Inclined Plane Homework: Calculating Speeds at Photogates 1 and 2

[anton717]

Homework Statement

Photogate 1 is at the 25.2 cm mark, and photogate 2 is at the 66.4 cm mark along the track, Assume
* The bottom of the track is at U = 0 J (potential energy is zero).
* The total mass of the cart is 214 g.
* The track is totally frictionless

Suppose the cart is released at rest at some point above photogate 2 (the higher point on the track). If the cart passes through photogate 2 at speed 0.497 m/s, at what speed will the cart pass through photogate 1?
v1 = ____________ m/s

Homework Equations

I am thinking about mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)
also vf^2=vi^2+2ad might be helpful?

The Attempt at a Solution

I went ahead and did PEi+KEi=PEf+KEf
and got a wrong answer for Vf (which is v1)? ( I got something like vf= 0.998 )

What have I done wrong? Did I not consider the acceleration?
Well, I tried that way a=gsin(x), I found the a to be 1.143 m/s^2 and after that I am a bit confused with 2 different heights and lack of info to find Vf (v1)... (may be I m not getting it)
Would be awesome if someone would help me with the concept or answer.
Thanks,

 

[Orodruin]

What did you use for h when trying to apply conservation of energy?

Also, your statement is missing information in the inclination of the plane.

 

[Suraj M]

that way a=gsin(x)

I don't see you mentioning $x$ anywhere before,

 

[Suraj M]

I am thinking about mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)

I doubt, you understand what $h$ and $h'$ actually mean. If you just substituted the value of the mark at which the photogates are present, you are wrong, h should be vertical to the ground, seeing the title of this thread i doubt if it is the vertical height that you have used

 

[anton717]

What did you use for h when trying to apply conservation of energy?

Also, your statement is missing information in the inclination of the plane.

Sorry :/, I forgot to write this : " Suppose the inclined plane is tipped at angle θ= 6.7 o. The 0 cm mark is at the bottom. "

 

[anton717]

I don't see you mentioning $x$ anywhere before,

It's Theta but I don't have the symbol so I just wrote x but then I realized that I did not even write the what x is equal to. Sorry. x = 6.7* so it's a=gsin(6.7). My fault.

 

[anton717]

I doubt, you understand what $h$ and $h'$ actually mean. If you just substituted the value of the mark at which the photogates are present, you are wrong, h should be vertical to the ground, seeing the title of this thread i doubt if it is the vertical height that you have used

Okay. Read the question carefully. It gives you 2 heights. But never gives you the actual height of the inclined plane. If you want to find the actual height then you need the total distance of the hypotenuse (at least) but its not given either. Those H' and H are taken from H=dsin(x) and H'=d'(sin(x)) are the heights where each photogate is located. I wrote this because I thought finding P.E and K.E energy at the point where each photogate is located would help me. But the answer I got was wrong...

 

[Suraj M]

mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)

This equation is correct, you should get the right answer from this equation. What should the answer be? Are you confident about your calculations?

 

[Suraj M]

gives you the actual height of the inclined plane

Which you don't need to know as they have already given you the velocity attained.

found the a to be 1.143 m/s^2 and after that I am a bit confused

You can do it this way too.
You don't need to be confused, you can assume you're new x-axis parallel to the inclined plane, then you're solution reduces to

vf^2=vi^2+2ad might be helpful?

 

[anton717]

This equation is correct, you should get the right answer from this equation. What should the answer be? Are you confident about your calculations?

I said it somewhere in the main post that I got about 0.998 ( got it yesterday, don't really remember the exact #) but it says its wrong ( its online and 1 try is left :p).
Yes I am pretty sure I calculated it correctly. I found each height, (converted it to m and kg) so I don't see the mistake in conversion too.

So according to you the change in PE + change in KE =0 formula is correct right? (formula that is mentioned above)
Then what's wrong...

The thing why I am not sure about vf^2=vi^2 +2ad is that the actual initial v is 0 since it leaves the inc.plane from rest. AND the "a" found is acceleration for the total distance of the inc.plane. It's not "a" from Photogate 2 to Photogate 1 . If you know what I mean. If we assume so the v final (photogate 1) will be much higher...which I think is wrong...

 

[Orodruin]

So according to you the change in PE + change in KE =0 formula is correct right? (formula that is mentioned above)
Then what's wrong...

Please write out your full computation with all numbers. If you do not, then we are just stabbing in the dark as to where you might have gone wrong.

 

[anton717]

Guys, forget it. I got the right answer. I made a mistake on one little thing.
Absolutely gutted.

 

[Suraj M]

I made a mistake on one little thing. .

And what was that?

 

[anton717]

And what was that?

basically the equation dKE + dPE = 0 (d as delta , change) is correct, but when I was plugging in number for given V (at photogate 2) I plugged in its mass by mistake ...
Its embarrassing lol