Shoot a Rocket off a Cart and through the Hoop

  • Thread starter Phoenixtears
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    Cart Rocket
In summary: V0 is the initial velocity, Vf is the final velocity, and at the point of launch, V0+at=Vf. So the acceleration is 7.5*F/(V0+at).
  • #1
Phoenixtears
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Homework Statement


A 380 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 7.5 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. At what horizontal distance left of the loop should you launch?
____m



Homework Equations



Rolling force= (coeffictient of rolling force(Mu))*(N)
F=Ma
Delta-x= V0*t + .5a(t^2)
Vf^2= V0^2 + 2ax
Vf= V0 + at

The Attempt at a Solution



I don't know how to attack this problem, but this is how I started. I first drew a force diagram that involves normal, weight, and rolling friction (involving the cart). But then I assumed that a completely separate force must be used for the rocket, because I'm assuming that the rocket is not strapped to the cart. So, that would involve normal, weight, thrust, and drag. Then I found the Mu first (even though it is technically known, I've seen problems that use weird coeffictients). In other words, 7.5= (9.8*380)*(Mu). Mu= .002- this involves the cart. I drew a table for delta-x, inital V, final V, acceleration, and time. I know that I solving for horizontal x and that I have verticle x (20). But what else is there to solve the problem. Technically rolling friction wouldn't be involved at all, correct? But then why should we know it. I can't seem to get from the 20 meters to anything else. Any suggestions?

Thanks in advance!

~Phoenix
 
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  • #2
New thoughts:

I believe that we have the initial velocities for hor. and vert.. Hori: 3 because that is the constant velocity we are given, and the vert. one is 0 because the cart is not moving vertically.

Also, can we say that we can figure out the acceleration? Because F=Ma, 7.5=380a, then a=.01974. But this can be said of neither vertical nor horizontal. So what does this acceleration represent?

I don't know where to go from here... is this information correct?
 
  • #3
You need to take into account the acceleration due to gravity.
 

Related to Shoot a Rocket off a Cart and through the Hoop

What is the purpose of shooting a rocket off a cart and through the hoop?

The purpose of this experiment is to demonstrate the principles of Newton's laws of motion and projectile motion.

What materials are needed for this experiment?

The materials needed include a cart, a rocket, a hoop, a launchpad, and a measuring tape.

How do you set up the experiment?

First, place the launchpad on a flat surface and secure it in place. Then, set up the hoop at a suitable height and distance from the launchpad. Next, load the rocket onto the cart and position it at the launchpad. Make sure the cart is aligned with the hoop. Finally, ignite the rocket and release the cart at the same time.

What are the safety precautions for this experiment?

This experiment should be conducted in a safe and open area, away from any obstacles. Always wear safety goggles and gloves when handling the rocket. Keep a safe distance from the launchpad and do not stand in front of the cart during the launch.

What factors can affect the success of the experiment?

The success of the experiment can be affected by various factors such as the angle and height of the launch, the weight of the rocket, and any external forces or wind. It is important to control these variables to achieve accurate results.

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