physics help

i'm having trouble answer this question:

"assume that the angle of inclination of the inclined plane is 25 degrees, the initial velocity of the puck is 8m/s [33 degrees up the inclined plane], and Pn is of the same vertical level as P0, find the velocity of the puck half way up the inclined plane."

edit: if this question does not get answered in the next 2 hours, then don't bother answering it, cause i'll just go to sleep and hand in the hwk tmr, and get a zero on this question

I'd help you, but your question makes no sense.

the wording is exactly the same from the question o.O

this is gr 12 physics by the way (i guess my teacher just made up the question without thinking about it)

Doesn't sound like the entire question.

on the sheet there are 20+ questions, basically we had to first use the air table which is raised at an angle, and pushing the puck upwards, and it will come down because of the gravity. Most of the questions are related to that, this question is just the last one and the most challenging one.

basically you are suppose to assume that the air table is raised at 25 degree angle, and you push the puck at the speed of 8m/s at 33 degree angle from the bottom of the air table. and find velocity of that puck half way up the air table. acceleration due to gravity is 9.8 m/s^2 down, but it's going to be reduced because it's at an angle.

edit: my friend showed me the answer on MSN, i still don't get what he did, but i'll just copy and paste that for the answer, the answer could be wrong but it's better than getting a 0 on this question.

here's he's answer:

vy= 8 sin(33)
vy=4.36m/s

vx=8 cos (33)
vx=6.71m/s

gi = -9.8 m/s2 * sin(25°) = -4.
Total vertical incline: dy-max = (02 – 4.362) / 2(-4.14) = 2.29 m
Velocity of the puck half way up the incline:
vy = sqrt[4.362 + 2(-4.14)( 2.29 / 2)] = 3.08 m/s
angle= tan-1(3.08 / 6.71) = 24.7 degrees
v = sqrt(6.712 + 3.082) = 7.38 m/s[24.7 degrees up the inclined plane]

Tell me what Pn and P0 are supposed to be, because I don't see a use for momentum in this question.

Your question is missing so many holes, it's not even funny.
What's Pn?
What's P0?
How long is the inclined plane?
Where is the puck's starting position?
What's the coefficient of friction?

Tell me what Pn and P0 are supposed to be, because I don't see a use for momentum in this question.

P0, Pn, are points, p0 is the starting point, pn is the ending point, i don't think it has much significance, because every question, pn and p0 is at the same vertical level o.O.

i'm having trouble answer this question:

"assume that the angle of inclination of the inclined plane is 25 degrees, the initial velocity of the puck is 8m/s [33 degrees up the inclined plane], and Pn is of the same vertical level as P0, find the velocity of the puck half way up the inclined plane."

edit: if this question does not get answered in the next 2 hours, then don't bother answering it, cause i'll just go to sleep and hand in the hwk tmr, and get a zero on this question

1) This isn't even a question. There's nothing asked of you to find, and there isn't even a sentence with a question mark at the end.

2) It's been two hours, gtfo2bed.

1) This isn't even a question. There's nothing asked of you to find, and there isn't even a sentence with a question mark at the end.

2) It's been two hours, gtfo2bed.

find the velocity of the puck half way up the inclined plane

find the velocity of the puck half way up the inclined plane

Oh, sorry.
Though the information doesn't seem to be complete anyway.

Still, gtfo2bed

Oh, sorry.
Though the information doesn't seem to be complete anyway.

Still, gtfo2bed

it's only 1am here, my mark and my future is in jeopardy here so i can't sleep >.>

It's impossible to find the velocity of the puck, halfway up the incline if you don't know how far half way up is.

i guess half way is exactly half, as in 1/2

check my friend's answer and see if it's right

Half of fucking what? A 100km incline?

nvm i understand it know

after figuring out vy and ay, you can figure out time, by giving that dy=0 (because that's when the puck return back to the initial position)

and then after you figure out time, you could figure out the displacement for x and then question done!

that whole incline thing just change the acceleration due to gravity a bit, it's just using a different gravity. i got it!

O.o

the gravity doesn't change... does it?

Don't mind the m in the pic, I was thinking forces

The distance will be half of its total travel, that's the only number you'd have. Now, what the hell is the 33o, I assume it means that the puck stays in contact with the table, but is hit 33o upwards (8o with respect to the table)

All my calculations are with respect to the slanted table, not the horizon.

Let's find the total distance

So we have
Vi = 7.922 = 8*cos(8o)
Vf = 0
It is decelerating at a rate of a = -9.8*sin(25o)
Hopefully the m will cancel out

I only use the timeless formula:
vf^2 = vi^2 + 2*a*d

0^2 = 7.922^2 + 2*-9.8*sin(25o)*(d)<solve for d

d = 7.547

We now switch to another equation mindset to find the vf halfway up the total distance.
Again, Vi = 4.357, a= -9.8*sin(25o), d = above/2, and vf = what we're solving for

vf^2 = 7.922^2+2*a*d/2


2's cancel out

vf^2 = 62.760 + (-4.142)(7.547)
vf^2 = 31.380
vfx = 5.602 m/s

My answer is in respect to the table. To find out in respect to the initial velocity you'd have to solve that all again in y terms (super bakery show fast forward)

vi=8*sin(8o)
vf = 0
a = -9.8cos(25o)
d = ?

solve for d
d = .0698

*super bakery show later*

vi = 8*sin(8o)
vf = ?
a = -9.8cos(25o)
d = .0698/2

vfy = .787

so this is all in respect to the table.

To double check my work, tan^-1( .787/5.602) is indeed 7.99, close enough to 8.