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fiasKO [112]
2 years ago
11

A student wish to measure the gravitational acceleration g. She does it by releasing a small lead ball from rest and measures th

e time t it takes the ball to drop a distance s. g can then be found using the relation t2. She measures t using an automated digital timer which manual states that it measures drop times up to 500 ms with an uncertainty of 3 ms. She sets the system up such that s 100.0 t 0.3 mm (measured using a measurement tape), and the timer reads that the drop time is 144 1 mS. . Calculate her best estimate of g and its uncertainty . Discuss how she could set up her experiment such that she would achieve a better precision in a single drop, without needing a more precise timer or being able to measure distances with higher precision
Physics
1 answer:
Goryan [66]2 years ago
5 0

Answer:

(9.64 +- 0.86) m/s^2

Explanation:

The generic motion equation for constant acceleration is

x = X0 + v0 * t + \frac{1}{2}*a * t^2

Where

X0: initial position

v0: initial speed

a: acceleration

t: time

If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

x = \frac{1}{2}*a * t^2

And the acceleration can be obtained as:

a = 2*\frac{x}{t^2}

Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

g = 2*\frac{100}{144^2} = 9.64e-3 \frac{mm}{ms^2} = 9.64 \frac{m}{s^2}

For the uncertainty we have to calculate the relative uncertainties first

For the distance (100 * 0.3)/100 = 0.3%

For the time (100 * 3)/144 = 2.08%

For multiplications or divisions the relative uncertainties are added

0.3% + 2.08% + 2.08% = 4.46%

We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.

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gregori [183]
Refer to the diagram shown below.

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The time, t, for the bolt to fall a known distance obeys the equation 
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Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).

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