How Far Does a Goalie Slide After Catching a Puck?

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A hockey player sends a .4kg puck at 80 m/s. A stationary golie cathces the puck. If the coefficient of friction of the golie on the ice is .05 and his mass is 90 kg how far does he slide on the ice before he comes to a stop.

I have found
normal force = mg
= 885.92N
and then friction = .05(885.92N)
= 44.296N
and using conservation of momentum i found the velocity after the collision will be 0.35m/s
but now I am stuck
please help.

 

[cepheid]

Welcome to PF Loading!

It's good that you've shown an attempt, but in the future please use the template provided.

A hockey player sends a .4kg puck at 80 m/s. A stationary golie cathces the puck. If the coefficient of friction of the golie on the ice is .05 and his mass is 90 kg how far does he slide on the ice before he comes to a stop.

I have found
normal force = mg
= 885.92N
and then friction = .05(885.92N)
= 44.296N
and using conservation of momentum i found the velocity after the collision will be 0.35m/s
but now I am stuck
please help.

So, you have the initial velocity. Given the frictional force that acts, what is the acceleration? So, how far will he get before he slides to a stop?

Hint: you can solve this just using kinematics. You have v_f, v_i, a, and you want d. Can you think of any equation with these things in them?

Note: haven't checked your actual numbers.

 

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okay so final velocity will be 0 and initial is .35m/s but then i don't have acceleration.
i also have to account for the friction.

eqn: Vfinal^2 - Vinitail^2 = 2a x d

 

[haruspex]

okay so final velocity will be 0 and initial is .35m/s but then i don't have acceleration.
i also have to account for the friction.

eqn: Vfinal^2 - Vinitail^2 = 2a x d

You know the mass, so you know the normal force. You know the coefficient of friction, so you know the frictional force. Doesn't that give you the acceleration?

 

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but that's giving me the acceleration due to gravity...man I am so confused abt this question

 

[cepheid]

but that's giving me the acceleration due to gravity...man I am so confused abt this question

No, you computed the friction force in your original post. The friction acts horizontally. It accelerates the player horizontally, to a stop. We're not talking about acceleration due to gravity. We're talking about acceleration resulting from that frictional force.

So, for friction: you have the force (F), you have the mass (m). How can you find the acceleration (a)? It's like one of the most fundamental laws of mechanics that will let you do so...

 

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Friction= (m x a)x the coefficient ?

 

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okay so this is what i have upto now:
mass of puck = .4kg
Speed of puck = 80m/s
Mass of goalie= 90kg and he is at rest
μ = 0.05

i found the common velocity after the colision to be 0.35 m/s
due to vertical equilibrium i stated that Fg=Fn
Fn=mg
Fn=885.92N
and then Friction = μ x Fn
Friction = 44.296N

 

[haruspex]

So the acceleration is?

 

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9.8?

 

[Chestermiller]

This is good so far. Now, make a list of the horizontal forces acting on the guy. There should only be one item on this list. This is the net force acting on the guy. So, if you apply Newton's second law to the guy, what is his mass times horizontal acceleration? What is his initial velocity? You need to assume that, when his horizontal velocity slows to zero, static friction takes over, and he remains stopped.

 

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oh ok...ahah wow...thanks for all the help guys :D