question dealing with momentum?

[teclo]

hi, i imagine the solution to this is rather simple. i don't know the answer to the problem, so i can't be sure if my answer is incorrect or not. i imagine it is. anyone who could offer any insight, i would be most appreciative of.

a .4 kg hockey puck is traveling with a velocity of <20,0>. when it is at <5,3> m it is hit by a hockey stick in the y direction and the stick breaks. sometime later, it is at <13, 21> m. later weights were piled on a hockey stick until it broke. it withstood about 1000 N of force.

how much time (t(contact)) was the stick in contact with the puck?

i determined .018 s

i doubt thats right. i assumed that the x component of the velocity would not change. in that case it took .4 seconds for the puck to reach the new position. if that is correct the new velocity is <20,45> m/s. so dV is <0,45> m/s. .4 kg * <0, 45> m/s = <0,1000> N * dt . since i don't know how to divide vectors (or if i can) i droped the directiong and used magnitude. i then divided giving me the answer. i have a feeling there's something i'm not understanding, but i'm not sure.

i'm kind of disappointed, i want to major in physics but i'm already confused in chapter one! we haven't done any problems in class, so i'm kind of taking a stab in the dark =/

[Tide]

The force on the hockey stick while interacting with the puck is equal to the change in momentum divided the time of contact.

[jamesrc]

Don't get discouraged; you got the right answer:

Don't worry about dividing vectors, just consider x and y components separately. Your reasoning through the problem was perfectly fine.

[needhelpperson]

hi, i imagine the solution to this is rather simple. i don't know the answer to the problem, so i can't be sure if my answer is incorrect or not. i imagine it is. anyone who could offer any insight, i would be most appreciative of.

a .4 kg hockey puck is traveling with a velocity of <20,0>. when it is at <5,3> m it is hit by a hockey stick in the y direction and the stick breaks. sometime later, it is at <13, 21> m. later weights were piled on a hockey stick until it broke. it withstood about 1000 N of force.

how much time (t(contact)) was the stick in contact with the puck?

i determined .018 s

i doubt thats right. i assumed that the x component of the velocity would not change. in that case it took .4 seconds for the puck to reach the new position. if that is correct the new velocity is <20,45> m/s. so dV is <0,45> m/s. .4 kg * <0, 45> m/s = <0,1000> N * dt . since i don't know how to divide vectors (or if i can) i droped the directiong and used magnitude. i then divided giving me the answer. i have a feeling there's something i'm not understanding, but i'm not sure.

i'm kind of disappointed, i want to major in physics but i'm already confused in chapter one! we haven't done any problems in class, so i'm kind of taking a stab in the dark =/

Just curious, to get the actual velocity, don't you have to use pythagoras's theorem?

45m/s^2 + 20^2 = vf^2

mvf - mvi = Ft

vf*.4kg - .4kg*20 = 1000N*t

t = .012 s

correct me if i'm wrong...

[jamesrc]

Just curious, to get the actual velocity, don't you have to use pythagoras's theorem?

45m/s^2 + 20^2 = vf^2

mvf - mvi = Ft

vf*.4kg - .4kg*20 = 1000N*t

t = .012 s

correct me if i'm wrong...

No, the impulse is a vector too. There was no impulse in the x-direction.

F\Delta t = m(v_f-v_i)

F = 1000N (estimated from the experiment; the direction of this force is in the +y direction as given in the problem)
vf = 45 m/s (calculated from the geometry of the trajectory)
vi = 0 (initially at rest in the y-direction)

This vector equation was reduced to a scalar one by considering +y to be + and -y to be -.

Solve to find &Delta;t = 18 ms

[teclo]

hurray i was right after all!! maybe i will make it to grad school.

on a side note, we have yet to cover acceleration. i'm using the book matter and interactions vol 1, which is actually really interesting (i just wish it was set up more like my integral calculus book ala homework problems)! hopefully one day i can return the help! thanks again, you guys rock.

[Mathechyst]

I calculate .0184s by solving the equation {\Delta}y_1+{\Delta}y_2=18 where

{\Delta}y_1=\frac{1}{2}at^2=1250t^2

is the distance traveled while the puck is in contact with the hockey stick and

{\Delta}y_2=vt=2500t(0.4-t)=1000t-2500t^2

is the distance traveled while the puck is not in contact with the hockey stick.

[jamesrc]

You know what, I agree that your way is more accurate if you consider the motion of the puck as a constant acceleration during the contact time. My impression was that using the impulse to find the answer was more in the spirit of the problem. The reason for the slight discrepancy is that v was calculated as 18/.4 = 45 without considering the movement during the impulse.