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Please help - Motion of a hockey puck

by Demmy
Tags: hockey, motion, puck
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Demmy
#1
Aug28-12, 11:41 AM
P: 12
1. The problem statement, all variables and given/known data
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.


2. Relevant equations



3. The attempt at a solution
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jedishrfu
#2
Aug28-12, 11:49 AM
P: 2,804
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).
Demmy
#3
Aug28-12, 11:53 AM
P: 12
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

Demmy
#4
Aug28-12, 12:17 PM
P: 12
Please help - Motion of a hockey puck

Btw, here are the answers

a) 3.13m, 3.13m/s
b) 21.9m, 6.25m/s
jedishrfu
#5
Aug28-12, 10:10 PM
P: 2,804
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
Demmy
#6
Aug28-12, 11:11 PM
P: 12
ahm, sir, where would I get my vinitial?
azizlwl
#7
Aug28-12, 11:22 PM
P: 963
Quote Quote by Demmy View Post
ahm, sir, where would I get my vinitial?
Apply Newton's first law of motion.
Jakeus314
#8
Aug28-12, 11:25 PM
P: 48
Should V initial at 5s be any different than V at 2s? Can you find V at 2s?
Demmy
#9
Aug28-12, 11:30 PM
P: 12
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.
Demmy
#10
Aug28-12, 11:31 PM
P: 12
@Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir
Demmy
#11
Aug28-12, 11:37 PM
P: 12
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.
Jakeus314
#12
Aug28-12, 11:37 PM
P: 48
Quote Quote by Demmy View Post
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?
Jakeus314
#13
Aug28-12, 11:38 PM
P: 48
Quote Quote by Demmy View Post
@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.
Jakeus314
#14
Aug28-12, 11:42 PM
P: 48
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.
Demmy
#15
Aug28-12, 11:49 PM
P: 12
ahm sir, sorry bout the convertion thingy, I don't get is sir.
Demmy
#16
Aug28-12, 11:53 PM
P: 12
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
Jakeus314
#17
Aug28-12, 11:54 PM
P: 48
Quote Quote by Demmy View Post
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?
Jakeus314
#18
Aug28-12, 11:57 PM
P: 48
Quote Quote by Demmy View Post
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?


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