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How fast does the rocket leave the ramp?

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A 1200 kg rocket car is placed at the bottom of a 100 metre long ramp inclined at 10 degrees. The rocket is turned on and it exerts a force of 8300 N for 5.7 seconds, and then it shuts down. Find the speed with which the rocket car leaves the ramp (assume frictionless)


    2. Relevant equations

    http://img98.imageshack.us/img98/9425/54fl1.png

    3. The attempt at a solution
    i missed a class involving kinematics and dynamics together so im not entirely sure how to solve it
     
  2. jcsd
  3. Feb 27, 2008 #2
    So you DO understand kinematics, it's the dynamics with Newton's laws?

    So here you have a force being exerted of 8300 Newtons. Your equation there, which was effectively the most important equation in physics from about 1666 to the late 1800s(damn you Maxwell)says the NET force is equal to the mass times the acceleration

    Now this rocket car has two forces acting on it. One is the force of gravity, acting down the ramp, can you find that? It's not JUST mg, it's the component of mg acting down the ramp

    The second force is the rocket, which is just 8300 Newtons, neato

    So then Fgravity+Frocket=Fnet (Frocket is obviously positive, Fgravity, acting in the opposite direction, will be negative)

    Once you find Fnet, you know that and mass, so you can find acceleration. So then you'll have a constant acceleration and you'll know for how long, and you can solve that type of problem I think. THEN the rocket cuts off, and you can find the speed when it cuts off and the distance it has traveled to that point. I'm guessing it'll still be less than a 100 m so you'll have to then look at JUST the force of gravity(which will be the same from earlier)and find that acceleration(which will be negative this time)

    Then you'll have acceleration again, and you'll know how much farther to the end of the ramp so you'll have distance, and you'll know the initial speed, so another of those

    Don't be intimidated, it's just two problems that are like the ones you've done a million times before, the only difference is you have to use F=ma to find your own acceleration this time for the two different parts(rocket on and off)
     
  4. Feb 27, 2008 #3
    thanks it took me 15 or so minutes but i got it :smile:
     
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