A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x

Question

A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball

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RobertKer 6 months 2021-07-20T02:52:33+00:00 1 Answers 4 views 0

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    2021-07-20T02:53:41+00:00

    Answer:u=-22 m/s

    Explanation:

    Given

    mass of puck m=0.06\ kg

    Average force f_{avg}=1.5\times 10^3\ N

    time of contact t=1.2ms=1.2\times 10^{-3}\ s

    puck leaves with a velocity of v=8\ m/s

    We know impulse is F_{avg}\Delta t[tex]=\text{change in momentum}[/tex]

    therefore

    1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i

    P_i=0.06\times 8-1.8

    P_i=0.48-1.8=-1.32\ kg-m/s

    Final momentum P_f=m\times v_f

    P_f=0.06\times 8

    P_f=0.48\ kg-m/s

    Impulse on the ball =F_{avg}\Delta t

    Impulse=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s

    Initial velocity is given by

    u=\frac{P_i}{m}=\frac{-1.32}{0.06}

    u=-22\ m/s

    i.e. initially ball is moving towards -x-axis

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