i need help on a science project it deals with a car going down a ramp and then lauching an egg 10ft help me!!

Question

9ft ramp 3 ft after ramp to stop and has to com to a complete stop before launching an egg over a 2 1\2 ft wall and to a target 10ft away please help me!!!

Answer 1

The car rolls down a ramp that is 9 ft long and angled at 30 degrees from horizontal. The car carries an egg and some kind of launcher.

3 ft from the bottom of the ramp, the car must stop.

After stopping the car then launches an egg over a 2 ½ foot wall to land on a target 10 feet away.
Where is the wall?

When you let me know the location of the wall I will adjust. For now I will place the wall 5 feet from the target towards the launch point.

Here's a picture
http://i142.photobucket.com/albums/r88/odu83/ballistic.jpg


First, the car will roll down a ramp and stop three feet past the bottom. What will stop the car? I would also suggest that you drag something that will slow the car to a stop at the appropriate point.
So figure out the coefficient of friction of what you drag. It could even be strips of old car tire or bicycle tire tube that makes a trailer behind the car. Set it up so you can adjust the surface area and add small weights to it since the coefficient of friction is related to the normal force.

First, look at what happens when the car rolls down the ramp:

The loss in potential energy of the total mass of the car plus egg plus trailer will be M, and
PE=M*g*h, where h in this case is the sin(30)*9 feet (the sine of 30 is .5, so h=4.5 feet) You may also want to adjust h for the center of mass of the car and trailer since that will be more accurate.

While the trailer is on the ramp, the normal force on the trailer will be cos(30)*m*g, where m is the mass of the trailer that is creating the friction so the car will stop
Multiply this by the coefficient of friction to compute the work done by friction, which is the amount of energy that will be removed from the system and dissipated as heat.
Let’s call this fwork(ramp)= µ*cos(30)*m*g*9.

Now we can compute the Kinetic energy of the car at the bottom of the ramp, let’s call it KE1:
PE-fwork(ramp)=KE1

With KE1 in hand, the next step is to compute the frictional work necessary to stop the car 3 feet from the bottom of the ramp:
Since a stopped car has KE=0, then
KE1-fwork(flat)=0, where fwork(flat)= µ*m*g*3. note that KE1= fwork(flat)
So
PE=fwork(ramp)+ fwork(flat)
Bringing in the variables:
M*g*h= µ* cos(30)*m*g*9+ µ*m*g*3
Simplify
M*h= µ*m*(cos(30)*9+3)
Or
µ*m=M*h/(cos(30)*9+3)
Given all this, now you can experiment with m and µ to make it all work.

I conceived an experiment. For each trailer you choose, you can compute µ by releasing the car down the ramp and measuring the stopping distance from the bottom of the ramp.

Have two different weights for the trailer m1 and m2. The stopping distances will be d1 and d2.
µ1=M*h/((cos(30)*9+d1)*m1)
µ2=M*h/((cos(30)*9+d2)*m2)

Average the two to compute µ. I actually recommend that you run the experiment until you find the m that will stop the car at d=3 ft. If you need to, create more trailers to find what will work best (including helping the car roll straight).



Now that you have the car stopped, the next step is to launch the egg.
The key variables are the muzzle velocity of the egg and the angle al of the cannon.
If I were building from scratch I would make a slide for the egg with a spring-loaded launcher. I would allow the angle and the compression of the spring variables. Let’s get back to that in a moment.

The requirements are that the egg flies 10 feet and land on a target. It must also fly over a 2-1/2 foot wall that is 5 feet before the target (again, adjust to the actual position since I assumed the wall position)
Since you are launching an egg, the muzzle velocity has to be low enough that the force of launch doesn’t break the egg. Just a note of caution.

Here we use the equations of motion for each component of the flight.

Horizontally, the flight time to target is t1, where 10=vm*cos(al)*t1
So t1=10/(vm*cos(al))
Note that vm is the muzzle velocity

Vertically, the flight time t1 is expressed as
0=h+vm*sin(al)*t1-.5*g*t1^2
Where h is the height of the muzzle.


The egg must also clear the wall, so t2*cos(al)*vm=5 and 2.5
These four equations must all be satisfied for the egg to clear the wall and hit the target. It looks complicated, huh?

Start by setting al= 45 degrees and see if there is a valid vm. This is the minimum vm.
Sin(45)=cos(45)=sqrt(2)/2
So t1=20/(vm*sqrt(2))
And 0=h+vm*sqrt(2)*t1/2-.5*g*t1^2
0=h+10-g*100/vm^2
So vm=sqrt(100*g/(h+10))

Now, does the egg clear the wall?

Let’s assume the h is about 3 inches or .25 feet
Vm= 17.7 ft/sec or 12 mph – that’s nice and slow
Check
t1=20/(17.7*sqrt(2))
t1=.8 seconds

0=.25+
17.7*.8*sqrt(2)/2-.5*32*.8^2

=0.018 close enough

t2=10/(sqrt(2)*17.7)
t2=.4

2.5<.25+17.7*sqrt(2)*.4/2-
.5*32*.4^2

2.5< 2.69 ft so it will clear the wall by 2.28 inches

Let’s get back to the launch. When the spring of the cannon releases there will be a recoil of the car. It would be nice if the car recoiled a minimum. A typical chicken egg weighs about 54 grams or .12 lbs.
The momentum after the launch will be
M-me*vc=me*vm*cos(al)
Or me/(M-me)=vc/(vm*cos(al))
From above and including the chicken egg weight (note that each ratio is unit-less, so we can use weights on the left)
(.12/(M-.12))*
17.7*sqrt(2)/2=vc


If the car weighs 100 times the egg, then M=12lbs
And vc= .06*(17.7*sqrt(2))/11.88
So vc=.12 ft/sec
Why do you care? Because if you look at muzzle velocity of the cannon without subtracting the backward velocity of the car, you will be slightly off. So the real velocity of the egg with respect to the floor will be the muzzle velocity of the cannon minus vc. Also note that vc has an inversely proportional relationship to M-me. So the lighter the car, the greater the vc.

Answer 2

Gravity and Inertia i presumed approximately this. in case you have a loose falling merchandise, you already know how briskly it features momentum, and it will proceed to earnings that momentum till slowed by way of friction. If there is not any friction then the motor vehicle might proceed to improve up on the fee of wich an merchandise possibilities up velocity because of the gravitational pull of the earth.