# Projectile Motion Question HELP

1. Dec 15, 2011

### duhfluteplaye

Projectile Motion Question!! HELP!!

1. The problem statement, all variables and given/known data

There is a marble (15g) in a spring shooter that is shot at 30 m\s at 55 degrees. The ball is on the spring shooter for .01s.

2. Relevant equations

What is the change in momentum right as it is shot?
What is the force?
What is the Vi,y?
What is the Vi,x?
How high is the apogee?
What is the total delta X?
What is the gravitational potential energy at the apogee?
What is the kinetic energy?
What is the spring constant?
What is the rebounce force?
What is the force to reload?
What is the work done on the marble?

3. The attempt at a solution
well i got the Vi, y and x by doing 30sing55 and 30cos55.
I'm guessing the change in momentum is .015(30sin55) because it's only in the y direction.

2. Dec 15, 2011

### Staff: Mentor

Re: Projectile Motion Question!! HELP!!

You'll need to show some more effort on all the parts of the question before we can see how to help you.

Regarding the change in momentum, since there is a velocity change in both x and y directions there is momentum change for both (components). You can simply consider the change in speed rather than deal with the individual components.

3. Dec 15, 2011

### duhfluteplaye

Re: Projectile Motion Question!! HELP!!

well for change in momentum i did .015*30sin 55 and then .015*30cos55 and got both components.
For the force i did change in momentum/time so .015*30sin55 and then divided by .01.
For the apogee i did vf^2=vi^2+2adeltax so 30sin55 squared+2(9.81) and then solved for delta x.
for the total x i found time by using vf=vi+at and then multiplied by the velocity in the x direction 30cos55.
for the potential i did .015*9.81*height of the apogee.
I believe that kinetic and potential are the same so i said kinetic was the same as potential.
For the spring constant i've found the force to be 36.86 (Step 2 above) and i'm lost as to how to find x.
I do not understand what they mean by rebounce force. Maybe Newton's 3rd law? so it would equal the kinetic energy?
I also do not understand what they mean by force to reload. (force applied to compress spring same displacement maybe?

Sorry I didn't give enough information before. We just got a new teacher and apparently our old teacher taught us everything wrong (she was a biology major) so now I have about 10 hours to relearn everything before our final. :P

4. Dec 15, 2011

### Staff: Mentor

Re: Projectile Motion Question!! HELP!!

So what is the total change in momentum? As I said, you can skip the components and use the change in speed in the calculation. Works out the same, but it's much simpler.
You need to use the total momentum change, not just one component. The spring is providing a momentum change in BOTH vertical and horizontal directions, so the force is necessarily not just in the vertical direction.
Show your results. What value did you use for acceleration? What did you get for time? For x?
No, the gravitational potential energy considers only motion in the vertical direction. The marble has kinetic energy in the horizontal direction, too. Why not use the initial speed of the marble to find KE?
I suspect that the force you found above (ignoring for a moment the fact that it disregarded part of the momentum) would be the AVERAGE force applied by the spring over the period of time that it is in contact with the marble. So it may be of limited use in finding the spring constant without knowing the total displacement (as you are hinting).

Perhaps you might consider that the marble and spring will form a mass-spring system which will have a natural oscillation period. From maximum compression to when the marble loses contact with the spring (at the equilibrium point of the motion) will be a quarter period. You should have a formula for the period of a spring-mass oscillator.
I have never heard the term "rebounce force" before. So I'm a tad mystified, too. If it's a force, it'll have Newtons as units.
Yes, that's right. For that you'll need to know the distance that the spring needs to be compressed.

5. Dec 15, 2011

### duhfluteplaye

Re: Projectile Motion Question!! HELP!!

I got .45 kgm/s for change in momentum.
I got (after just using the initial velocity instead of both components) 45N for the force.
I got 30.72 m for the height of the apogee.
For acceleration i used 9.81 since it is free fall. I got 5 s for time (2.5 up to the apogee and the same amount back down). I got 86.16m for the total x.
For potential i got 4.52 J and for kinetic i got 6.75 J.
We have never done anything like that with spring constants. The absolute only thing we have ever learned/talked about was conservation of energy and we received the formula .5kx^2 (I'm in high school physics).
I'm assuming I would find distance of compression by using the spring constant found above?

6. Dec 16, 2011

### Staff: Mentor

Re: Projectile Motion Question!! HELP!!

That looks good so far!
I see. The work done on the marble to give it its KE will be the average force (that you calculated above) multiplied by the distance traveled while the spring goes from compressed to relaxed (w = Fav*Δx). That should let you calculate the Δx for the spring.

The average force is the NET average force that's applied to the marble. It will be the sum of the spring's force and the component of the gravitational force acting along the line of motion (much like a block on a slope problem). The actual force delivered by the spring needs to be larger than Fav by that amount in order to make the net come out to Fav. Let's call this "actual spring average force" Fs. I leave it to you to work out the value of the gravitational force component that needs to be added to Fav to yield Fs.

For the spring constant, the force supplied by the spring at full compression will be twice the average force (springs are linear: F = k*Δx, so if the average force over the whole motion of the spring is Fs, then it's 2*Fs at max compression and zero when it's relaxed). So given the maximum compression Δx from above, you have 2Fs = Δx*k. So you can find k. Whew!