3.1.5 Measurement of g

Question 1

Compare methods of measuring g:

1. Light gates and
2. Using motion sensors.

Answer 1

Similarities:
- Both give values which can be used to calculate g
-Both eliminate the human reaction time, and so are both more accurate than using a stopcock.
Differences:
- Light gates connected to a data logger measures the time it takes for an interrupt (or double interrupt) card attached to a model vehicle or falling object to move between two points
- A motion sensor records displacement at reg intervals, to create a displacement time graph;

Question 2

Name three methods determine acceleration of free fall.

Answer 2

1. Using a trapdoor and electromagnet;
2. Using light gates and a timer (which can have a light beam that is interrupted);
3. Using an accelerometer measures it directly;

Question 3

Why must the object in free fall being used to measure acceleration NOT have reached terminal velocity?

Answer 3

Because the forces (drag and weight) have balances, and the object has stopped accelerating: there is no acceleration to measure;

Question 4

Name the two types of approach that can be used to measure acceleration (not the methods of measuring it). What are the similarities of these approaches?

Answer 4

1. Direct approaches:
   - an accelerometer;
   - timing the falling of an object and using SUVAT equations of motion;
2. Indirect approaches:
   - measuring time taken for a pendulum to complete a full swing (because this is dependant on g);

ALL OF THESE APPROACHES INCOLVE SUVAT (acc due to gravity)

Question 5

Name the process, which one must go through when utilising an electromagnet and trapdoor to measure g.

Answer 5

1. The electromagnet supports a sleet ball;
2. The current going through the electromagnet is turned off and so the ball falls at the same time that an electronic timer is triggered;
3. The current has been broken and so the timer shall start;
   3.The timer stops once the ball breaks the circuit again, when it hits the trap door.
4. A mean value for t is taken;
5. S (displacement) is measured;
6. s=ut+1/2at^2 is rearranged to find a, giving:
   g=2s/t^2, as u is zero and a = g.
7. The numbers from he experiment are plugged in, giving a value for g.

Question 6

How would one find a value of g from a graph in the trapdoor experiment? Include values as to what the axis on the graph would entail - and why they would show this.

Answer 6

1. From the trap door method of finding g,
   s=1/2gt^2, which is in the form of y=mx+c;
2. Therefore, x axis has t^2 and y axis has s - units, m;
3. The gradient of this graph is g/2, which means that in order to find g, one would find the gradient and double it.
4. This is one of the only arrangements that allows us to find g from a graph, and is why the axisies are labelled in such a way.

Question 7

Name three sources of uncertainty in an electromagnet and light gate method of measuring g.

Answer 7

1. If the electromagnet is too strong, there will be a delay in releasing the ball after it is triggered - so the steel ball must only just be supported;
2. If the distance of the ball is too large, or the ball too small - the ball will begin to reach terminal velocity as drag catches up;
3. The displacement must be measured accurately from the bottom of the ball, and the ball dropped from the top of the trap door;

Question 8

In utilising light gates to measure g, what is the process that one would go through? What dies this method assume?

Answer 8

1. Connect light gate to data logger & computer with data logging software;
2. Measure the time taken for a piece of card to travel through the light gate as it falls;
3. Add blue tack or weight the card with paper clips to ensure that it falls evenly. One can also draw a line on the card from which to evenly drop it each time.
4. Use a ruler to measure the displacement of the card.
5. Find the velocity of the card using s=d/t
   The method assumes that velocity is constant when passing through the light gate.

Question 9

How would one find a value of g from a graph in the light gate experiment? Include values as to what the axis on the graph would entail - and why they would show this.

Answer 9

v^2=u^2+2as

u=0, a=g, so

2g=v^2/s
By varying the length from which the card is dropped, v^2 is on the y axis and s is on the x
The gradient = 2g, so g is a half of the gradient (compared the trap door experiment, where g is double the gradient).