# Finding horizontal distance

1. Feb 18, 2014

### get_physical

1. The problem statement, all variables and given/known data
http://s27.postimg.org/pqmx061qb/Screen_Shot_2014_02_17_at_11_42_12_PM.png [Broken]

2. Relevant equations

vix=vfx
ax=0
viy=0

3. The attempt at a solution
i found that t = 0.639s

not sure where to go from there

Last edited by a moderator: May 6, 2017
2. Feb 18, 2014

### tiny-tim

hi get_physical!

(try using the X2 button just above the Reply box )
if you know what vi,x is, that should give you R

3. Feb 18, 2014

### lendav_rott

How would you find the 2 block mass velocity after the collision? Think about the conservation of momentum.

4. Feb 18, 2014

### get_physical

m1v1 = m2v2
2*v1= 7*v2

i don't know v1 how to find v2?

5. Feb 18, 2014

### get_physical

I also tried using energy conservation formula:

mgh = mgh + 0.5mv^2
v = 9.165m/s
split it into x and y components, i get vix = 9.165 because Viy = 0 (at the very top)

but i did not get the right answer for R

6. Feb 18, 2014

First find the velocity of the first block.

7. Feb 18, 2014

### get_physical

some hints on how to find that?

8. Feb 18, 2014

The block falls a distance of 20m,the assuming the acceleration is 10m/s2,you can find the velocity after it has reached the lowest part of the ramp.As the ramp is frictionless, and the 2nd part is horizontal,the speed gained not change in the horizontal part so it will stick the other block with that velocity.

9. Feb 18, 2014

What is that 7?It should be 5 kg

10. Feb 18, 2014

### get_physical

but they are sticking together so the mass of both of them is 7

11. Feb 18, 2014

### get_physical

wouldn't we just be finding the vertical velocity in this case then?
shouldn't d be the hypotenuse?

12. Feb 18, 2014

Oh.Sorry,never studied momentum.

13. Feb 18, 2014

### get_physical

So how come we can use d= 20m when that is the height and not what the block actually travelled?

if we use d = 20 wouldn't we be finding the vertical component of the velocity???

14. Feb 18, 2014

### lendav_rott

Well, say you have a similar ramp that goes to some 20m high. You need to move a box on a shelf (20m high :D :D ) and another box with the same mass is already on the shelf. The potential energy the other box on the shelf has, is exactly the same as the work you have to do to get the 1st box up on the shelf.

In this scenario, the block is at the height of 22m and slides down to 2m. It makes no difference how much it actually manages to travel, the change in potential energy is what you are after. Why does an object hit the floor harder and harder if you drop it from higher and higher? Its potential energy increases the higher you bring it, the same potential energy becomes its kinetic energy as it falls down and at the point of impact, its kinetic energy is equal to the potential energy it used to have.

Last edited: Feb 18, 2014
15. Feb 18, 2014

### get_physical

Yes, I understand that part, I thought i was using d=20m with one of the kinetics formula to find vi.

I used mgh = mgh+ 1/2mv^2 to find the velocity. Is that right? Because I still got it wrong.

16. Feb 19, 2014

That means potential energy =potential energy+kinetic energy.?

Potential energy difference between the top and the bottom of the ramp should be equal to kinetic energy gained.
So $mg20-mg2=mg18=\frac{1}{2}mv^2$
$mg18=\frac{1}{2}mv^2$

17. Feb 19, 2014

There are no other forces here,just gravity.So horizontal or vertical speed does not matter.What matters is the speed gained due to gravity.

18. Feb 19, 2014

### lendav_rott

mgΔh = mv2/2

if you assume mgh = mgh + more energy, then energy is created from nothing.

About the velocities - same way I could ask you, why were you able to calculate the time it takes for the blocks to fall down? They are travelling horizontally aswell, why did you calculate the time from freefall distance? (the result is correct, but do you know why it is? )

Last edited: Feb 19, 2014
19. Feb 19, 2014

### get_physical

But the mass are different. At the beginning, there's only PE, but at the bottom there is both PE and KE as some of the PE has been converted to KE?

mgh = mgh + 0.5*m*v^2
2*9.8*22 = 7*9.8*2 + 0.5*7*v^2

Where did I go wrong?

20. Feb 19, 2014

### lendav_rott

Aha, but the sliding block has the kinetic energy of mgΔh only just (read: millionths of a second) before the collision. Therefore the kinetic energy of the sliding block is mgΔh. Equate the two and you will find its velocity. (When it is about to connect with the other block, its velocity will only have 1 component, which is the horizontal velocity, the one you are after)

You are observing only the 2 kg block before the collision, don't forget that. Conservation of momentum kicks in after the collision, there is no need to consider the 2 blockmass before hand.

(I have to say, I tried to find the velocity of the 2kg block and I get that it is 14√2 m/s , which is a bit scary, but considering there is no friction I suppose it is possible.)

Last edited: Feb 19, 2014