Page 71 - text book form physics kssm 2020

Chapter 2 Force and Motion I

3. Calculate the acceleration of the trolley from the ticker Two elastic strings
tape obtained and record it in Table 2.11.
4. Plot a graph of acceleration, a against the reciprocal of mass,
1 and state the relationship between acceleration, a and Two trolleys
m
mass, m.
Figure 2.52
Results:

Table 2.11
Mass u / cm s –1 v / cm s –1 t / s a / cm s –2
1 trolley, m
2 trolleys, 2m
3 trolleys, 3m

The results of Activity 2.13 shows that the acceleration of
F – Force
an object depends on the applied force and the mass of the object.
m – Mass
a a – Acceleration
Acceleration is directly
proportional to the applied a ∝ F
force when the mass of an m constant
object is fixed Combining the two
0 F relationships:
Figure 2.53 Acceleration-force graph F
a ∝
Acceleration is inversely m
a Therefore, F ∝ ma
proportional to mass 1
a ∝
of an object when a m
constant force is applied F constant
1 on the object
–
0 m
Figure 2.54 Acceleration-reciprocal of mass graph
The relationship between force, F, mass, m and
acceleration, a for an object in motion is File
F ∝ ma
(v – u) Expression of Newton’s Change of momentum = mv – mu
F ∝ m (mv – mu)
t Second Law of Motion Rate of change of momentum =
(mv – mu) t
F ∝
t
In S.I. units, 1 N is the force that
Newton’s Second Law of Motion states that –2
the rate of change of momentum is directly produces an acceleration of 1 m s
proportional to the force and acts in the direction when applied on a mass of 1 kg.
of the applied force. From the relationship As such, –2
F ∝ ma 1 N = k × 1 kg × 1 m s
F = kma, k is a constant k = 1
Therefore, F = ma

65
2.6.1 65
2.6.1