# Frictionless puck goes up hill then falls. What's V0?

1. Mar 19, 2013

### Zsmitty3

1. A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.2m from the foot of the cliff with no air resistance. What was the speed v0 at the bottom of the hill? There is also a diagram showing the hill. The puck starts at the bottom and travels 15.3m to the top. The cliff then drops off vertically straight down 8.5m. The puck lands 6.2m from where it went airborn off the cliff.

2. I think this is a PE KE question.

3. PE is zero initially which means the total energy is KE there. KE=1/2mv^2 so if we can find the total energy initially we can solve for v. BUT, we have no mass for that equations. So i tried setting up something like KEi+PEi=KEf+PEf since no energy is lost from friction or any other external force. I'm not sure where to go from there. PE is also going to be 0 after it falls of the cliff as well. PE just before it falls is at a max which is mgh, but still we have no mass given.

2. Mar 19, 2013

### Simon Bridge

Is that 15.sm along the surface to the top of the hill or is the hill 15.3m high?
How high is the cliff? What angle does the puck go over the cliff at?

Conservation of energy is the way to go - don't worry about the mass, just call it "m".
... that's a good start - you want to write the KEs and PEs in terms of variables for masses and velocities and heights, then stand back and see what leaps out at you :)

3. Mar 20, 2013

### Zsmitty3

Sorry

It's a long problem and the diagram makes it easier. It goes 15.3 along the ice as it rises up the hill. The cliff is 8.5 m. No angle from the cliff is given. But the Adj. is 8.5 and the Opp. is 6.2 (I think) when it lands. So Tan^-1 O/A=θ which means the angle is 63°? So:

mg h(i)+.5mv^2 (i) = mgh (f)= .5mv^2 (f)

would all the m's cancel leaving:

gh + .5v^2 = gh + .5mv^2

(9.8)(0) + .5v(i)^2 = (9.8)*(?)+ .5v(f)^2

Really V inital should be equal to V final right? Because there is no friction and PE is 0 when its on the ground in both locations?

So basically KE is going to equal the PE at the top of the hill. The PE of the top of the hill is mgh. m is unknown. g is 9.8. h is 8.5. So PE=m(9.8)(8.5). Thats where I get stuck. PE intial = PE final and KE initial= KE final. PE will both be 0. So KE=KE. AHHHHHH!

4. Mar 20, 2013

### Zsmitty3

Kinematic

Also can you not just solve for the time the puck is in the air using kinematics. In the y-direction: v2 = 02+2(9.8)(8.5) leaving v=12.907362

Then solve for t

12.907362=0+9.8t

t=1.317s that the puck is in the air

so it's also that long in the air in the x-direction.

so 6.2= 0+v(i)*(1.317)+0

v=4.7076 m/s ???

5. Mar 20, 2013

### haruspex

That doesn't seem to be the angle of slope of the hill. I assume it's a straight ramp, and it starts at the same height that it finishes, so you can calculate the slope from height and the hypotenuse.
Having done that, you need to use energy to calculate the speed at top of hill, then determine the horizontal and vertical components of that. Then you can use the kinematic equations.

Edit: I meant, I assume that the puck starts at the same height as it finishes (after descending from the cliff).

Last edited: Mar 20, 2013
6. Mar 20, 2013

### Zsmitty3

The slope is not straight. The base of the slope is 15.3 m. Here is the diagram.

7. Mar 20, 2013

### Zsmitty3

I made the mistake of assuming that the v at the top of the hill would equal the initial v due to no friction but I'm wrong because gravity correct?

So PE at the top of the hill is going to just be gh since the m cancels? So 9.8*8.5=83.3 PE at the top of the hill. So since PE is 0 at the bottom the KE must be 83.3. So 83.3=.5mv^2 but the m cancels so 83.3=.5v^2. Which means v= sqrt(83.3/.5). V=12.91 m/s??

8. Mar 20, 2013

### haruspex

Yes.
If its initial KE were only equal to the gain in PE then when it gets to the top of the hill it won't be moving. It will need some remaining horizontal speed to reach its landing point.
Since the top of the hill is horizontal, I would suggest working backwards. How long does it take to hit the ground? How fast then does it need to be moving when it becomes airborne to reach its landing point?
Btw, there is one piece of information missing in the OP. It ought to state that the slope of the hill decreases slowly enough that the puck does not become airborne until it passes the top.

9. Mar 20, 2013

### Zsmitty3

True.

So if it starts 8.5 m up and lands 6.2 m away from the edge of the cliff you would just find the time in the y-direction. To do that V final is going to equal zero once it hits the ground. So use
V2=Vo2+2a(X-Xo)

V2= 02+2(9.8)(8.5)
V=12.91 at the top of the cliff

the solve for t using v=vo+at
12.92=0+9.8t
t=1.317s

so now go to x-direction velocity while airborn using x=xo+vot+1/2at2

602=0+vo(1.317)+1/2*0(frictionless)(1.317^2)
Vo=4.7076 m/s at the top of the hill

so if we make the top of the hill "final"

.5mv2+mgh(0)= .5m4.70762+m9.8*8.5

m's cancel

.5v2+0=.5(4.7076)2+9.8(8.5)

.5v2=11.0807+83.3
V2=188.7614
V=13.739 m/s

Is this correct? It would make since because Vo is higher that V after it's gone up the hill.

10. Mar 20, 2013

### haruspex

You mean, at the bottom of the cliff.
6.2, but that's the right result.
Looks fine.

11. Mar 20, 2013

Thanks!