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3 years ago

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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 forc

e 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

1 answer:

kramer3 years ago

8 0

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 $F_{avg}\Delta t$ $=\text{change in momentum}$

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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Likurg_2 [28]

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1)

We can use the following equation:

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2)

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4)

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