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Homework Help: Speeding Bullet

  1. Nov 23, 2003 #1
    A 50-g bullet is shot into the 2-kg can of a ballistic pendulum. The can rises to a height of 1.3m Determine the speed of the bullet just before the collision.

    Don't know how to begin.
     
  2. jcsd
  3. Nov 23, 2003 #2
    I know you say that you don't know how to begin, but I'm hoping you see that this is a momentum problem.

    IMO momentum problems are best done by drawing a free body diagrams: one immediatly before an event, one immediatly after (or during) an event and one at the end of the event as felt be the whole system (ie a baseball just prior to being hit by a bat, a baseball while (or just after) being hit by the bat, and the final distance traveled by the baseball). Once you have a good set of FBD's analyze the energies at each point of the overall event.

    That should get you going.
     
  4. Nov 23, 2003 #3
    Since the collision is completely inelastic, momentum is conserved.
     
  5. Nov 23, 2003 #4

    HallsofIvy

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    Science Advisor

    Conservation of energy. (You appear to be doing problems from a chapter on "conservation of energy!)

    Initially, the bullet with mass 50 g= 0.05 kg has (unknown) speed v. It's kinetic energy is (1/2)(0.05)v2= 0.025v2. Taking the height of the bullet and block at the moment of impact to be 0 potential energy, since the block is not moving, the total energy of bullet and block is the kinetic energy of the bullet: 0.025v2.

    The bullet and block together rise to a height 1.3 m above the base height, and have 0 speed there. Their potential energy is (0.05+2)(9.8)(1.3)= 26.117 Joules and is the total energy.

    Solve 0.025v2= 26.117.
     
  6. Nov 23, 2003 #5
    I think if something hits and sticks, energy is not conserved.
     
  7. Nov 24, 2003 #6

    joc

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    you'd need both momentum and energy equations for this question. letting

    Mb = mass of bullet,
    M = total mass of can and bullet,
    Vb = velocity of bullet just before collision,
    V = velocity of bullet+can immediately after collision,
    h = height of rise,

    we can write 2 equations:

    (momentum) MbVb = MV

    (energy) (1/2)MV^2 = Mgh

    solving for Vb, we get

    Vb = MV/Mb
    Vb = (M/Mb) sqrt(2gh)
    Vb = 2.05/0.05 rt(2x9.81x1.3)
    Vb = 207 ms^-1
     
    Last edited: Nov 24, 2003
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