Rolling Marble on Table Top Poinci: A marble is rolling on top of a table, going

Question

Rolling Marble on Table Top Poinci: A marble is rolling on top of a table, going along the +x-axis. Its positions at a fixed time interval, At = 0.19 s, is shown with ×'s on the graph below 65 60 50 45 40 35 30 25 20 15 10 5 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100105|11512012SL301 35140 Horizontal position, x, (cm) What is the speed of the marble just before it rolls off the table? 101.2cm/s The horizontal speed is constant. Use several observations to calculate the time interval accurately Submit Answer Incorrect. Tries 6/10 Previous Tries How long does it take for the marble to fall from the top of the table to the floor? Where along the x-axis does the marble hit the floor? Neglect any air resistance NOTE: Both answers must be correct to get credit Submit Answer Tries 0/10 At what angle does the marble hit the floor? Use +x-axis to be 0 degree going counter-clockwise Submit Answer Tries 0/10 What is the position of the marble 0.069 s after it rolls over the edge? Put the x coordinate in the first box below and the y-coordinate in the second box Submit Answer Tries 0/10

Explanation / Answer

Measuring the distance on such a graph isn't very accurate, but...
I've used these results to continue:
x1 to x3 = 0.45 m
x1 to x5 = 0.90 m

Now, using the formula
d = (Vi + Vf) t / 2
where
d = distance
Vi = initial velocity
Vf = final velocity
t = time
you can write 3 equations:

0.45 = (Vb + Va) 0.38 / 2
0.90 = (Vc + Va) 0.76 / 2
0.45 = (Vc + Vb) 0.38 / 2
where
Va = velocity at x1
Vb = velocity at x3
Vc = velocity at x5

Solving this system of 3 equations with 3 unknowns will give
Va = 1.184 m/s
Vb = 1.184 m/s
Vc = 1.184 m/s <- - - - - - - - - - - ANSWER speed just before falling from table

To find the time it needs to hit the floor, use this formula:
h = at²/2
where
h = height = -0.6 m (minus because falling down)
a = acceleration by gravity = -9.8 m/s²
t = time = ?
so
-0.6 = -9.8 t² / 2
t² = 0.122
t = 0.35 s <- - - - - - - - - - - ANSWER time to hit the floor

Multiplying this time by the horizontal velocity will give the horizontal range, so
r = 0.35 * 1.184
r = 0.414 m
Adding the length of the table gives:
x = 0.95 + 0.414
x = 1.364 m <- - - - - - - - - - - ANSWER where marble hits the floor

To find the angle at which it hits the floor, you first need to find its vertical velocity.
(Vf)² = (Vi)² + 2ad
where
Vf = final vertical velocity = ?
Vi = initial vertical velocity = 0 m/s
a = acceleration by gravity = -9.8 m/s²
d = distance = -0.6 m
so
(Vf)² = 0² + 2(-9.8)(-0.6)
(Vf)² = 11.76
Vf = -3.43 m/s (falling down, so taking negative square root)

The angle will be:
tan() = (Vv / Vh)
where
= angle
Vv = vertical velocity = -3.43 m/s
Vh = horizontal velocity = 1.184 m/s
so
tan() = (-3.43 / 1.184)
= -70.95° <- - - - - - - - - - - ANSWER angle at impact
(or positive angle: = 289.04°)

And finally, the position at t = 0.069 s

x = 1.364 + 1.184 * 0.069
x = 1.446 m < - - - - - - - - - ANSWER x-position

y = 0.6 - 9.8 * 0.069² / 2
y = 0.577 m < - - - - - - - - - ANSWER y-position

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