Calculating Horizontal Distance: Solving for Time and Velocity

[get_physical]

Homework Statement

http://s27.postimg.org/pqmx061qb/Screen_Shot_2014_02_17_at_11_42_12_PM.png

Homework Equations

vix=vfx
ax=0
viy=0

The Attempt at a Solution

i found that t = 0.639s

not sure where to go from there

 

[tiny-tim]

hi get_physical!

(try using the X₂ button just above the Reply box )

i found that t = 0.639s

if you know what v_i,x is, that should give you R

 

[lendav_rott]

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

 

[get_physical]

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

m1v1 = m2v2
2*v1= 7*v2

i don't know v1 how to find v2?

 

[get_physical]

hi get_physical!

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

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

 

[adjacent]

m1v1 = m2v2
2*v1= 7*v2

i don't know v1 how to find v2?

First find the velocity of the first block.

 

[get_physical]

First find the velocity of the first block.

some hints on how to find that?

 

[adjacent]

some hints on how to find that?

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.

 

[adjacent]

m1v1 = m2v2
2*v1= 7*v2

i don't know v1 how to find v2?

What is that 7?It should be 5 kg

 

[get_physical]

What is that 7?It should be 5 kg

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

 

[get_physical]

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.

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

 

[adjacent]

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

Oh.Sorry,never studied momentum.

 

[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?

 

[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.

 

[get_physical]

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.

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.

 

[adjacent]

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.

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 $mg20-mg2=mg18=\frac{1}{2}mv^2$
$mg18=\frac{1}{2}mv^2$

 

[adjacent]

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?

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.

 

[lendav_rott]

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.

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 traveling horizontally aswell, why did you calculate the time from freefall distance? (the result is correct, but do you know why it is? )

 

[get_physical]

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 $mg20-mg2=mg18=\frac{1}{2}mv^2$
$mg18=\frac{1}{2}mv^2$

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?

 

[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.)

 

[get_physical]

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 m/s , which is a bit scary, but considering there is no friction I suppose it is possible.)

Thank you for your reply.

At the very beginning, it says the 2kg block is at rest so v=0 so no KE, only PE. Isn't it?
I don't really get why the kinetic energy of the sliding block is mgΔh. The PE of the block at the very beginning is 2*9.8*22

When it has joined with the other block, they have both potential and kinetic energy as it is moving and the potential energy is now at h=2. Also, the m = 7kg now. Isn't it?

Thank you so much.

 

[lendav_rott]

At the very beginning, it says the 2kg block is at rest so v=0 so no KE, only PE. Isn't it?
Correct

I don't really get why the kinetic energy of the sliding block is mgΔh. The PE of the block at the very beginning is 2*9.8*22
When it is still 2m from the ground, it hasn't quite used up its potential has it? You might think that if the surface had friction then the block might come to a stop while still on the ramp where it has a potential energy of mg(2m) again. Even then,at the beginning of the slide it starts with the "total energy pool" of mg(22m) , but it comes to a stop while still on the ramp because of friction as friction too does work against the slide, depleting the block's "energy pool" to keep sliding.

In our current scenario, however, there is no friction - the block does start out with mg(22m) potential energy. As I said, we are only observing what is happening to the sliding block. We want to know its kinetic energy right before it collides with the the other block.

The conservation of energy states that energy can only transform from one state to another, it cannot be created out of nothing. Since there are no other factors (friction) to consider we are only left with Potential -> Kinetic -> Potential.

Perhaps the confusing part for you is "how come the amount of energy it has obtained doesn't depend on how much a distance it travels"? Well, as it slides Down it gains kinetic energy by coming closer and closer to ground 0, but if it travels only horizontally and there is no friction to act against it, then it maintains its current kinetic energy forever if you want it to.

Therefore, its kinetic energy before collision is mg(20m). Kinetic energy is also equal to mv²/2 - find the sliding block's velocity before collision.

Use the newly found velocity and use the law of conservation of momentum to find the velocity of the 2-block-mass.

Also sorry about a little calculation error - I find its velocity before collision to be 14√2 m/s not 14

When it has joined with the other block, they have both potential and kinetic energy as it is moving and the potential energy is now at h=2. Also, the m = 7kg now. Isn't it?
Once the 2-block-mass is moving it has no potential energy, it will have transformed into kinetic energy.
Yes, m = 7kg

 

[get_physical]

Yes! That's what I've been doing all along!

mg(22) = mg (2) + KE
HOWEVER, I was using the mass of BOTH blocks for the RIGHT side.

I think I get it now. Thanks!

 

[adjacent]

Therefore, its kinetic energy before collision is mg(20m).

Ah.A little mistake by me.I thought 20 was the top distance from ground to top ramp.Sorry

 

[BvU]

Hello Get. I think you interpreted

mgh = mgh+ 1/2mv^2

ok, but somehow v = 9.2 m/s is an incorrect result there !?