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Homework Help: Bullet on a Block(Momentum)

  1. May 10, 2008 #1
    1. The problem statement, all variables and given/known data
    The muzzle velocity of a gun found at a crime scene is tested by firing the 100 g bullet
    into a block of mass 9.9 kg. The block is initially at rest on a frictional surface,
    coefficient of 0.1. The bullet sticks into the block and the combination slides a total
    distance of 4.5 metres along the surface. Calculate the muzzle velocity of the gun.

    2. Relevant equations
    This collision is inelastic.
    Fy = 0 = Fn = Fg = mg(Force in y)
    Ff = uFn = (.1)(9.9kg)(9.8m/s^2) = 9.703N (Friction Force)
    Fx = ma = (-Ff) + (Fa)

    vx = dx/t

    mBvB + mLvL = v'(mB + mL) (B - bullet, L - block)
    mBvB = v'(mB + mL) (vL = 0)

    3. The attempt at a solution
    I know the frictional force, if I could find a way to figure out the v' in the momentum equation, I could figure everything out. Now I know the displacement, if there was some way I could find the time, or even acceleration. I think I might have ideas on how to taclke the problems, but I can't get too far, I am still going to try, if anyone could help that would very well appreciated. Thank You.

    I am missing a peice of the puzzle to finish the question.

    1. The problem statement, all variables and given/known data
    Two titanium spheres approach each other head-on with the same speed and collide
    elastically. After the collision, one of the spheres, whose mass is 300 g, remains at rest.
    What is the mass of the other sphere?

    2. Relevant equations
    Momentum Conserved

    m1v1 + m2v2 = m1v1' + m2v2' (v2' = 0, v1 = v2, m1 = .3kg)
    v(m1+m2) = m2v2'

    m1v1^2 + m2v^2 = m2v2'^2(should I factor?)
    v^2(m1 + m2) = m2v2'^2

    3. The attempt at a solution
    Right now I have a feeling that I should be isolating for m2 or something so I can cancel it out when I sub into another equation.
    Last edited: May 10, 2008
  2. jcsd
  3. May 10, 2008 #2

    Doc Al

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    Staff: Mentor

    What's the work done by friction? How does that relate to the post-collision speed?
  4. May 10, 2008 #3
    The work done by friction is opposite to that of the applied force. Since the friction is overcome, then work done will be opposite force of friction * the displacement correct.

    W = F*d = 9.702N * 4.5m = 1/2mv^2

    I think thats how I can figure out the velocity post collison.

    Heres what I work out:

    v = sqrt(2*Fd/mT) = 8.73m/s... So I get a post collision velocity of the combination of masses of 8.73m/s.

    Solving for v1 = v'(mB+mL) / mB = 8.73m/s(10kg) / .1kg = 837.2m/s...for a bullet his answer sounds about right. So thank you very much.

    Just one more to go.
  5. May 10, 2008 #4

    Doc Al

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    Staff: Mentor

    Be sure when you calculate the friction force and the kinetic energy that you use the mass of the entire "bullet + block". Otherwise, looks good.
  6. May 10, 2008 #5

    Doc Al

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    Staff: Mentor

    Careful. While the initial speeds are the same, they move in opposite directions.

    Good. Combine this with the momentum equation (after you correct it) and the speeds will cancel out.
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