12.3 CHAPTER 12. FORCE,MOMENTUM AND IMPULSE
- Shown below are sections of the four tickertapes obtained. The tapes are marked with
the letters A, B, C, D, etc. A is the first dot, B isthe second dot and so on. The distance
between each dot is alsoshown.
�� � � � � �
Tape 1
5mm 9mm 13mm 17mm 21mm 25mm
��� � � � �
Tape 2
3mm 10mm 17mm 24mm 31mm 38mm
�� � � � � �
Tape 3
2mm13mm 24mm 35mm 46mm 57mm
�� � � � � �
Tape 4
9mm 24mm 39mm 54mm 69mm 84mm
Tapes are not drawn to scale.
A B C D E F G
A B C D E F G
A B C D E F G
A B C D E F G
Instructions:
- Use each tape to calculate the instantaneousvelocity (in m·s−^1 ) of the trolley at pointsB
and F (remember to convert the distances to m first!). Use these velocities to calculate the
trolley’s acceleration ineach case. - Tabulate the mass and corresponding acceleration values as calculated in each case. En-
sure that each column and row in your table is appropriately labelled. - Draw a graph of acceleration vs. mass, usinga scale of 1 cm = 1 m·s−^2 on the y-axis and
1 cm = 1 kg on the x-axis. - Use your graph to read off the acceleration of the trolley if its mass is5 kg.
- Write down a conclusion for the experiment.
You will have noted in the investigation above that the heavier the trolley is, the slower it moved. The
acceleration is inversely proportional to the mass. In mathematical terms:
a∝
1
m
In a similar investigationwhere the mass is kept constant, but the appliedforce is varied, you willfind
that the bigger the forceis, the faster the object will move. The acceleration of the trolley is therefore
directly proportional to the resultant force. In mathematical terms:
a∝ F.
Rearranging the above equations, we get a∝FmOR F = ma
Newton formulated hissecond law as follows: