1. CHAPTER 2 : FORCE & MOTION
ANSWER (d) (i) the first tape
2.1 Linear motion
Question 1
(a) Scalar quantity (ii) the last tape
(b) Arrow:
(e) acceleration
(f)
(c) 700
(d) Zero
2.2 Motion Graph
Question 2 Question 4
(a) A.c current (a)(i) velocity
(b) (i) 0.02 s
(ii) constant velocity (ii)
(iii)
(b)(i) acceleration
(b)(ii)
Question 3
(a) Ac current (b)(iii)
(b) Tape chart
Question 5
(a) Displacement
Time
(b) R : from AB
(c)
(ii)
(c) Constant acceleration
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2. Question 6
(a) 15 m (e) mPuP = (mP + mQ)v
(b) Velocity (2)(1) = (2 + m) (0.4)
(c) (i) constant velocity m = (2 – 0.8)/ 0.4
(ii) at rest = 3 kg
Question 7 Question 10
(a)(i) acceleration (a) quantity of matter in object
(a)(ii) displacement (b) (i) equal but opposite direction
(b) (ii) 0 = MV + mv
MV = - mv
Section of Type of motion
(iii) principle of conservation of
the graph of the car
momentum
(c) (1200)v = (4)(60) = 240
Constant
OA v = 240/1200 = 0.2 m s-1
acceleration
Constant velocity
AB Question 11
(a) Principle of conservation of
momentum
Question 8
(b) Momentum of the man (forward)
(a)(i) constant velocity
is equal with the momentum of
(a)(ii) constant acceleration and
the boat (backward) but in
constant velocity after 2.01 pm
opposite direction
(b)(i) zero
(c) 0 = (50 x 2) + (20)v
(b)(ii)the car moves with constant
v = - (100)/20 = - 5 m s-1
velocity
(d) Rocket
(c)
Question 12
(a) Mass x velocity
(b) Momentum = 0.08 x 100
= 8.0 kg ms-1
(c) Decrease
(d) To lengthen the time of impact /
to reduce the impulsive force
Question 13
(a) Inelastic collision
2.4 Momentum (b) (1200)(30 ) + (1000)( – 20)
=
(a) Inelastic collision = (1200 + 1000)v
v = 7.27 m s-1
.
(b) (c) Inertia
(d) The larger the mass and
velocity, the higher the
c) mv = 2 x 100 = 2 x 1 = 2 kg ms-1
= = .
momentum.
Momentum = mass x velocity
.
(d)
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3. Truck with higher momentum Question 17
is difficult to stop or higher (a) The position between the two
impact if accident occurs balls are equal.
The distance between the two
2.6 Impulsive force balls increase.
Question 14 (b) Gravitational force
(a) Change of momentum (c) (i) gravitational acceleration, g
(b) (i) Force on surface B is larger (ii) 10 m/s2
than surface A (iii) Mass does not affect g
(ii) Time of impact on surface (d) (i) Velocity decreases
A is longer than on surface (ii) moves against gravitational
B force.
(iii) Constant/unchanged
(iv) The shorter the time of Question 18
impact, the larger the (a) Gravitational force
force //force Inversely (b) Surface area of the feather is
proportional to the time larger
of impact (c) Velocity increases constantly /
(v) Sponge/mattress constant acceleration
(c) Higher mass high velocity, Final velocity is constant.
high impact / high momentum / Final velocity for water droplet
high impulsive force is higher than final velocity for
feather.
Question 15 (d) Final velocity is inversely
(a)(i) gravitational potential energy proportional to surface area
(ii) kinetic energy (e) Graph
(b)(i) mgh = ½ mv2
v2 = 2gh = 2 x 10 x 2.5
v = 7.07 m s-1
(ii)
(c)(i) soft/spongy
Lengthen the time of impact
Reduce the impulsive force
(ii) use parachute
To reduce velocity/momentum
Question 16
(a) Impulsive force
(b) High impulsive force/short
impact time
(c) 0.05 x 5 = 0.25 kg m/s
(d) Use sponge, mattress / soft
material 2.9 Equilibrium of forces
2.8 Gravitational Force
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4. Question 19 (ii) R = mg – ma / R < mg
(a) Diagram
Question 22
(a) Force is anything that can
move a stationary object //
stop a moving object //
change direction / shape /
speed of an object.
(b) Label force
(b) F = ma
a = 5 / 2 = 2.5 m s-2
(c) 5 N but in opposite direction
Question 19(spm 08) (c) (i) Fx = 120 cos 30⁰ = 104.4 N
(a)(i) magnitude: equal (ii) FY = 120 sin 30⁰ = 60 N
Direction: opposite
(ii) zero
Question 23
(iii) equilibrium of forces
(a) Force is anything that can
(b)(i) Acceleration
move a stationary object //
(ii) there is resultant force, force is
stop a moving object // change
directly proportional to
direction / shape / speed of an
acceleration
object.
Question 20
(b) Diagram
(a) 650 = 300 + P
P = 350 N
(b) The bicycle moves with
constant velocity. The
resultant force equals zero.
(c) Velocity increases / accelerate
(d) Inertia
(e) The time of impact is short. (c) Fx = 1500 x cos 60⁰ = 750 N
High impulsive force. (d) (i) Fx : to make car move
forward // overcome
Question 21 frictional force
(a) 500 N
(b) 1. At rest (ii) Fy : to lift the car off the
2. moving down or up with ground // to move the car
constant velocity upwards// to overcome
(c) F = R – mg the weight of the car
(d) There is a resultant force acts
upward.
R = mg + ma
(e)(i) decreases Question 24
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5. (a) 500 N (ii) energy lost to heat/frictional
(b) (i) increased force
(ii) decreased
(iii) Unchanged Question 28
(c) Resultant force = 0 (a) Gravitational potential energy
(d) R = (46)(10) = 460 N (b) (i) E = mgh = (30)(10)(2.5)
F = mg - R = ma = 750 J
50a = 500 – 460 (ii) E = ½ mv2 = ½ (30)(4)(4)
a = 0.8 ms-2 = 240 J
(c) (i) Energy consumes in (b)(i) is
Question 25 larger than in (b)(ii)
(a) Attraction force (ii) energy lost to heat /
(b) (i) same magnitude, opposite frictional force
direction
(ii) equal in magnitude, Question 29
opposite direction (a) (i) Diagram 29.2 is further than
(c) (i) zero in diagram 29.1
(ii) equal (ii) decreases
(d) The net force is zero, at rest (b) Streamline
The net force is zero, moves (c) W = 70 x 10 = 700 J
with constant velocity. (d) (i) kinetic energy to
(e) Equilibrium of forces gravitational potential
energy to kinetic energy
2.10 Energy (ii) sound / heat
Question 26
(a) To gain maximum kinetic Question 30
energy before he begins to (a) (i) Trolley Q has less mass
jump. (ii) Newton’s second law
Kinetic energy increases with (b)
velocity.
(b) To gain elastic potential energy
from the pole.
Change to gravitational
potential energy.
(c) mgh = 60 x 10 x 52 = 3 120 J
(d) 10 m s-2
(e) To lengthen the time of impact
between the athlete and the (c) Velocity of the trolley
mattress. decreases
So reduce the impulsive force (d) Trolley Q moves faster along
frictionless slope
Question 27 (e)(i) mgh = 0.5 x 10 x 0.7
(a) Work = force x displacement = 3.5 J
(b) (i) 220 x 0.5 = 110 J
(ii) 20 x 10 x 0.5 = 100 J (ii) Total energy at the top of
(c) (i) Work done in 4(b)(i) is larger the slope:
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6. Kinetic energy +
gravitational potential energy
= 3.5 + 3.0 = 6.5 J
2.11 Elasticity
Question 31
(a) (i) Newton, N
(ii) e is directly proportional to
W
(iii) Hooke’s Law
(b) (i) 12 N
(c) Gradient = 12 = 240 N/m
0.05
(d) E = ½ Fx = ½ x 10 x 0.04 = 0.2 J
(e)
Question 32
(a) (i) Hooke’s Law
(b) (i) elastic potential energy
(ii) E = ½ x 60 x 0.08 = 2.4 J
(c) (i) 80 N
(ii) Spring Q
The elastic limit for spring Q
is 100 N which more than 80
N.
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