Help with basic physics problem.

[jameslat]

Homework Statement

If a tennis ball with a mass of 57 g is shot out of a spring cannon at a 45 degree angle and goes 10m find the answer to the following questions:

What is the velocity.
What spring constant must the cannon have.
What would be the constant converted into lbs/in.

Homework Equations

FS=-kx
vf^2=vi^2 + 2ax
x=(vi^2(sin^2(theta)))/-a
F=ma

The Attempt at a Solution

ok i started off with finding the vi
10=vi^2(.5)/9.8
vi^2=100
vi=10

and with that then i can now try and find the acceleration of the spring right?
from 0 to 10m/sec

vf^2=vi^2+2ax
10^2=0+2ax
50=ax
50/x=a
first question is this x referring to x as how far i must pull back the string or the distance that i will be shooting at? (10 m)

f=ma
F=-kx
so
ma=-kx
.057(50/x)=-kx
2.85/x=-kx
2.85=-kx^2

good so far?

then at this point I am supposed to guess a K and use that.
so a 20 lbs/in has 2.2597 N/m
( http://www.boltscience.com/pages/convert.htm )

2.85/2.2587=x^2
so x = 1.12 m
so i would have to pull in back a meter?

if i set k to 31
2.85/31=x^2
x=.30 which is abt a foot right?
but 31 N/m is like almost 300 lbs/in spring!

what am i doing wrong?

Thanks for any input.
-James

 

[cupid.callin]

ok i started off with finding the vi
10=vi^2(.5)/9.8
vi^2=100
vi=10

Isnt that .5 wrong? and you even didnt used it. Its sin2θ

vf^2=vi^2+2ax
10^2=0+2ax
50=ax
50/x=a
first question is this x referring to x as how far i must pull back the string or the distance that i will be shooting at? (10 m)

You cannot used it as spring force is not constant and therefore nor is accleration. Use energy conservation

 

[jameslat]

so should Vi=9.9 then? (if sin(45) = 1)

ok how could i do that then? Like what formula would I need to use?

Could i use ma=-kx?

 

[cupid.callin]

so should Vi=9.9 then? (if sin(45) = 1)

velocity is 9.9 but still sin(45) remains 1/√2 ... theformula is this: R = (u² sin2θ)/g

ok how could i do that then? Like what formula would I need to use?

Could i use ma=-kx?

well it would be (1/2)mv² = (1/2)kx²

but still you will have two unknowns.

Are you sure you are not missing some detail of question?

 

[jameslat]

This isn't like a normal home work problem.
It's like one of those that your teacher gives at the end of the week that is super hard for some extra credit.

All I was told was that "If someone has a spring cannon that could shoot 10 m at a 45 degree angle what type of spring would need to be in this cannon, and how far back should they pull it."

Basically I am to try and make up a spring poundage, just as long as it works.

with
(1/2)mv2 = (1/2)kx2

i know the bass, and i know the end velocity, could i just make up a k until i find out a realistic x?
(i'm also assuming the v is 9.9 velocity right?)

thanks again.

 

[SammyS]

Homework Statement

If a tennis ball with a mass of 57 g is shot out of a spring cannon at a 45 degree angle and goes 10m find the answer to the following questions:

What is the velocity.
What spring constant must the cannon have.
What would be the constant converted into lbs/in.
...

-James

Hi James.

The projectile is launched at a 45° angle.  → (v₀)_x = (v₀)_y = (v₀)/√(2) .

The time to reach max height is (v₀)_y /g, so that the time from launch to impact (assuming this is at the same elevation as the launch) is: t = 2·(v₀)_y /g .

Since the horizontal component of velocity, v_x , is constant, v_x = (v₀)_x. The distance traveled horizontally (in units of meters) is: 10 = (v₀)_x·t .

Substitute the previous expression for t into the above expression for horizontal distance. Plug (v₀)/√(2) in for both (v₀)_x and (v₀)_y , then solve for v₀ .

For the spring constant, use conservation of energy as cupid.callin mentioned. Do this by comparing energy immediately before and immediately after launch. However, to find the spring constant, you must know how much the spring was compressed. I don't see this anywhere in your post!

Your spring constant likely will be expressed in units of N/m i.e. Newtons/meter. Convert Newtons to lbs. and meters to feet.

 

[jameslat]

so i should have my work converted to lbs/ft?

can i ask you just off the top of your head what would be a probable estimate of what lbs/ft spring i would need?

ok so let's just say 20 lbs/in converted = 240 lbs/ft
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2=240x^2
2.793/240=x
so x=.011?

however this answer doesn't make sense to me because you said i have to know how much it was compressed already?

thanks for any input.

 

[SammyS]

so i should have my work converted to lbs/ft?

can i ask you just off the top of your head what would be a probable estimate of what lbs/ft spring i would need?

ok so let's just say 20 lbs/in converted = 240 lbs/ft
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2=240x^2
2.793/240=x
so x=.011?

however this answer doesn't make sense to me because you said i have to know how much it was compressed already?

thanks for any input.

[STRIKE]The most recent value that I saw you had for the launch velocity was incorrect. [/STRIKE] Correction: Your answer WAS correct. Sorry! DUH

Physicists seldom work out problems in lbs & ft. or inches.

 

[jameslat]

So your saying 9.9 m/sec is wrong to begin with?
Just a reminder I'm a HS student and my teacher just got done teaching us basic 2-d kinematics and now is moving into springs and forces and half the equations that r being presented to me I never even seen before in my life.

"Physicists seldom work out problems in lbs & ft. or inches.
Report Post Reply With Quote"
are you saying i should keep them into N/m then?

I'm good at solving math problems, but sometimes (like when I don't know what to solve for) I get stuck.

Is this answer even doable in a real world way? Or do you think my teacher just gave us an impossible problem?

How would you solve it? (please don't give an answer but tell me step by step what you would do and give the formula that you use for each step)

thanks again.

 

[jameslat]

ok the fact is I'm trying to build one of these at home and I want to find out what type of spring to buy.
and my math didn't make any sense when it said 247 pound/in spring so i was hoping to find the right answer with you guys.

But i guess to my dilemma no one can help me unless if I have more information.

Just answer me this, would a 20 lbs/in 2.7 N/m work to shoot a tennis ball 10 yards away at a 45 degree angle? I'd only be able to pull it back a max of 1 foot.

 

[SammyS]

So your saying 9.9 m/sec is wrong to begin with?
Just a reminder I'm a HS student and my teacher just got done teaching us basic 2-d kinematics and now is moving into springs and forces and half the equations that r being presented to me I never even seen before in my life.

"Physicists seldom work out problems in lbs & ft. or inches.
Report Post Reply With Quote"
are you saying i should keep them into N/m then?

I'm good at solving math problems, but sometimes (like when I don't know what to solve for) I get stuck.

Is this answer even doable in a real world way? Or do you think my teacher just gave us an impossible problem?

How would you solve it? (please don't give an answer but tell me step by step what you would do and give the formula that you use for each step)

thanks again.

Our messages crossed paths! I did edit my last message to say you were correct.

Just estimating, I would think that k = 50 to 200 N/m would be reasonable. I figured pulling back about 0.3 meters (about 1 foot) would be reasonable - that gives k in the low 50's. A sling-shot might be a good example. Probably would take less than a 1 foot pull for 10 meters with a tennis ball, That's how I came up with the range of 50 - 200 N/m.

Convert that to lb/in or lb/ft & I'll get back to you.

 

[cupid.callin]

According to your teacher;s question ... he's asking you both K and x

So in my opinion, find K for average springs available (or even use any K which is not too small or large) and find x for it!

 

[jameslat]

ok 50 N/m = 442.53 lbs/in (and vice versa)

@cupid.callin
so with (1/2)mv2 = (1/2)kx2
and
mass=.057kg
V=9.9 m/s
k=50
I'd be write to do the equation and get
(1/2).057*9.9^2 = (1/2)50x^2
x=.334 Would this be accurate?

Thanks again (things are starting to get clearer)

 

[SammyS]

ok 50 N/m = 442.53 lbs/in (and vice versa)
...

I get that 50 N/m = 3.43 lbs/ft = 0.286 lbs/in

Spoiler

So if you want to pull the spring somewhere between 3 to 12 inches, the spring should be somewhere in the 1/4 to 1 lbs/in range.

1 N ≈ 0.22481 lbs

1 m ≈ 39.370 in

 

[jameslat]

http://www.boltscience.com/pages/convert.htm

why is it when I try and input my work with this calculator (above)
I get a huge error (like 300 lbs/in)?also would this be right then?
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2 = 25x^2
x=.3342

so I would only need to pull it back .334 of a meter with a .286 lbs/in spring and it will travel 10 m according to the above launch variables?

 

[SammyS]

http://www.boltscience.com/pages/convert.htm

why is it when I try and input my work with this calculator (above)
I get a huge error (like 300 lbs/in)?

That website converts Torque. N×m ←→ ft×lbs (or in×lbs)

Torque is in units of force × length.
Spring constant is in units of force / length.

also would this be right then?
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2 = 25x^2
x=.3342

so I would only need to pull it back .334 of a meter with a .286 lbs/in spring and it will travel 10 m according to the above launch variables?

No, the 0.286 lbs/ft is for a 12 inch pull.

A spring with 4 times that, ≈ 1.1 or 1.15 lbs/in is for a 3 inch pull.

Depending upon the design of your launcher, the spring must also propel some moving parts to 9.9 m/s. Also, the calculation is for an ideal spring, so the mass of the spring will also come into play, so will friction in the launcher mechanism. Since a tennis ball has such a small mass, these other factors may overwhelm the energy needed to launch the ball at the proper speed.

Isn't this getting complicated? !

A rubber band (under stretching conditions) acts much like a spring. (I still like the slingshot idea.)

 

[jameslat]

what if i shaved my tennis balls (don't laugh) and lubed the pipe with a silicon based oil? (like they use for cleaning gun barrels, like waxing it) that would minimize friction wouldn't it?

and yes it is getting just a bit.
so pulling it back 12 inches gives a .285 lbs.

So if a spring says it's 20 lbs/in and it's 1 foot long then i need to pull it back the whole foot to get the 20 pounds of force?

 

[jameslat]

so if i have a 20 lbs/in spring that is one foot long i need to push it 1 foot back to get my 20 pounds of force?

 

[SammyS]

so if i have a 20 lbs/in spring that is one foot long i need to push it 1 foot back to get my 20 pounds of force?

You pull it back 1 inch to get 20lbs, 2 inches for 40 lbs, etc.

inch × (lbs/inch) → lbs .

 

[jameslat]

ok so tell me if i do this right
if i have a spring that has 1 lbs/in which equals 175.12 N/m
and if I put that into this equation.
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2=.5(175.12)x^2
then that x=.1786
which is roughly 7.03 inches.

Is this all correct?

 

[cupid.callin]

i shaved my tennis balls

LOL. Nice one!

_________________________________
Why do you need to convert spring constant?
Its best to use anything in SI
Did your teacher ask you to do that?

 

[jameslat]

i thought i had to convert it to N/m
much like you have to convert 57g to .057kg right?

 

[SammyS]

ok so tell me if i do this right
if i have a spring that has 1 lbs/in which equals 175.12 N/m
and if I put that into this equation.
(1/2)mv2 = (1/2)kx2
.5(.057)9.9^2=.5(175.12)x^2
then that x=.1786
which is roughly 7.03 inches.

Is this all correct?

Looks good.

 

[jameslat]

MOST EXCELLENT!
Thank you all so much for your time and patients:P