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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

Respuesta :

Answer:u=-22 m/s

Explanation:

Given

mass of puck [tex]m=0.06\ kg[/tex]

Average force [tex]f_{avg}=1.5\times 10^3\ N[/tex]

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

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

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

therefore

[tex]1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i[/tex]

[tex]P_i=0.06\times 8-1.8[/tex]

[tex]P_i=0.48-1.8=-1.32\ kg-m/s[/tex]

Final momentum [tex]P_f=m\times v_f[/tex]

[tex]P_f=0.06\times 8[/tex]

[tex]P_f=0.48\ kg-m/s[/tex]

Impulse on the ball [tex]=F_{avg}\Delta t[/tex]

Impulse[tex]=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s[/tex]

Initial velocity is given by

[tex]u=\frac{P_i}{m}=\frac{-1.32}{0.06}[/tex]

[tex]u=-22\ m/s[/tex]

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