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20 December, 10:48

how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 * 10-3 s. Find (a) the final momentum of the ball, (b) the impulse imparted to the golf ball, and (c) the average force exerted on the ball by the golf club.

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  1. 20 December, 13:15
    0
    Answer: a) 12857.1 m/s/s b) 578.6 N

    Explanation:

    Impulse = change in momentum

    Ft = mV2 - mV1

    V = AT, 45 /.0035 = 12857.1 m/s/s

    (b).045 x 12857.1 = 578.6 N
  2. 20 December, 14:43
    0
    a) Final Momentum = P2 = 2.025 kgm/s

    b) Impulse = J = 2.025 kgm/s

    c) Favg = 578.57 N

    Explanation:

    Mass of the golf ball = m = 0.045 kg

    Initial speed = V1 = 0 m/s

    Final Speed = V2 = 45 m/s

    Time = t = 3.5 * 10^-3

    a) Final Momentum = P2 = mV2 = (0.045) (45) = 2.025 kgm/s

    b) Impulse = J = change in momentum = ∆P = mV2 - mV1

    J = (0.045) (45) - (0.045) (0) = 2.025 kgm/s

    c) Impulse = Favg * t

    Favg = J/t = 2.025/3.5 * 10^-3

    Favg = 578.57 N
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