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Linear Momentum Problem

  1. Apr 7, 2005 #1
    A 0.010-kg bullet traveling horizontally at 400.0 m/s strikes a 4.0-kg block of wood sitting at the edge of a table. The bullet is lodged into the wood. If the table height is 1.2 m, how far from the table does the block hit the floor?

    this is what i did:

    lets the mass of bullet = m2
    the mass of block of wood = m1
    velocity of the block of wood before the collision take place = v1
    velocity of the bullet = v2
    velocity of the block and bullet after the collision take place = vf
    momenta of the block = p1
    momenta of the bullet = p2

    the momentum of block and bullet after collision equals the sum of momenta of the block and the bullet before the collision:

    pf = pi

    The initial momentum is the sum of the momenta of the block and bullet

    pf = p1 + p2

    Since the block is initially at rest, p1 = 0. Then

    pf = p2
    (m1 + m2)(vf) = (m2)(v2)

    Solving for vf:

    vf = [(0.010)(400.0)]/(4.0+0.010) = 0.9975 m/s

    And I am kind of stuck here because I'm not sure how to calculate the distance from the table to the point that the block hits the floor. If it is the projectile motion formula, I guess I'm jsut not sure what to use and what to plug in.
  2. jcsd
  3. Apr 7, 2005 #2
    If I am interpreting the question correctly then vf is the initial horizontal velocity of the block after it beings to fall. To work out the horizontal distance between the table and the point at which the block hits the ground you can use x_h = v_h*t, where t is the time taken for the block to hit the ground and the horizontal velocity, v_h is constant(v_h = v_f). To find the value of t consider the block's veritcal motion. The initial vertical velocity is zero, acceleration is 9.8 m/s² down and the vertical displacement is 1.2 metres down. Just plug in the values, using correct signs, into s = ut + (1/2)at², solve for t and substitute back into x_h = v_h * t.
  4. Apr 7, 2005 #3
    this helps a whole lot. Thank You! But what does the s in the final equation represent?
  5. Apr 7, 2005 #4
    s is often used as displacement, which is what Benny is using it to represent.

    [tex] s = ut + \frac {1} {2} at^2 = v_o t + \frac {1} {2} at^2 = x [/tex]
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