1. The problem statement, all variables and given/known data
I was looking at the problem discussed in the thread below.
http://ift.tt/TDxvbR
A block is initially kicked in the direction perpendicular to the downslope of the plane with initial speed V.
The problem asks for the speed of the block down the plane after a long time if the coeff of friction is [itex] tan \theta [/itex].
2. Relevant equations
The thread I posted above mentions using the fact that [itex] \sqrt{v_x^2 + v_y^2} + v_y = C[/itex], where C is a constant, since the speed gained in the y-direction will be equal to the speed lost overall.
C is initially V. Eventually [itex] v_x [/itex] goes to zero and [itex] v_y [/itex] goes to [itex] v_f [/itex]. So
[itex] v_f + v_f = V [/itex] and the speed down the plane after a long period of time is [itex] \frac{V}{2} [/itex]
However, I am interested if this problem can be solved in another way, using the equations of motion.
3. The attempt at a solution
If [itex] \gamma = arctan \frac{v_y}{v_x} [/itex], then the equations of motion for the x (perpendicular to the downslope) and y (down the slope) directions respectively will be
[itex] -mg\,sin\theta\, cos \gamma = m \ddot x[/itex]
[itex] mg\,sin\theta - mg\,sin\theta\, sin\gamma = m \ddot y[/itex]
These simplify to
[itex] \frac {-gsin\theta}{\sqrt{\frac{v_y^2}{v_x^2} + 1}} = \ddot x [/itex]
[itex] gsin\theta \,(1-\frac{\frac{v_y}{v_x}}{\sqrt{\frac{v_y^2}{v_x^2} +1 }}) = \ddot y [/itex]
Also, [itex] v_y = \dot y [/itex] and [itex] v_x = \dot x [/itex]. The total speed, V = [itex] v_y^2 + v_x^2 [/itex], is constant too. So we further can further simplify the equations to these:
[tex] \frac{-gsin\theta}{V} \dot x = \ddot x [/tex]
[tex] gsin\theta \,(1-\frac{\dot y}{V}) = \ddot y [/tex]
I am wondering how to proceed in solving these equations. I am trying to find [itex] \dot y [/itex].
I was looking at the problem discussed in the thread below.
http://ift.tt/TDxvbR
A block is initially kicked in the direction perpendicular to the downslope of the plane with initial speed V.
The problem asks for the speed of the block down the plane after a long time if the coeff of friction is [itex] tan \theta [/itex].
2. Relevant equations
The thread I posted above mentions using the fact that [itex] \sqrt{v_x^2 + v_y^2} + v_y = C[/itex], where C is a constant, since the speed gained in the y-direction will be equal to the speed lost overall.
C is initially V. Eventually [itex] v_x [/itex] goes to zero and [itex] v_y [/itex] goes to [itex] v_f [/itex]. So
[itex] v_f + v_f = V [/itex] and the speed down the plane after a long period of time is [itex] \frac{V}{2} [/itex]
However, I am interested if this problem can be solved in another way, using the equations of motion.
3. The attempt at a solution
If [itex] \gamma = arctan \frac{v_y}{v_x} [/itex], then the equations of motion for the x (perpendicular to the downslope) and y (down the slope) directions respectively will be
[itex] -mg\,sin\theta\, cos \gamma = m \ddot x[/itex]
[itex] mg\,sin\theta - mg\,sin\theta\, sin\gamma = m \ddot y[/itex]
These simplify to
[itex] \frac {-gsin\theta}{\sqrt{\frac{v_y^2}{v_x^2} + 1}} = \ddot x [/itex]
[itex] gsin\theta \,(1-\frac{\frac{v_y}{v_x}}{\sqrt{\frac{v_y^2}{v_x^2} +1 }}) = \ddot y [/itex]
Also, [itex] v_y = \dot y [/itex] and [itex] v_x = \dot x [/itex]. The total speed, V = [itex] v_y^2 + v_x^2 [/itex], is constant too. So we further can further simplify the equations to these:
[tex] \frac{-gsin\theta}{V} \dot x = \ddot x [/tex]
[tex] gsin\theta \,(1-\frac{\dot y}{V}) = \ddot y [/tex]
I am wondering how to proceed in solving these equations. I am trying to find [itex] \dot y [/itex].
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