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Basic momentum and magnitude questions!

  1. Dec 9, 2009 #1
    1. A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend
    with a constant force of 47 N. How far must the student be pushed, start-
    ing from rest, so that her final kinetic energy is 354 J ?
    Answer in units of m.

    for this i did 354/2 which i got 177 then i divided by 47 . is that correct?

    2 (part 1 of 2) A 4.78 g bullet moving at 510.8 m/s penetrates a tree trunk to a depth of 4.93 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet.
    Answer in units of N.

    for this i did KE=1/2mv^2 and i found the answer and did f=ma. is that right??

    3. (part 2 of 2) b) Assuming that the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment the bullets stops moving.
    Answer in units of s.

    for this one i just did vf^2=vi^2+2at. is that correct??

  2. jcsd
  3. Dec 9, 2009 #2
    The first part is wrong. You need to use the following formula:

    [tex]Work = Force * Distance[/tex]

    So it should just be 354/47, giving you 7.5 meters.

    I can't tell from what you've written if you're right on part 2 or not, so here's how you do it. You need to set the kinetic energy of the bullet equal to the work done to stop it, which you can relate to force -- the answer you're looking for. Mathematically:

    [tex]\frac{1}2mv^2 = W = Fd[/tex]

    You know the mass and the velocity of the bullet. You know the distance d it travels inside the trunk. Solve for F and you should get about 12,650 newtons (but check my work).

    For part 3....knowing the force it took to stop the bullet, you can figure out the acceleration from F = ma. Once you know the acceleration, you put it into the following kinematics equation:

    [tex]x = v_it + \frac{1}2at^2[/tex]

    where t is the time you need to solve for. You know x (distance bullet travels inside tree), you know initial velocity (510.8 m/s), and you know the acceleration that you calculated above. Solve for time t, but you will need the following quadratic equation:

    [tex]1323106.69t^2 + 510.8t - .0493 = 0[/tex]

    You will get two values: 8 x 10^-5 seconds or -4.66 x 10^-4 seconds. You have to dismiss the second value on physical grounds (can't have negative time). So you should get get about 8 x 10^-5 seconds. Again, check my work, but that answer makes sense intuitively. The bullet should stop really quickly given the enormous force opposing its motion.
  4. Dec 9, 2009 #3
    ohh i see. okay i understand part 1 completly but not part 2.
    so i have to find time right? where did you get 1323106.69 from like i got 2646213. hmm. im confused.lol
  5. Dec 9, 2009 #4
    It's just divided by two to get a simpler constant in front of the t^2.
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