Kinetic energy / projectile motion question

[Daniiel]

Homework Statement

[PLAIN]http://img690.imageshack.us/img690/5756/12281060.jpg

Homework Equations

k = 0.5 mv^2
projectile motion

The Attempt at a Solution

Ok so for part one i just did some algebra and got 1:1 as the ratio, no problems there
For part two i treated it as a projectile after it leaves the roof with v = root (2g)
with that i used y = v sin theta t - 0.5gt^2
to find t
so -3 = root (19.6) sin240 - 4.9 t^2
root (19.6) sin240 = -3.834
simplified and changed to a quadratic 4.9t^2 + 3.834t - 3 = 0
found t with the quadratic formula, t= 0.4836
then subbed into Vy = V sin theta - gt
find Vy = -8.57

So that's the velocity of the y component
Do i have to add that to the velocity of the y component of the initial velocity?
because I've tried it both ways, I'm a bit confused, my tutor told me to add both the inital and final y components, but he didnt treat it as projectile motion, he did it a different way
But let's just say 8.57 is Vy

Vx = v sin 60 = 3.834

the velocity = the magnitude of 3.834i + 8.57j = 9.39

then K = .5 * 5 * 9.39^2 = 220.43

Kinetic rotational energy = 5g

so 220.43/ 5g = 4.5

so the ration is 4.5:1 ?

when i find the ratio by adding the inital Vy and the final Vy i get 6.7:1 or somthing like that

is this right?
my tutor for 7.04:1, but i noticed afew errors so I am not sure if its correct, it's sure a cleaner answer than 4.5 though

 

[Doc Al]

Since all you care about is the kinetic energy, not the speed, stick to energy methods and you'll get the answer cleanly with little calculation.

You are correct that the ratio of KE/KEr is 1:1 while it's rolling. So what's the KE as it leaves the roof? How much energy does it pick up as it falls 3 m? What's the final KE?

 

[Daniiel]

As is leaves the roof KE = 0.5 x 5 x 2g = 5g ?
KEf = 0.5 x 5 x (2 x g x 3) v^2 =2gh h = 3
= 15g
15g:5g = 3:1?

sheeeeet
is that right?

 

[Doc Al]

As is leaves the roof KE = 0.5 x 5 x 2g = 5g ?

Write it as 0.5 x mgh₁ (where h₁ = 1 m).

KEf = 0.5 x 5 x (2 x g x 3) v^2 = root(2gh) h = 3
= 15g
15g:5g = 3:1?

The amount of energy gained during the fall is mgh₂ (where h₂ = 3 m).

So what's KE_f? How does that compare to KE_r?

 

[Daniiel]

ohh
sorry, so its 20g:5g 4:1

mgh2 = 15 mgh1 = 5
add them both because mgh2 is the amount of energy gained, then compair with initial energy?

aw man
massive over think of that question
if you saw the amount of work i did you would laugh

 

[Daniiel]

oh wait sorry again

196:49 4:1 ?

 

[Doc Al]

oh wait sorry again

196:49 4:1 ?

No. Answer the questions in my last post, one by one.

 

[Daniiel]

The amount of energy gained during the fall is mgh₂ (where h₂ = 3 m).

So what's KE_f? How does that compare to KE_r?

ok so, KEr = KEi and KEr doesn't change, so KErf is = 5g
is the 5g part right? and the v^2= 2gh?

KEf = KEi + KE

KEf = 0.5mgh1 + mgh2 = 24.5 + 147 =171.5

171.5:49
3.5:1

im kinda confused now, how come we are 0.5mgh1 instead of K = 0.5 m v^2
and mgh2 is the potential energy of the fall right?

 

[Daniiel]

i think it must be the v^2 = 2g that is stuffing me up
i just did it like
KE = 0.5mgh1 + mgh2 KEr = 0.5mgh1
KEr:KEf = 24.5:(24.5 + 147) = 1:7
which is what my tutor got kind of
which is weird because i saw he made mistake with his trig early onv^2 = vi^2 + 2ah
so before falling v^2 = 2ah = 2g
after falling v^2 = 2a + 2ah = 8g

so KEi = KEr = 0.5 x 5 x 2g = 5g
KEf = (0.5 x 8g x 5) + (5 x 9.8 x 3) = 343

KEf:KEr = 343:49 = 7:1
that kind seems right to me
im kinda confused why i had to add the gravitational potential energy to the kinetic energy
is it because (0.5 x 5 x 8g) covers the diagonal KE and mgh2 the vertical? or is that just completely wrong?

 

[Daniiel]

sorry to be rude and post again Doc Al, but was 1:7 right?
this is from another question but if somthing is show like
r>>R does that mean r is directly proportional to R?

 

[Doc Al]

i think it must be the v^2 = 2g that is stuffing me up
i just did it like
KE = 0.5mgh1 + mgh2 KEr = 0.5mgh1
KEr:KEf = 24.5:(24.5 + 147) = 1:7

This is exactly right.

KE = 0.5mgh₁ + mgh₂ = 3.5 mg (plugging in the values for h)
KEr = 0.5mgh₁ = 0.5 mg (plugging in the values for h)

which is what my tutor got kind of
which is weird because i saw he made mistake with his trig early on

There's no need for any trig at all.

this is from another question but if somthing is show like
r>>R does that mean r is directly proportional to R?

The notation r >> R means that r is much greater than R.

 

[Daniiel]

thanks a bunch Doc Al, now to stress over the final 3 questoins ;) and start some math

was my second way of getting 7:1 just a fluke?
when i did v^2 = vi^2 + 2ah and so on?

 

[Doc Al]

was my second way of getting 7:1 just a fluke?
when i did v^2 = vi^2 + 2ah and so on?

Sorry, but I don't really understand what you did. If you explain it step by step, I'll take a look.

To solve this kinematically, you'd have to find the speed as it leaves the roof. (How would you find the acceleration as its rolling?) Once it leaves the roof it's a projectile. Find the vertical and horizontal components of the speed. Then solve for the final vertical component of speed. Then combine them to find KEf. (Sounds like a lot of work!)

 

[Daniiel]

haha yea
good 4 pages
with errors on the first 3
butt with this other thing

I have this formula in my physics textbook in the kinetic energy part, it
v^2 = vi^2 + 2ah
where vi = initial velocity and in this case a = g
so i applied it to this question and got
before falling off the edge
vi = 0 h = 1
therefore v^2 = 2a = 2g

after falling off the roof
vi = 2g from what i found as the velocity before it falls
h = 3 a = g
v^2 = 2g + 2g3 = 8g

so back to the ratio
Before the fall:

KEi = KEr = 0.5 x 5 x 2g = 5g
KEr = 5g so KEr after the fall = 5g because it doesn't change as a it falls through space

After the fall v^2= 8g what i found by putting into that equation before
KEf = 0.5mv^2 + mgh where h = 3
=(0.5 x 8g x 5) + (5 x 9.8 x 3) = 343

so KEf = 343 and KEr = 5g = 49
KEf:KEr = 343:49 = 7:1

but looking at it the way you told me to makes much more sense and is a lot easier
but i don't get why its 0.5 mgh1 instead of 0.5mv^2
is it because its sin30 mgh ?

 

[Doc Al]

I have this formula in my physics textbook in the kinetic energy part, it
v^2 = vi^2 + 2ah
where vi = initial velocity and in this case a = g
so i applied it to this question and got
before falling off the edge
vi = 0 h = 1
therefore v^2 = 2a = 2g

Careful: While it's rolling down the roof, a ≠ g. (a = g for free fall, but rolling isn't free fall.)

after falling off the roof
vi = 2g from what i found as the velocity before it falls
h = 3 a = g
v^2 = 2g + 2g3 = 8g

Your starting v² isn't right, so your final won't be right either.

so back to the ratio
Before the fall:

KEi = KEr = 0.5 x 5 x 2g = 5g
KEr = 5g so KEr after the fall = 5g because it doesn't change as a it falls through space

After the fall v^2= 8g what i found by putting into that equation before
KEf = 0.5mv^2 + mgh where h = 3
=(0.5 x 8g x 5) + (5 x 9.8 x 3) = 343

If v^2= 8g, KEf would equal 0.5 x m x v^2 = (0.5 x 5 x 8g) = 20 g. (Note: This is incorrect, since v^2 does not equal 8g.)

so KEf = 343 and KEr = 5g = 49
KEf:KEr = 343:49 = 7:1

Seems like you made a few errors that canceled each other.

but looking at it the way you told me to makes much more sense and is a lot easier
but i don't get why its 0.5 mgh1 instead of 0.5mv^2
is it because its sin30 mgh ?

KE = 0.5 mgh₁ because the full change in potential energy equals mgh₁ and half goes to KE and the other half goes to KEr. (This is conservation of energy.)

Of course, KE also equals 0.5mv^2. The correct value of v^2 (after rolling down the roof) is gh₁, not 2gh₁.

 

[Daniiel]

oh fair enough
thanks heaps Doc Al