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Kinetic energy of a bullet

  1. Oct 7, 2003 #1
    A bullet with a mass of 4.90 g and a speed of 609 m/s penetrates a tree to a depth of 4.07 cm. Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped.

    I thought I could use the equation time = distance divided by velocity, but that's not right. Apparently the answer is 1.33e-4 and I can't figure out how to get it!
  2. jcsd
  3. Oct 7, 2003 #2


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    You CAN'T use "time = distance divided by velocity" because there is no constant velocity.

    My first thought was to use "energy". At first the bullet has kinetic energy (1/2)m v2= (1/2)(0.0049)(609)2= 909 Joules (I converted mass from grams to kg). At the end it has no kinetic energy. The "constant friction force" must have done work equal to that, and since the force IS constant, that is "force times distance". You can use that to find the force: F*0.0407= 909 so
    F= 909/0.0407= 22326 Newtons. Once you know the force, you can find the acceleration (f= ma): 22326= 0.0049a so a= 22326/.0049= 4556278 m/s2, and then determine the time for the velocity to reduce to 0: 4556278 t= 609 so t= 609/4556278= 0.00134= 1.34 x 10-4 seconds.

    Another way, MUCH simpler (I thought of it AFTER I did the above!) is to use "average velocity". As long as the acceleration is constant, which it is here because the friction force is constant, the average velocity is just "(initial velocity+ final velocity)/2" which, here, is (609+ 0)/2= 304.5 m/s.
    At that average velocity, the bullet will require 0.0407/304.5 =
    0.000134= 1.34 x 10-4 seconds to penetrate 0.0407 m.
    (And we didn't need to use the mass!)
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