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A very small steel marble is shown rolling at a constant speed on a horizontal t

ID: 2270818 • Letter: A

Question


A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at M, falls, and hits the ground at P. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from L to P.

A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at M, falls, and hits the ground at P. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from L to P.

Explanation / Answer

p(t) = 0.5 * a * t^2 + v0 * t + p0
where
p(t) is position at time t
a is acceleration
v0 is initial velocity at time 0
p0 is position at time 0

We'll use the fall from M to P to compute v0.

Vertical acceleration due to gravity = -9.8 m/s^2 = -980 cm/s^2
Let t = 0 when the marble is at point L.
horizonatal
px(t) = 100 = 0.5 * 0 * t^2 + v0 * t + 50 = v0 * t + 50
vertical
py(t) = 0 = 0.5 * -980 * t^2 + 0 * t + 70 = -490 * t^2 + 70

Two equations. Two Unknowns. Solve for t and v0.

we get . t=0.376sec v0 = 132.97cm/sec

To find the time need to travel from point L to M use v0 from above
px(t) = 50 = 0.5 * 0 * t^2 + v0 * t + 10 = v0 * t + 10

we get t= 0.30 sec

Add the two times for the total time from L to P.

so total time = 0.30+0.376 = 0.676sec

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