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Please help - Motion of a hockey puck

  • Thread starter Demmy
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  • #1
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Please help -- Motion of a hockey puck

Homework Statement


Hi guys, can you please help me answer this problem?


A hockey puck with mass 0.160 kg is at rest at the origin on the horizontal, frictionless surface of the rink. At time t=0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; he continues to apply this force until t=2.00 s. (a) What are the position and speed of the puck at t=2.00 s? (b) If the same force is again applied at t=5.00 s, what are the position and speed of the puck at t=7.00s


I got the 1st one, but I couldn't get the second one.


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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welcome to PF!

You need to show your work before we can help. So why not show us how you solved for (a) and where you got stuck for (b).
 
  • #3
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Given: F=0.250 N
m=0.160 kg
F=ma

a=F/m

leads to:

a=0.250/0.160

i got a= 1.5625

then I used the formula a=(2s)/(t^2) to get the distance

then I got s=3.13 m



then, couldn't solve for b, please help
 
  • #4
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Btw, here are the answers

a) 3.13m, 3.13m/s
b) 21.9m, 6.25m/s
 
  • #5
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extend (a) to 5 secs what is dist and velocity then these become the initial conditions in your eqns.

so at 2 secs the force ends and the puck glides along at a constant velocity (assume no friction) for 3 secs (so you're now at 5 sec mark)

vfinal = vinitial + a * t where t =7 - 5

dfinal= dinitial + vinitial * t + 1/2 a * t
 
  • #6
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ahm, sir, where would I get my vinitial?
 
  • #7
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ahm, sir, where would I get my vinitial?
Apply Newton's first law of motion.
 
  • #8
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Should V initial at 5s be any different than V at 2s? Can you find V at 2s?
 
  • #9
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sir, I got the final velocity

I converted a= 1.563 m/(s^2) to m/s

so I got v= 3.13 m/s

then I used it as my vinitial to get my vfinal using the equation you gave me, my vfinal= 6.256 m/s

but when I tried to substitute that to the 2nd equation you gave me, I didn't got the correct answer.
 
  • #10
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@Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir
 
  • #11
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I have another question, I got letter a) but why is the unit I got is m/(s^2) the answer should be m/s, but I got the right answer. so I just randomly thought about converting the acceleration to velocity, because the unit of acceleration is m/(s^2) and the velocity is m/s, but if I do that in a) the unit would be correct but the answer would be wrong.
 
  • #12
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sir, I got the final velocity

I converted a= 1.563 m/(s^2) to m/s

so I got v= 3.13 m/s

then I used it as my vinitial to get my vfinal using the equation you gave me, my vfinal= 6.256 m/s

but when I tried to substitute that to the 2nd equation you gave me, I didn't got the correct answer.
try not to say you converted an acceleration in to a velocity... You may find a velocity using an acceleration etc.

V = V0 + a*t
V from pt A is good for V0 for pt B why?
 
  • #13
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@Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir
Good.. Because no force is acting on the puck from 2s to 5s.
 
  • #14
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The position in pt B comes from three distances traveled while moving. Pt A distance contributes some, then some more while moving at a constant speed, then more while accelerating again. Find the position for pt B from those pieces and you'll do fine.
 
  • #15
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ahm sir, sorry bout the convertion thingy, I don't get is sir.
 
  • #16
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I tried to find the distance for 5s, so I used the same acceleration


so my equation is (1.563*(5^2))/2

then I got 19.5375
 
  • #17
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ahm sir, sorry bout the convertion thingy, I don't get is sir.
Maybe I'm just being picky. When people say it that way I start thinking they're missing some algebra/units/equation concepts.

Is English not your first language?
 
  • #18
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I tried to find the distance for 5s, so I used the same acceleration


so my equation is (1.563*(5^2))/2

then I got 19.5375
5^2 is an error here. Has the puck been accelerating for 5 seconds?
 
  • #19
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Check my post 14 again. Do pt B by adding three things. Pt A answer, d=r*t during the 3 seconds of no acceleration, and last add the distance traveled while accelerating for 2 seconds with a starting velocity of ptA velocity.
 
  • #20
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I have another question, I got letter a) but why is the unit I got is m/(s^2) the answer should be m/s, but I got the right answer. so I just randomly thought about converting the acceleration to velocity, because the unit of acceleration is m/(s^2) and the velocity is m/s, but if I do that in a) the unit would be correct but the answer would be wrong.
I can't say for sure why you got acceleration units.

Do unit analysis to see what it should be, given the values you added or multiplied.
 
  • #21
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no sir, english is my third language
 
  • #22
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it didn't say that the the puck has stop moving, it just said that same force is applied at t=5s
 
  • #23
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4,731


you have three states in this problem

1) at t0 for 2 secs puck accelerates to velocity v2 and travels dist d2

2) at t2 to t5 puck glides at constant velocity (ie v2) distance d5 = d2 + v2 * (3 secs)

3) at t5 puck accelerates to velocity v7 = v2 * a * (2 secs) and dist d7 = d5 + v2 * (2 secs) + 1/2 a * (2 secs)
 
  • #24
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sir, I tried all of your equation I didn't got the correct answer.

and I have a question in number 2 in the equation "d5 = d2 + v2 * (3 secs)" isn't that suppose to be "a" or acceleration instead of "v" velocity? because I checked it.
 
  • #25
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it didn't say that the the puck has stop moving, it just said that same force is applied at t=5s
Correct. The puck does not stop moving.
 

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