Question

A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.8m/s along the length (1.75m) of the table at one end, by the time it has reached the other end, the puck has drifted a 2.5cm to the right but still has a velocity component along the length of 3.8m/s. She correctly concludes that the table is not level and correctly calculates its inclination from the given information. What is the angle?

I have tried to solve it using newtons second law by taking the mgcos0 =ma(along the side of the table with 0.025m ) but no answer

Answer 1

i think you drew the diagram incorrectly

Because there is no friction were not interested in the y-axis forces,

\(\sf \sum F_x=F_{gx}=ma\)

\(\sf \sum F_x=sin\theta mg=ma\rightarrow a=gsin\theta \)

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Find out how long it took to slide on the length:

\(\sf \Delta s=v_o\Delta t\rightarrow \Delta t=\dfrac{\Delta s}{v_o}=\dfrac{1.75~m}{3.8~m/s}=0.46~s\)

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Now back to the displacement in the direction perpendicular

\(\sf \Delta x=v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf \Delta x=v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf a_x=\dfrac{2\Delta x}{(\Delta t)^2}\rightarrow gsin\theta=\dfrac{2\Delta x}{(\Delta t)^2}\)

\(\sf \theta=arcsin (\dfrac{2\Delta x}{g(\Delta t)^2})=arcsin(\dfrac{2*(0.025~m)}{9.8~m/s^2*(0.46~sec)^2})=1.38^o\)