Free Fall & Measuring g | CAIE 9702 Paper 2 Notes

Paper 2

Physics9702

Free Fall & Measuring g

means the only force is gravity. Near Earth's surface, where the field is treated as uniform, every object in free fall has the same acceleration,

g = 9.81 \text{ m s}^{-2}

, whatever its mass.

Free fall is not new physics: it is SUVAT with

a = g

already known.

SUVAT in free fall

In a vacuum a feather and a hammer land together. With no air, mass makes no difference to the acceleration.
“Dropped” means $u = 0$ . If downward is chosen as positive, the equations then shorten to $v = gt$ , $s = \tfrac{1}{2}gt^2$ and $v^2 = 2gs$ .

A feather and a hammer side by side inside a glass vacuum chamber. — Remove the air and mass no longer changes the fall: the feather and hammer have the same downward acceleration g.

Free fall is SUVAT with $a = g = 9.81 \text{ m s}^{-2}$ and, if the object is dropped, $u = 0$ . No new equations are needed.

The experiment: measuring g

The syllabus asks you to describe an experiment that measures $g$ from a falling object. The standard set-up uses an electromagnet, a steel ball and an electronic timer.

An electromagnet holds a steel ball at a measured height $h$ above a trapdoor.
Switching the current off releases the ball and starts the timer at the same instant.
The ball hits the trapdoor, which stops the timer.
Record $h$ and $t$ , then repeat for several heights.

A steel ball below an electromagnet, a vertical height scale, a trapdoor switch and an electronic timer. — Switching off the electromagnet releases the ball and starts the timer. The trapdoor stops the timer after the ball falls the measured height h.

Independent variable: $h$ . Dependent variable: $t$ . Keep the same ball, release system and landing switch so the timing method is unchanged.
The ball is dropped, so $h = \tfrac{1}{2}gt^2$ . Plotting $h$ against $t^2$ (not against $t$ ) turns this into a straight line through the origin with gradient $g/2$ .
So $g = 2 \times \text{gradient}$ . Using a graph of many readings, instead of one reading, reduces the effect of random error. Repeat each height and average the times before plotting.

Plot h against t² and the points lie on a straight line through the origin. gradient = g/2 ≈ 4.9 m s⁻², so g = 2 × gradient ≈ 9.8 m s⁻².

Worked example

Reading g from the graph

On the graph above, the line through the origin passes through (0.40 s², 1.96 m). Find g.

Gradient: $\dfrac{1.96}{0.40} = 4.9 \text{ m s}^{-2}$ .
$g = 2 \times 4.9 = 9.8 \text{ m s}^{-2}$ .

Answer

g \approx 9.8 \text{ m s}^{-2}

Your turnOriginal experiment check · 4 marks

A student plots $h$ against $t^2$ and obtains a gradient of 4.76 m s⁻². Calculate the experimental value of $g$ , then state one reason for repeating each height.

Show worked answer

From $h=\tfrac12gt^2$ , the gradient is $g/2$ .
$g=2(4.76)=9.52\ \text{m s}^{-2}$ .

Answer

g = 9.52 m s⁻². Repeating allows a mean time to be calculated, reducing the effect of random timing variation.

Common mistake

Residual magnetism may slow the ball just after release. This can make the measured fall time too large and the calculated value of

g

too small. Repeating does not remove a systematic effect that shifts every reading in the same direction.

Free Fall & Measuring g

means the only force is gravity. Near Earth's surface, where the field is treated as uniform, every object in free fall has the same acceleration,

g = 9.81 \text{ m s}^{-2}

, whatever its mass.

Free fall is not new physics: it is SUVAT with

a = g

already known.

SUVAT in free fall

In a vacuum a feather and a hammer land together. With no air, mass makes no difference to the acceleration.
“Dropped” means $u = 0$ . If downward is chosen as positive, the equations then shorten to $v = gt$ , $s = \tfrac{1}{2}gt^2$ and $v^2 = 2gs$ .

Free fall is SUVAT with $a = g = 9.81 \text{ m s}^{-2}$ and, if the object is dropped, $u = 0$ . No new equations are needed.

The experiment: measuring g

The syllabus asks you to describe an experiment that measures $g$ from a falling object. The standard set-up uses an electromagnet, a steel ball and an electronic timer.

An electromagnet holds a steel ball at a measured height $h$ above a trapdoor.
Switching the current off releases the ball and starts the timer at the same instant.
The ball hits the trapdoor, which stops the timer.
Record $h$ and $t$ , then repeat for several heights.

Independent variable: $h$ . Dependent variable: $t$ . Keep the same ball, release system and landing switch so the timing method is unchanged.
The ball is dropped, so $h = \tfrac{1}{2}gt^2$ . Plotting $h$ against $t^2$ (not against $t$ ) turns this into a straight line through the origin with gradient $g/2$ .
So $g = 2 \times \text{gradient}$ . Using a graph of many readings, instead of one reading, reduces the effect of random error. Repeat each height and average the times before plotting.

Plot h against t² and the points lie on a straight line through the origin. gradient = g/2 ≈ 4.9 m s⁻², so g = 2 × gradient ≈ 9.8 m s⁻².

Worked example

Reading g from the graph

On the graph above, the line through the origin passes through (0.40 s², 1.96 m). Find g.

Gradient: $\dfrac{1.96}{0.40} = 4.9 \text{ m s}^{-2}$ .
$g = 2 \times 4.9 = 9.8 \text{ m s}^{-2}$ .

Answer

g \approx 9.8 \text{ m s}^{-2}

Your turnOriginal experiment check · 4 marks

A student plots $h$ against $t^2$ and obtains a gradient of 4.76 m s⁻². Calculate the experimental value of $g$ , then state one reason for repeating each height.

Show worked answer

From $h=\tfrac12gt^2$ , the gradient is $g/2$ .
$g=2(4.76)=9.52\ \text{m s}^{-2}$ .

Answer

g = 9.52 m s⁻². Repeating allows a mean time to be calculated, reducing the effect of random timing variation.

Common mistake

Residual magnetism may slow the ball just after release. This can make the measured fall time too large and the calculated value of

g

too small. Repeating does not remove a systematic effect that shifts every reading in the same direction.