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natali 33 [55]
3 years ago
10

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
Physics
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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This problem involves Newton's universal law of gravitation and the equation to follow would be.

F = GM₁M₂/r²  

Given: M₁ = 0.890 Kg;  M₂ = 0.890 Kg;  F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²

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temperature and mass

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3 years ago
What type of friction is using chalk in the summer to draw on the ground in Copley square?
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A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
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u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a =  average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
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Also
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Answer:
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