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You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle 0 with the horizontal. Choose your origin at one of the two comers that touch the floor, the x axis pointing along the bottom edge of the board, the y axis pointing up the slope, and the z axis normal to the board. You now kick a frictionless puck that is resting at O so that it slides across the board with initial velocity (v_ox, v_oy,0). Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from O when it does so.

Answer :

opudodennis

Answer:

[tex]\frac {2v_{ox}v_{oy}}{gsin\theta}[/tex]

Explanation:

The rectangular board is represented as shown in the attached image

Here, m is mass, g is acceleration due to gravity and \theta is the angle of inclination

The vertical component of the force on particle is represented as

[tex]mgcos(90-\theta)-mgsin\theta[/tex]

Acceleration along y-axis is given by

[tex]a=-gsin\theta[/tex]

Distance moved by puck in x-direction is

[tex]x=v_{oy}t +0.5(at^{2})= v_{oy}t +0.5(-gsin\theta) t^{2}= v_{oy}t -0.5-gsin\theta t^{2}[/tex]

When the puck returns to the floor level, total distance covered is zero hence

[tex]0= v_{oy}t -0.5-gsin\theta t^{2}[/tex] and making t the subject

[tex]v_{oy}t =0.5-gsin\theta t^{2}[/tex]

[tex]t=\frac {2v_{oy}}{gsin\theta}[/tex]

To find the distance of the puck from origin

[tex]x=v_{ox}\frac {2v_{oy}}{gsin\theta}=\frac {2v_{ox}v_{oy}}{gsin\theta}[/tex]

Distance from origin is [tex]\frac {2v_{ox}v_{oy}}{gsin\theta}[/tex]

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