Titled reference frame, N2L with position and velocity

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion on an inclined plane. The scenario describes a ball thrown with an initial speed at an angle above the incline, requiring the application of Newton's second law to analyze its motion and determine its position over time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem, questioning the clarity of the question and the implications of the initial velocity. There are attempts to derive the equations of motion and to understand the components of velocity and acceleration along and perpendicular to the incline.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some have provided insights into the equations of motion, while others are seeking clarification on the reasoning behind certain steps. There is no explicit consensus, but productive exchanges are occurring regarding the setup and approach to the problem.

Contextual Notes

Participants note potential ambiguities in the problem statement and the need to clarify the definitions of the axes used in the analysis. There is also mention of the importance of distinguishing between the components of motion along the incline and the effects of gravity.

  • #31
Oblio said:
will i be setting vy to zero and then solving a free fall?

I don't understand... the question asks you to find the range in the x direction I believe...
 
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  • #32
using kinematics?
 
  • #33
Oblio said:
using kinematics?

yes... what is the displacement along the x direction and y direction... in terms of time...
 
  • #34
d = v0*t + (1/2)at^2
 
  • #35
Oblio said:
d = v0*t + (1/2)at^2

Yes. Now use the values we've calculated for v0 and a for this problem... for the x- direction... and the y-direction...
 
  • #36
dx = vocos\varthetat + (1/2)(-g)t^2sin\phi

dy = vosin\varthetat + (1/2)(-g)t^2cos\phi
 
Last edited:
  • #37
i have a feeling this is wrong for dy...
 
Last edited:
  • #38
Oblio said:
i have a feeling this is wrong for dy...

Looks right to me. So now you want to find the range...
 
  • #39
range?
 
  • #40
Oblio said:
range?

the question asks for the range...
 
  • #41
which is really just the displacement up the plane?
 
  • #42
Oblio said:
which is really just the displacement up the plane?

yes, maximum displacement... the object is thrown at an angle, then hits the incline eventually... the displacement up the plane when it hits is the range...
 
  • #43
is this the same kinematics equation?
 
  • #44
Oblio said:
dx = vocos\varthetat + (1/2)(-g)t^2sin\phi

dy = vosin\varthetat + (1/2)(-g)t^2cos\phi

use these equations to find dx when dy = 0... that gives the range.
 
  • #45
solve for... vo? and insert into dx?
 
  • #46
Oblio said:
solve for... vo? and insert into dx?

no, you want to eliminate t... so solve for t and substitute into dx.
 
  • #47
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?
 
  • #48
Oblio said:
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?

hmm... close but some mistakes... can you post what you got for t?
 
  • #49
t = vosin(theta) / (-1/2)gsin(phi)
 
  • #50
Oblio said:
t = vosin(theta) / (-1/2)gsin(phi)

setting dy = 0, and solving for t will give you something different... it should be cos(phi) and the - shouldn't be there.
 
  • #51
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
 
  • #52
Oblio said:
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

cool. plugging that into dx should work.
 
  • #53
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
 
  • #54
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)

you missed t^2... didn't square t.
 
  • #55
oops.
man I am bad..
 
  • #56
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2
 
  • #57
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2

careful the 1/2 and g are squared...
 
  • #58
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
 
  • #59
Oblio said:
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)

Looks good... try to simplify and use a trig identity to get it to look like the formula they gave for the range...
 
  • #60
will trig identities apply since theta and phi are present?
 

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