Question

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 1

Answer:

$\frac {2v_{ox}v_{oy}}{gsin\theta}$

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

$mgcos(90-\theta)-mgsin\theta$

Acceleration along y-axis is given by

$a=-gsin\theta$

Distance moved by puck in x-direction is

$x=v_{oy}t +0.5(at^{2})= v_{oy}t +0.5(-gsin\theta) t^{2}= v_{oy}t -0.5-gsin\theta t^{2}$

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

$0= v_{oy}t -0.5-gsin\theta t^{2}$ and making t the subject

$v_{oy}t =0.5-gsin\theta t^{2}$

$t=\frac {2v_{oy}}{gsin\theta}$

To find the distance of the puck from origin

$x=v_{ox}\frac {2v_{oy}}{gsin\theta}=\frac {2v_{ox}v_{oy}}{gsin\theta}$

Distance from origin is $\frac {2v_{ox}v_{oy}}{gsin\theta}$