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Physics Form 4 Chapter 2 Force and Motion I

4. The ticker timer is switched on and the trolleys are pulled down the slope using the elastic string
that is stretched at a constant length in order to provide a constant force.
5. The tape is then analysed for the acceleration of the trolley. The value is recorded.
6. The experiment is repeated by using 2 then 3 elastic strings pulled to the same length.
Results:
Table 2.9
Number of elastic string,
1. 1 2 3
F / unit
Initial velocity, u / cm s –1
Final velocity, v / cm s –1

Acceleration, a / cm s –2
Chapter
2 2. The graph plotted is a straight line passing through the Acceleration, a / cm s –2
origin, therefore a α F.
Conclusion: 3
If a mass of an object is constant, the bigger the force, the
bigger is its acceleration. The hypothesis is accepted.
2

1

0
Force, F /N
Figure 2.6

Newton’s Second Law of Motion SPM Tips
1. From the experiments, If a is directly proportional If a is inversely
1
(a) a ∝ m when F is constant. to F, proportional to m,
a
(b) a ∝ F, when m is constant a
Combining both,
F = kma, where k = is a constant.

2. By defining 1 newton (N) as the unit force F m
that causes an object with a mass of 1 kg to a ∝ F
1
accelerate 1 m s -–1 and substituting into the a ∝ m
equation,
F = kma
where 1 = k (1)(1) 3. Newton’s Second Law of Motion describes the
Therefore, k = 1 relationship between acceleration of an object
with this, and the force applied to it. Newton’s Second
Law of Motion states that when a net external
force acts on an object, the acceleration of
the object is directly proportional to the net
force and has a magnitude that is inversely
proportional to its mass.

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