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Kinetic Energy, Force, and Collisions

  1. Nov 19, 2006 #1
    A 49 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically and returns to its original height. The time interval for a round trip is 7.00 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
    A.4490 N----b. 3360 N------------c. 7850 N-----d. 6720 N------e. 5590 N

    If t_total = 7.00 s and time of collision is 0.5 ms*(1 s/1000 ms) = 5*10^-4 s, then 7.00 s – 5*10^-4 s = 6.9995 s, which is the time that is not dedicated to the contact time.

    Since the ball rebounds, it takes 6.9995 s/2 = 3.49975 s for the ball to travel from rest to the plate another 3.49975 s for the ball to travel from the plate back to the original height.

    F = m*(delta v)/t_contact

    To find the ball’s velocity, I though of the conservation of energy,
    Where v = sqrt(2*g*h).

    I also thought of the kinematic equation v_f = v_o + a*t in which a = 9.8 m/s^2 if the ball is dropped.

    Now, would v = 9.8 m/s^2*(3.49975 s) = 34.29755 m/s??? (same value if h = 0.5a*t^2 used)

    F = (.049 kg)*(34.29755 m/s)/(5*10^-4 s) = 3361.2 J ??????




    A 18 g bullet is shot vertically into an 10 kg block. The block lifts upward 9 mm. The bullet penetrates the block in a time interval of 0.001 s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:

    a.490 J
    b.0.88 J
    c.250 J
    d.330 J
    e.0.0016 J

    First off, I was puzzled at why a time interval was given. “Does it possibly have something to do with the impulse?” I thought.

    Here is what I ended up doing. Please check to see if the setup is correct.

    Inelastic collision: m_bullet*v = (m_bullet + m_block)*v’

    For block: For conservation of energy, mgh = 0.5*m*v^2 ??
    v = sqrt(2*g*h) = sqrt(2*9.8*0.009 m) = 0.42 m/s

    So v_block = v’?

    V_bullet = [(m_bullet + m_block)*v’]/[ m_bullet] = [(0.018 kg + 10 kg)*0.42 m/s]/[0.018 kg] = 233.753 m/s

    KE = (mv^2)/2 = 0.5*(0.018 kg)*(233.753 m/s)^2 = 492 J (choice a.)

    Thanks.
     
  2. jcsd
  3. Nov 19, 2006 #2

    PhanthomJay

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    Yes, nice work on both. In the first problem, i wouldn't worry about that 0.5ms in comparison to the 7000ms flight time . In question 2, the impact time is not needed, but what really is not needed is the assumption that the force on the bullet is constant, since it most definitely is not.
     
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