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Impulse/Change in Momentum Problem

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A major league catcher catches a fastball moving at 95mi/h and his hand and glove recoils 10.0 cm in bringing the ball to rest. If it took 0.00470 seconds to bring the ball (with a mass of 250g) to rest in the glove,

    What is the magnitude and direction of the change in momentum of the ball?
    What is the average force that the ball exerts on the hand and glove?

    2. Relevant equations

    p = mv Δ(mv) = F*t 95.0mi/hr = 42.5 m/s m = 250g = 0.25 kg F=ma

    3. The attempt at a solution

    I attempted to answer the first question by finding the initial momentum

    P = (0.25)(42.5) = 10.625

    I don't know where to go from here
  2. jcsd
  3. Jan 6, 2009 #2
    Try J=F*Δt

    J=Impulse aka the answer to the first question
    Δt= Change in time

    You have Δt but still need force.



    You have mass but still need acceleration


    Δv=change in velocity
  4. Jan 7, 2009 #3
    actually... you dont need acceleration

    Recall Impulse : F x Δt = ΔP [Change in momentum] = mΔv

    therefore to find Force we just F = ΔP / Δt = [(m_1)(v_i) - (m_1)(v_f)] / Δt

    essentially anyways... make sure to take into account all momentums... anyways.. u have ur initial momentum of the ball which is ur ball travelling at 95 miles an hour?

    so intial momentum = (42.5 m/s)(0.25kg) = 10.625 Joules

    assuming momentum is conserved, and the ball isnt bouncing backwards, pf = pi therefore, magnitude and direction is equal and opposite

    part 2. force exerted on the glove

    F x Δt = ΔP = mΔv
    F = [(m_1)(v_i) - (m_1)(v_f)] / Δt - Final velocity is zero so we're left P_i
    = [(0.25kg)(42.5m/s)] / 0.00470seconds
    = 2260.6 N -- after sig figs adjustment ---> 2300 N

    Therefore, force exerted on the glove is 2300 N =)

    hope that helps, that was fun ^^ more fun then kinematics = = blech
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