservation of energy: energy stored in the spring = kinetic energy on release. (You'll get some losses - noise, friction so that's really a > sign).

The ballistic equation just have acceleration (gravity) acting on the down component of the velocity... however, drag (air resistance) acts against the whole motion, and tends to depend on a power of the velocity .... Then it gets complicated... spin can create lift, for example.

But you can have a lot of fun comparing theory with practise ... will your projectile reach terminal velocity? for eg. What is the relationship between mass and range?

BTW: If you can - a useful way to check the muzzle speeds is to use a photogate.
You could even build one from a LED and a photodiode, but you need some way to record the waveform... one trick is to turn your computer's sound card into a digital oscilloscope: monitor the wave via the mic socket.

7. Aug 13, 2012

### Sony95

Thank you so much for that very fine explanation, it was really helpful. I really appreciate that fine detailed explanation. I understood why the ball didn't hit in the same spot all the time. Its due to air resistance, may be friction and the energy from the spring is given out in different forms. Its good that I understood what was going on because it was quite confusing. No, but I don't have to asses sound waves and muzzle speeds for this experiment as it not required. But I am sure, it will be fun and I will try to do it after I hand the assignmnent in. Once again thank you very much.
Sony

8. Aug 13, 2012

### Simon Bridge

<sigh> OH well, my fault: I'd hoped you'd try the calculation hints. But I wasn't clear.

You have said:
... which cannot be true.

Conservation of energy: $$\frac{1}{2}kx^2=\frac{1}{2}mv^2 \Rightarrow v=x\sqrt{\frac{k}{m}}$$... that's a bit simpler than yours and assumes $v_i=0$

But yours also has muzzle-velocity inversely depending on projectile mass.

This means that the greater the mass, the lower the muzzle-velocity, at the same compression, even in the absence of air resistance or friction or any of the other things. You should see a reduced range with higher mass.

What is the assignment?

9. Aug 14, 2012

### Sony95

Ok I'll get my head around that concept of conservation of momentum andtry to add it in my assignment. Beacuse right now we are doing projectile motion. It is a highschool physics assignment.

10. Aug 14, 2012

### Simon Bridge

Conservation of energy :)

That's HS level - and important to ballistics (projectile motion) ... the initial, vertical, kinetic energy gets exchanged for gravitational potential energy - when $mgy=\frac{1}{2}mv_{y0}^2$ then y is the maximum height reached ;)

But I meant - what is the assignment supposed to have you investigate?
Were you supposed to determine the muzzle-velocity, experimentally, from the range? $v_0=\sqrt{gR}$ for 45deg elevation :)
You can ace the assignment by determining k for the spring experimentally and comparing the two results?

Last edited: Aug 14, 2012
11. Aug 15, 2012

### Sony95

ok it is supposed to investigate how the weight of an object affects the time of flight.
Yes I were supposed to determine the Initial velocity and k, the spring constant I did. I actually did another equation which seems to give reasonable Vi. By the way, the assignment is due tommorow.
Thank you

12. Aug 15, 2012

### Simon Bridge

No worries and good luck ... and if you find yourself tinkering with the experiment after the assignment is over: there is no hope for you ;) you're a scientist.

13. Aug 15, 2012

### Sony95

Thank you. And ha ha i am not that smart to be a scientist. :shy:
U seem really smart, I bet ur a scientist :)

14. Aug 15, 2012

### Simon Bridge

Huh, I just know how to avoid the dumber mistakes because, guess why ... makes me look smarter than I am ;) "Scientist" is something you are not something you do.

15. Sep 8, 2012

### Sony95

Hello again, sorry for the very late reply, was busy with other subjects. I have a very high achievement in the physics assignment I was doing. :)
Thank you so much for helping me.

16. Sep 10, 2012

Well done.