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JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
1
2. FORCE AND MOTION
2.1 ANALYSING LINEAR MOTION
Distance and displacement
1. Types of physical quantity:
(i) Scalar quantity: ………………………………………………………………….
(ii) Vector quantity: …………………………………………………………………
2. The difference between distance and displacement:
(i) Distance: …………………………………………………………………………
(ii) Displacement: ……………………………………………………………………
3. Distance always longer than displacement.
4. Example: The following diagram shows the location of Johor Bahru and Desaru. You can travel by car using existing road via Kota Tinggi, or travel by a small plane along straight path. Calculate how far it is from Johor Bahru to Desaru if you travelled by: a. The car
b. The plane
Solution:
Hands-on Activity 2.2 pg 10 of the practical book. Idea of distance and displacement, speed and velocity.
Speed and velocity
1. Speed is ..…………………………………………………………………………………
2. Velocity is: ..……………………………………………………………………………...
3. Average of speed: ………………………………………………………………………
4. Average of velocity: ……………………………………………………………………...
Only have magnitude
Have both magnitude and direction
length of the path taken
distance of an object from a point in a certain direction
the distance traveled per unit time or rate of change of distance
the speed in a given direction or rate of change of displacement
total distance traveled, s (m) , v = s m s-1
time taken, t (s) t
displacement, s (m) , v = s ms-1
Time taken, t (s) t
Kota Tinggi
The path traveled by the plane is shorter than traveled by the car.
a. by car = 41 + 53 = 94 km b. by plane = 60 km
So, Distance = 94 km Displacement = 60 km
60 km
41 km 53 km
Desaru Johor Bahru
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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5. Example:
An aeroplane flies from A to B, which is located 300 km east of A. Upon reaching B, the aero plane then flies to C, which is located 400 km north. The total time of flight is 4 hours. Calculate
i. The speed of the aeroplane ii. The velocity of the aeroplane
Solution:
Acceleration and deceleration
1. Study the phenomenon below;
Observation: ………………………………………………………………………………
2. Acceleration is, ……………………………………………………………………….
Then, a =
3. Example of acceleration;
20 m s-1 0 m s-1 40 m s-1
The velocity of the car increases.
the rate of change of velocity
Final velocity – initial velocity Time of change
Or, a = v – u t
20 m s-1 0 m s-1 40 m s-1
A B C t = 2 s t = 2 s
C
A B 300 km
i. Speed = Distance Time
= 300 + 400 4 = 175 km h-1
400 km
ii. velocity = displacement time (Determine the displacement denoted by AC and its direction)
= 125 km h-1 (in the direction of 0530) A 300 km
= . 500 . 4
B
C
400 km
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Calculate the acceleration of car;
i) from A to B
ii) From B to C
4. Deceleration happens ...…………………………………………………………………
………………………………………………………………………………………………
5. Example of deceleration;
A lorry is moving at 30 m s-1, when suddenly the driver steps on the brakes and it stop 5 seconds later. Calculate the deceleration of lorry.
Analyzing of motion
1. Linear motion can be studied in the laboratory using a ticker timer and a ticker tape.
Refer text book photo picture 2.4 page 26.
(i) Determination of time:
(ii) Determination of displacement as the length of ticker tape over a period of time. x y
(iii) Determine the type of motion;
………………………………………………………………………………………..
...……….……………………………………………………………………………..
.……………………………………………………………………………………..
aAB = 20 – 0 = 10 m s-2 2 aBC = 40 – 20 = 10 m s-2
2
when the velocity of an object decreases, In calculations, a
will be negative
Answer : v = 0 m s-1, u = 30 m s-1, t = 5 s Then , a = 0 – 30 = -6 m s-2
5
. . . . . . . . the frequency of the ticker timer = 50 Hz ( 50 ticks in 1 second) so, 1 tick = 1 second = 0.02 seconds 50
xy = displacement over time t measure by ruler
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
Uniform velocity Acceleration Acceleration, then deceleration
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
4
(iv) Determination of velocity
displacement = ……………………… time = ………………………………..
Velocity, v =
(v) Determine the acceleration
The equation of motion
1. The important symbols : ………………………………………………………………..
………………………………………………………………………………………………
2. The list of important formula;
3. Example 1 : A car travelling with a velocity of 10 m s-1 accelerates uniformly at a rate of
3 m s-2 for 20 s. Calculate the displacement of the car while it is accelerating.
Length/cm
8 7 6 5 4 3 2 1 0
u
v a = = = =
v – u t 40.0 – 15.0 .. 5(0.2) 25.0 1.0 25.0 cm s-2
ticks s : displacement, v : final velocity
u : initial velocity, t : time, a : acceleration
1. tvus )(21
+= 2. t
uva −=
3. atuv += 4. 2
21 atuts +=
5. asuv 222 +=
given : u = 10 m s-1 , a = 3 m s-2 , t = 20 s. s = ?
s = ut + ½ at2
s = (10)(20) + ½ (3)(20)2
= 800 m
s = 800 m
. . . . . . . . 12.6 cm 7 x 0.02 = 0.14 s 12.6 = 90.0 cm s-1 0.14
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Example 2 : A van that is travelling with velocity 16 m s-1 decelerates until it comes to rest. If the distance travelled is 8 m, calculate the deceleration of the van.
Exercise 2.1
1. Figure 2.1 shows a tape chart consisting of 5-tick strip. Describe the motion represented by AB and BC. In each case, determine the ; A to B acceleration, BC uniform velocity
(a) displacement
(b) average velocity Figure 2.1
(c) acceleration 2. A car moving with constant velocity of 40 ms-1 . The driver saw and obstacle in front
and he immediately stepped on the brake pedal and managed to stop the car in 8 s. The distance of the obstacle from the car when the driver spotted it was 180 m. How far is the obstacle from where the car has stopped?
given : u = 16 m s-1 , v = 0(rest) , s = 8 m a = ?
v2 = u2 + 2 as
02 = 162 + 2 a(8)
a = -16 ms-2
Length / cm
16
12
8
4
0 A B C Time/s
s = 4 + 8 + 12 + 16 + 16 + 16 = 72.0 cm
vaverage = )1.0(6
0.72
= 120.0 cm s-1
Note : v = 1.00.16 = 160 cm s-1
a = t
uv − = 5.0
40160 − u = 1.00.4 = 40.0 cm s-1
= 240 cm s-2 t = 5 (0.1) = 0.5 s
u = 40 ms-1 v = 0 t = 8 s s initial = 180 m (from car to obstacle when the driver start to step on the brake) sfinal = ? ( from car to obstacle when the stopped)
obstacle sinitial s sfinal
s = ( ) ( ) mtvu 160804021
21
=+=+
sfinal = sinitial – s = 180 – 160 = 20 m
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2.2 ANALYSING MOTION GRAPHS
The data of the motion of the car can be presented………………………………….
The displacement-time Graph
0m 100m 200m 300m 400m 500m displacement 0s 10s 20s 30s 40s 50s time
a) displacement (m) Graph analysis: ……………………………………………………………… ……………………………………………………………… time (s) ……………...………………………………………………
b) displacement (m) Graph analysis:
……..………………………………………………………… ………………………………………………………………… time (s) ……….………………………………………………………… c) displacement (m) Graph analysis:
…….…………………………………………………………… ………………………………………………………………… time (s) ..…………………………………………………………………
The object moves with uniform velocity for t seconds. After t seconds, the object returns to origin (reverse) with uniform velocity Total displacement is zero Graph is quadratic form . Displacement increases with time. Graph gradient increases uniformly The object moves with increasing velocity with uniform acceleration.
d) Displacement (m) Graph analysis:
…………………………….……………………………………… ……………………………………………..……………………… time (s) ………………………………………………………………………
…………………………………………………………………
in the form of graph called a motion graphs
Uniform displacement all the time
Graph gradient = velocity = 0
The object is stationary or is not moving
Displacement increases uniformly
Graph gradient is fixed
The object move with uniform velocity
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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The velocity-time Graph
c) v (m s-1) Graph analysis:
…………………………………..………………….
………………………………………………………
………………………………………………………
t1 t2 t (s)
e) displacement (m) Graph analysis:
…………………………………………………………..
…………………………………………………………..
………………………………………………………….. time (s)
………………………………………………………….. f) displacement (m) Graph analysis: A B ………………………………………………………….. …………………………………………………………..
…………………………………………………………… O C time (s)
a) v/ m s-1 Graph analysis:
…………………………………………………………..
…………………………………………………………..
……………………………………………………………
t t / s
b) v/ m s-1 Graph analysis:
………………………………………..………………..
…………………………………………………………
…………………………………………………………
t t / s …………………………………………………………
Its velocity increases uniformly
The graph has a constant gradient
The object moves with a uniform acceleration
The area under the graph is equal to the displacement, s of the moving object : s = ½ ( v x t)
Graph is quadratic form. Displacement increases with time. Graph gradient decreases uniformly The object moves with decreasing velocity, with uniform deceleration. OA = uniform velocity (positive – move ahead) AB = velocity is zero (rest) BC = uniform velocity (negative – reverse)
No change in velocity
Zero gradient the object moves with a constant velocity or
the acceleration is zero.
The area under the graph is equal to the displacement of the moving object :
s = v x t
The object moves with a uniform acceleration for t1 s
After t1 s, the object decelerates uniformly (negative
gradient ) until it comes to rest.
The area under the graph is equal to the displacement of
the moving object : s = ½ vt2
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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d) v (m s-1) Graph analysis:
...…………………………………..………………..
……………………………………………………..
………………………………………………………
t (s) ………………………………………………………
.……………………………………………………...
e) v (m s-1) Graph analysis:
………..…………………………..………………..
……….……………………………………………..
………………………………………………………
t (s) ………………………………………………………
Examples ………………………………………………………
1. s/m
20
10
-10
2.
O
P Q
R
S
0 2 4 6 8 t/s
O
P Q
R
v/m s-1
10 5
0 2 4 6 8 10 t/s
Calculate:- (i) acceleration,a over OP, PQ and QR (ii) Displacement
Solution : Given : VO = 0 m s-1, VP = 10 m s-1 ,
VQ = 10 m s-1 VR = 0 m s-1 tOP = 4 s tPQ = 4 s tQR = 2 s
(i) aOP = 22.5ms4
010 −=− aPQ= 2−=
− ms 041010
aQR = 2−−=− ms 5.02100
(ii) S = 70.0m10)(10)(421
=+
The shape of the graph is a curve
Its velocity increases with time.
The gradient of the graph increases.
The object moves with increasing acceleration.
The area under the graph is equal to the total displacement of the moving object. The shape of graph is a curve
Its velocity increases with time.
The gradient of the graph decreases uniformly.
The object moves with a decreasing acceleration.
The area under the graph is the total displacement of the moving object. Given : SOP = 20 m SOQ = 20 m SOR = 0 m SOS = - 10 m tOP = 2 s tPQ = 3 s tQR = 2 s tRS = 1 s
(i) VOP = 110ms220 −= VQR = 110ms
2200 −−=
−
VRS = 110ms1
010- −−=−
(ii) S = -10m
Calculate:- (i) Velocity over OP, QR and RS (ii) Displacement Solution :
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
9
Exercise 2.2
1. (a) s/m (b) s/m (c) s/m
10
t/s 0 2 4 t/s t/s
-5 -10
Figure 2.21
Describe and interpret the motion of a body which is represented by the displacement time graphs in Figure 2.21
2. Describe and interpret the motion of body which is represented by the velocity-time
graphs shown in figure 2.22. In each case, find the distance covered by the body and its displacement
(a) v/m s-1 (b) v/m s-1
10
t/s 0 2 4 t/s
-5 -10
Figure 2.22
a) The body remains at rest 5 m at the back of initial point b) The body start to move at 10 m in front of the initial point, then back to initial
point in 2 s. The body continue its motion backward for 10 m. The body move with uniform velocity.
c) The body move with increasing it velocity.
(a) The body moves with uniform velocity , 5 m s-1 backward. (b) The body start its motion with 10 m s-1 backward and stop at initial
point in 2 s, then continue moving forward with increasing velocity until 10 m s-1 in 2 s.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2.3 UNDERSTANDING INERTIA
Idea of inertia
1. ………………………………………………………………………………………………
2. ………………………………………………………………………………………………
3. ………………………………………………………………………………………………
Hand-on activity 2.5 in page 18 of the practical book to gain an idea of inertia
4. Meaning of inertia :
…………..………………………………………………………………………………….
………………………………………………………………………………………………
Mass and inertia
1. Refer to figure 2.14 of the text book, the child and an adult are given a push to swing.
(i) which one of them will be more difficult to be moved ……………………...
(ii) which one of them will be more difficult to stop? …………………………….
2. The relationship between mass and inertia :
……………………………….……………………………………………………………..
3. The larger mass ………………………………………………………………………….
………………………………………………………………………………………………
Effects of inertia
1. Positive effect : …………………………………………………………………………
(i) ………………………………………………………………………………………
(ii) ………………………………………………………………………………………
(iii) ………………………………………………………………………………………
2. Negative effect : ………………………………………………………………………….
(i) ……………………………………………………………………………………...
……………………………………………………………………………………..
(ii) ………………………………………………………………………………………
………………………………………………………………………………………
(iii) ………………………………………………………………………………………
………………………………………………………………………………………
(iv) ………………………………………………………………………………………
A pillion rider is hurled backwards when the motorcycle starts to move.
Bus passengers are thrust forward when the bus stop immediately.
Large vehicle are made to move or stopped with greater difficulty.
The inertia of an object is the tendency of the object to remain at rest or, if moving, to
continue its uniform motion in a straight line
An adult
An adult
The larger the mass, the larger its inertia.
have the tendency to remain its situation either at rest or in
moving.
Application of inertia
Drying off an umbrella by moving and stopping it quickly.
Building a floating drilling rig that has a big mass in order to be stable and safe.
To tight the loose hammer
We should take a precaution to ovoid the effect.
During a road accident, passengers are thrust forward when their
car is suddenly stopped.
Passengers are hurled backwards when the vehicle starts to move and are hurled
forward when it stops immediately.
A person with a heavier/larger body will find it move difficult to stop his movement.
A heavier vehicle will take a long time to stop.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Exercise 2.3
1. What is inertia? Does 2 kg rock have twice the inertia of 1 kg rock?
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
2.
Figure 2,3 A wooden dowel is fitted in a hole through a wooden block as shown in figure 2.31. Explain what happen when we
(a) strike the top of the dowel with a hammer,
………………………………………………………………………………………
………………………………………………………………………………………
(b) hit the end of the dowel on the floor. ………………………………………………………………………………………
……………………………………………………………………………………
2.4 ANALYSING MOMENTUM
Idea of momentum
1. When an object is moving, …...…………………………………………………………
2. The amount of momentum ...……………………………………………………………
3. Momentum is defined…………………………………………………………………….
………………………………………………………………………………………………
it has momentum.
depends on its mass and velocity.
as the product of its mass and its velocity, that is
Momentum, p = m x v Unit= kg m s-1
Inertia is the tendency of the object to remain at rest or, if moving, to continue its
uniform motion in a straight line.
Yes, the inertia increase with the mass increased.
A wooden block moves up of a wooden dowel.
A wooden block has inertia to remains at rest.
The wooden block move downward of a wooden dowel.
A wooden block has inertia to continue it motion.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Conservation of momentum
The principle of conservation of momentum :
………………………………………………………………………………………………………
………………………………………………………………………………………………………
1. Elastic collision .…………………………………………………………………………..
Before collision after collision
(mb + mg)
mg
vg = 0
mb
vb&g
Momentum = mbvb
Momentum = (mb+mg)vb&g
Starting position before she catches the ball
vb
Receiving a massive ball
mb vb
mg
vg
Momentum = mbvb
Momentum = - mgvg
Starting position before she throws the ball
Throwing a massive ball
In the absence of an external force, the total momentum of a system remains
unchanged.
The colliding objects move separately after collision.
Momentum : m1u1 + m2u2 = m1v1 + m2v2
m1 m2 m1
m2
u2 v2
u1
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2. Inelastic collision :………………………………………………………………………...
Before collision after collision
3. explosion : …….....…………………………………………………………………...
Before explosion after explosion
Example 1 :
Car A Car B
Car A of mass 100 kg travelling at 30 m s-1 collides with Car B of mass 90 kg travelling at 20 m s-1 in front of it. Car A and B move separately after collision. If Car A is still moving at 25 m s-1 after collision, determine the velocity of Car B after collision.
Solution :
Example 2 :
Car A of mass 100 kg travelling at 30 m s-1 collides with Car B of mass 90 kg travelling at 20 m s-1 in front of it. Car A is pulled by Car B after collision. Determine the common velocity of Car A and B after collision.
Solution :
m1 m2 m1 + m2
u2 = 0
u1 v
The colliding objects move together after collision. Momentum : m1u1 + m2u2 = (m1 + m2) v
The objects involved are in contact with each other before explosion and are separated after the explosion.
Momentum : (m1 + m2)u = m1 vv - m2 v2 Given : mA = 100 kg , uA = 30 m s-1, vA = 25 m s-1, mB = 90 kg,
uB = 20 m s-1 , vB = ? mAuA + mBuB = mAvA + mBvB (100)(30) + (90)(20) = (100)(25) + (90)(vB) vB = 25.56 m s-1
(m1 + m2), u = 0 v1 m2
v2
Given : mA = 100 kg , uA = 30 m s-1, mB = 90 kg, uB = 20 m s-1 , v(A+B) = ? mAuA + mBuB = (mA + mB ) v (B+A) (100)(30) + (90)(20) = (100 + 90) v (B+A) v(A + B) = 25.26 m s-1
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Example 3 :
A bullet of mass 2 g is shot from a gun of mass 1 kg with a velocity of 150 m s-1 . Calculate the velocity of the recoil of the gun after firing. Solution :
Exercise 2.4
1. An arrow of mass 150 g is shot into a wooden block of mass 450 g lying at rest on a smooth surface. At the moment of impact, the arrow is travelling horizontally at 15 ms-1. Calculate the common velocity after the impact.
2. A riffle of mass 5.0 kg fires a bullet of mass 50 g with a velocity of 80 m s-1 .Calculate the recoil velocity. Explain why the recoil velocity of a rifle is much less than the velocity of the bullet.
2.5 UNDERSTANDING THE EFFECT OF A FORCE
Idea of force
1. What will happen when force act to an object?
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
Given ; mb = 2 g = 0.002 kg, mg = 1 kg, u(g+b) = 0 , vb = 150 m s-1 vg = ?
0 = mgvg – mb vb, 0 = (1)(vg) – (0.002)(150), vg
= 0.3 m s-1
Force can make an object;
1. Move 2. Stop the moving
3. Change the shape of the object 4. Hold the object at rest
ma = 150 g mwb = 450 g m (a+wb) = 600 g va = 15 m s-1 vwb = 0 v(a+ wb) = ? mava + mwbvwb = m(a+wb)v(a+wb) , (0.15 x 15) + (0.450 x 0) = 0.6 v(a+ wb) v(a+ wb) = 3.75 m s-1
mr = 5.0 kg mb = 50 g vr = ? vb = 80 m s-1 mr vr = mb vb , ( 5.0 ) vr = ( 0.05)(80) vr = 0.8 m s-1
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Idea of balanced forces
1. An object is said to be in balance when it is:
………………………………………………………………………………………………
………………………………………………………………………………………………
2. Stationary object
……………………………… explanation :
………………………………………………
………………………………………………
……….……………………………………..
…………………………………………
3. An object moving with uniform velocity
…………………………….. explanation :
…..……………. …………… ……………………………………………..
……………………………………………..
……………………………………………..
……………………………… ………..…………………………………….
……………………………………………..
……………………………………………..
Idea of unbalanced forces
1. A body is said to be in unbalanced..……………………………………………………
2. ……………………….. Explanation;
………………………………………………
………………………………………………
………………………………………………
……… …….. ………………………………………………
Relationship between forces, mass and acceleration (F = ma)
Experiment 2.2 page 29. Aim : To investigate the relationship between acceleration and force applied on a constant mass.
Experiment 2.3 page 31 Aim: To investigate the relationship between mass and acceleration of an object under constant force.
Stationary object
Normal reaction, N
Frictional force Force, F Force , F = Friction
Resultant = F – Friction
= 0 (object is in equilibrium)
weight, w = mg Examples :
1.A car move at constant velocity.
2.A plane flying at constant velocity.
when it is moving in acceleration.
Resultant force
The ball move in acceleration
because the forces act are not balanced.
F > F’
F F’ So, the ball move in F direction
1. In a stationary state
2. Moving at uniform velocity
Normal reaction, N
Magnitude R = W but R acts in an
opposite direction to the weight.
( object is in equilibrium )
weight, w = mg
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
16
1. Refer to the result of experiment 2.2 and 2.3,
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
2. 1 Newton (F = 1 N) is defined as the force required to produce an acceleration of 1 m s-2 (a=1 m s-2) when it is acting on an object of mass 1 kg ( m = 1 kg)
So, …………………………………………………………………………………………
3. Example 1 : Calculate F, when a = 3 m s-2 and m = 1000 kg
Example 2 :
Calculate the acceleration, a of an object.
Exercise 2.5
1. A trolley of mass 30 kg is pulled along the ground by horizontal force of 50 N. The opposing frictional force is 20 N. Calculate the acceleration of the trolley.
2. A 1000 kg car is travelling at 72 km h-1 when the brakes are applied. It comes to a stop in
a distance of 40 m. What is the average braking force of the car?
mm == 2255 kkgg FF == 220000 NN
F = ma
F = ma
F = (1000)(3)
F = 3000 N F = ma 200 = 25 a a = 8.0 ms-2
it is found that; a ∝ F when m is constant and a ∝ 1/m when F is constant.
Therefore, a ∝ F/m
From a ∝ F/m,
F ∝ ma
Therefore, F = kma … k =constant =1
m = 30 kg , F = 50 N , Ff = 20 N , a = ? F – Ff = ma , 50 – 20 = 30 a a = 1.0 m s2 m = 1000 kg , u = 72 km h-1, v = 0, s = 40 m, F = ? Note : u = 72 km h-1 =20 m s-1 F = ma, v2 = u2 + 2as = 1000 x 5.0 0 = 202 + 2a(40) = 5000.0 N a = 5.0 m s2
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
17
2.6 ANALYSING IMPULSE AND IMPULSIVE FORCE
Impulse and impulsive force
1. Impulse is ……………………………………………………………………………….
2. Impulsive force is ………………………………………………………………………
………………………………………………………………………………………………
3. Formula of impulse and impulsive force:
Refer, F = ma
Example 1; v u
wall
If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 1 s
Impulse, Ft = and impulsive force, F =
Example 2; v u
Wall with a soft surface
If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 2 s
Impulse, Ft = and impulsive force, F =
4. The relationship between time of collision and impulsive force.
………………………………………………………………………………………………
………………………………………………………………………………………………
Exercise 2.6
The change of momentum
The large force that acts over a short period of time during
collision and explosion.
It is known that a = ( v – u ) / t
Therefore, F = m( v – u) t So, Ft = mv – mu , Unit = N s
Ft is defined as impulse, which is the change in momentum. F = mv – mu ,
t Ft = mv – mu Unit : newton (N) F is defined as impulsive force which is the rate of change of momentum over the short period of time
5(10) - (- 5(10)) 100 = 100 N
= 100 Ns 1
5(10) - (- 5(10)) 100 = 50 N
= 100 Ns 2
Impulsive force , F ∝ 1 / t
Therefore, F decreases when the time of collision increases ( refer to examples )
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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1. A force of 20 N is applied for 0.8 s when a football player throws a ball from the sideline. What is the impulse given to the ball?
2. A stuntman in a movie jumps from a tall building an falls toward the ground. A large
canvas bag filled with air used to break his fall. How is the impulsive force reduced? 2.7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES
Importance of safety features in vehicles
Safety features in vehicles
Crash resistant door pillars
Anti-lock brake system (ABS)
Traction control bumpers
Windscreen
Air bags
Head rest
Crumple zones
Reinforced passenger compartment
Fimpulse = Ft = 20 x 0.8 = 16.0 Ns
1. A large canvas bag will increase the time of collision. 2. When the time of collision increase the impulsive force will decrease.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Safety features Importance
Padded dashboard Increases the time interval of collision so the impulsive force produced during an impact is thereby reduced
Rubber bumper Absorb impact in minor accidents, thus prevents damage to the car.
Shatter-proof windscreen Prevents the windscreen from shattering
Air bag Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers.
Safety seat belt Prevents the passengers from being thrown out of the car. Slows down the forward movement of the passengers when the car stops abruptly /suddenly.
Side bar in doors Prevents the collapse of the front and back of the car into the passenger compartment. Also gives good protection from a side-on collision.
Exercise 2.7
1. By using physics concepts, explain the modifications to the bus that help to improve that safety of passengers and will be more comfortable.
- The absorber made by the elastic material : To absorb the effect of impact (hentaman) during it moving
- Made by the soft material of bumper : To increase the time during collision, then the impulsive force will be decreased.
- The passenger’s space made by the strength materials. : To decrease the risk trap to the passenger during accident. - Keep an air bag at the in front of dash board and in front of passengers
: Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers.
- Shatter-proof windscreen : Prevents the windscreen from shattering.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2.8 UNDERSTANDING GRAVITY
Carry out hands-on activity 2.8 on page 35 of the practical book.
Acceleration due to gravity.
1. An object will fall to the surface of the earth because………………………………...
2. The force of gravity also known ………………………………………………………...
3. When an object falls under the force of gravity only, ………………………………...
………………………………………………………………………………………………
4. The acceleration of objects falling freely ………………………………………………
5. The magnitude of the acceleration due to gravity depends ………………………...
………………………………………………………………………………………………
Gravitational field
1. The region around the earth is ………………………………………………………….
2. The object in gravitational field …………………………………………………………
3. The gravitational field strength is defined ……………………………………………..
4. The gravitational field strength, g can be calculated as;
5. At the surface of the earth,
…………….………………………………………………………………………………..
6. This means
……………………………………………………………………………………………..
7. Example 1. Can you estimate the gravitational force act to your body? mass = 60 kg, g = 9.8 N kg-1, F = ? Example 2,
A satellite of mass 600 kg in orbit experiences a gravitational force of 4800 N. Calculate the gravitational field strength.
It pulled by the force of gravity.
as earth’s gravitational force.
the object is said to be free
falling
is known as acceleration due to gravity.
on the strength of the gravitational field
.
the gravitational field of the earth.
is on the force of gravity.
as the gravitational force acting on a 1 kg mass.
. g = F . where, F : gravitational force m m : mass of an object
g = 9.8 N kg-1
that an object of mass 1 kg will experience a gravitational force of 9.8 N.
Solution : F = mg = (60) (9.8)
= 588.0 N
Given : m = 600 kg. F = 4800 N, g = ?
g = F = 4800 . = 8 N kg-1
m 600
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Example 3, A stone is released from rest and falls into a well. After 1.2 s, it hits the bottom of the well.
(a) What is the velocity of the stone when it hits the bottom?
(b) Calculate the depth of the well.
Weight
1. The weight of an object is defined ……………………………………………………..
2. For an object of mass m, the weight can be calculated as :
Example : The mass of a helicopter is 600 kg. What is the weight of the helicopter when it land on the peak of a mountain where the gravitational field is 9.78 N kg-1?
Exercise 2.8
1. Sketch the following graphs for an object that falling freely.
(a) Displacement-time graph, (b) Velocity-time graph (c) Acceleration-time graph
as the gravitational force acting on the object.
weight, W = mg
where, g = acceleration due to gravity.
W = mg
= 600 x 9.78
= 58 68 N
Given : u = 0 ms-1, t = 1.2 s, a = g = 9.8 ms-2
(a) v = ? v = u + at
= 0 + (9.8)(1.2)
= 11.76 ms-1
(b) Depth = s = ? s = ut + ½ at2
= (0)(1.2) + ½ (9.8)(1.2)2
= 7.0561 m
(a) s / m (b) v / m s-1 (c) a / m s2 t / s t / s t / s
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2. The following data was obtained from an experiment to measure the acceleration due to gravity.
Mass of steel bob = 200 g, distance covered = 3.0 m, time of fall = 0.79 s. Calculate the acceleration due to gravity of steel bob.
Give the explanation why your answer different with the constant of gravitational acceleration, g = 9.8 m s-2.
2.9 IDEA OF EQUILIBRIUM FORCES
An object is in equilibrium when :
1. ………………………………………………………………………………………………
2. ………………………………………………………………………………………………
stationary object
An object moving with uniform velocity
It is in a stationary state
It is moving with uniform velocity
Normal reaction, R Normal reaction, R
Weight, W=mg weight, W=mg
Magnitude of R = W Magnitude of R = mg cos θ R and W acts in opposite direction. And acts in opposite direction. So, Resultant force = W – R = 0 So ,Resultant force = mg cos θ – R = 0
( object in equilibrium ) ( object in equilibrium ) normal reaction, R friction force force, F Weight, W
Force , F = Frictional force Resultant force = F – Frictional force
= 0 (object in equilibrium)
m = 200 g s = 3.0 m t = 0.79 s u = 0 g = ? = 0.2 kg s = ut + ½ g t2 3.0 = 0 (0.79) + ½ g (0.792) g = 9.6 m s-2
The answer is less than the value of g because of the air frictional force.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Addition of Force
1. Addition of force is defined as ...……………………………………………………..
………………………………………………………………………………………………
………………………………………………………………………………………………
Examples : the forces are acting in one direction
F1 = 10 N
F2 = 5 N
Resultant force, F
Example : the forces are acting in opposite directions
F1 = 10 N
F2 = 5 N
Resultant force, F
Example : the forces are acting in different directions
F2 = 5 N
500 F
F1 = 10 N
Parallelogram method:
1. Draw to scale.
2. Draw the line parallel with F1 to the edge of F2, and the line parallel with F2 to the
edge of F1
3. Connect the diagonal of the parallelogram starting from the initial point.
4. Measure the length of the diagonal from the initial point as the value of the
resultant force.
a resultant force is a single force the
represents in magnitude and direction two or more forces acting on an object
F resultant = the total of forces (including the directions of the forces)
= F1 + F2 = 10 + 5 = 15 N
= F1 - F2 = 10 - 5 = 5 N
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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F2
F
F1
Triangle method
1. Draw to scale.
2. Displace one of the forces to the edge of another force.
3. Complete the triangle and measure the resultant force from the initial
point.
Example 1: During Sport Day two teams in tug of war competition pull with forces of
6000 N and 5300 N respectively. What is the value of the resultant force?
Are the two team in equilibrium?
Example 2: A boat in a river is pulled horizontally by two workmen. Workmen A
pulls with a force of 200 N while workmen while workmen B pulls with a
force of 300 N. The ropes used make an angle 250 with each other. Draw a
parallelogram and label the resultant force using scale of 1 cm : 50 N.
Determine the magnitude of resultant force.
Solution : Resultant force, F = 6000 – 5300 =700 N They were not in equilibrium Resultant force, F = 10.5 x 50 = 525 N
250
10.5 cm
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Resolution of a force
1. Resolution of a force is …………………………………………………………………
θ
Refer to trigonometric formula:
Example : The figure below shows Ali mopping the floor with a force 50 N at an angle of 600 to the floor.
F = 50 N
Example of resolution and combination of forces
reverse process of finding the resultant force
Fy F is the resultant force of Fx and Fy Therefore, F can be resolved
into Fx and Fy F
Vertical Component
Fx horizontal component
Cos θ = FFx , therefore Fx = F cos θ
Sin θ = FFy , therefore Fy = F sin θ
Fx
Fx = F cos θ = 50 cos 60 = 50 (0.5)
= 25 N Fy = F Sin θ = 50 sin 600
Fy = 50 (0.8660) = 43.3 N
F = mg sin 400 + 200
= 800(0.6427) + 200 = 514.2 + 200 = 714.2 N
mg = 800 N
600
F = ?
200 N
400
400
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Problem solving
1. When a system is in equilibrium, ……………………………………………………….
2. If all forces acting at one point are resolved into horizontal and vertical
components, ……………………………………………………………………………
3. Example 1; Show on a figure;
a) the direction of tension force, T of string b) the resultant force act to lamp c) calculate the magnitude of tension force, T a)
mlamp = 1.5 kg
Wlamp = 14.7 N
Exercise 2.9
1. Two force with magnitude 18 N and 6 N act along a straight line. With the aid of diagrams, determine the maximum possible value and the minimum possible value of the resultant force.
2. A football is kicked simultaneously by two players with force 220 N and 200 N respectively, as shown in Figure 2.9. Calculate the magnitude of the resultant force.
220 N
900
200 N
T b) T’ T 700 700
the resultant force is equal to zero.
the sum of each component is equal to zero.
(c ) T’ = 2T sin 700 Therefore, mlampg = 2T sin 700
T = 0lamp
2sin70
gm
= 02sin701.5(9.8) = 7.82 N
Fmaximum when both of forces act in same direction; Fmaximum = 18 + 6 18 N 24 N = 24 N 6 N Fminimum when the forces act in opposite direction ; Fminimum = 18 – 6 18 N 12 N = 12 N 6 N F = Resultant of Force F2 = 2202 + 2002 F = 297.32 N F
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2.10 UNDERSTANDING WORK, ENERGY AND EFFICIENCY
Work
1. Work is done, ……………………………………………………………………………..
………………………………………………………………………………………………
2. WORK is the product.…………………………………………………………………….
………………………………………………………………………………………………
3. The formulae of work;
4. Example 1;
Example 2;
80 N
600
s = 5 m
When a force that acts on an object moves the object through a
distance in the direction of the force.
of a force and the distance traveled in the direction of
the force.
WORK = FORCE X DISPLACEMENT
W = F x s
W : work in Joule/J
F : force in Newton/N
s : displacement in meter/m
W = Fs
If, F = 40 N and s = 2 m
Hence, W = 40 x 2
= 80 J
Force, F
s
W = Fs
= 80 cos 600 (5)
= 80 (0.5) (5)
= 200 J
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Example 3;
Example 4;
F = 600 N
S = 0.8 m
Energy
1. Energy is .................................................................................................................
2. Energy cannot be ....................................................................................................
3. Exist in various forms such as …………………...……………………………………
………………………………………………………………………………………………
4. Example of the energy transformation;
………………………………………………………………………………………………
………………………………………………………………………………………………
5. ………………………………………………………………………………………………
Example :
………………………………………………………………………………………………
T T
F = 30 N
h = 1.5 m W = F s = F h
= 30 (1.5)
= 45.0 J
W = F s = 600 x 0.8 = 480 J
It is the potential to do work.
created nor be destroyed.
potential energy, kinetic energy, electrical
energy, sound energy, nuclear energy, heat and chemical energy.
When we are running up a staircase the work done consists of energy change from
Chemical Energy à Kinetic Energy à Potential Energy
The energy quantity consumed is equal to the work done.
If 100 J of work is done, it means 100 J of energy is consumed.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Work done and the change in kinetic energy
1. Kinetic energy is …………………………………………………………………………
2. Refer to the figure above,
3. Example 1; A small car of mass 100 kg is moving along a flat road. The resultant force on the car is 200 N. a) What is its kinetic energy of the car after moving through 10 m? b) What is its velocity after moving through 10 m?
Work done and gravitational potential energy
h = 1.5 m
1. Gravitational potential energy is………………………………………………………...
………………………………………………………………………………………………
2. Refer to the figure above;
3. Example; If m = 10 kg
s
Force, F
Through, v2 = u2 +2as u = 0 and, as = ½ v2
energy of an object due to its position.
(possessed by an object due to its position in a gravitational field)
W = Fs = mg h where, F = mg So, Gravitational energy, Ep = mgh
energy of an object due to its motion.
Work = Fs
= mass
= m ( ½ v2)
The formulae of Kinetic energy, Ek = ½ mv2
Solution : Given : m = 100 kg , F = 200 N
a. Kinetic energy, Ek = Fs
= 200 x 10= 2000 J
b. Velocity, v à ½ mv2 = 2000
v = 6.32 m s-1
W = 10 (10) 1.5 = 150 J
Therefore Work done = 150J And, Ep = 150 J
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Principle of conservation of energy
Carry out hands-on activity 2.10 on page 38 of the practical book.
To show the principle of conservation of energy.
1. Energy cannot be ………………………………………………………………………
……………………………………………………………………………………………
2. Example : a thrown ball upwards will achieve a maximum height before changing its direction and falls
3. Example in calculation : A coconut falls from a tree from a height of 20 m. What is the velocity of coconut just before hitting the earth?
Power
1. Power is …………………………………………………………………………………
2. A weightlifter lifts 180 kg of weights from the floor to a height of 2 m above his head in a
time of 0.8 s. What is the power generated by the weightlifter during this time? g = 9.8 ms-2)
created or destroyed but can be changed from one form to
another form.
Maximum Potential energy
Kinetic energy decrease potential energy decrease and potential energy and kinetic energy Increase increase
Maximum kinetic energy
Given : h = 20 m, u = 0 , g = 9.8 ms-2 , v = ?
Ep = Ek
mgh = ½ mv2
m(9.8)(20) = ½mv2
v2 = 392, v = 19.8 m s-1
the rate of doing work.
Therefore, power, P = timetakenworkdone , so, P =
tW
Where, P : power in watt/W W : work in joule/J
t : time to do work in seconds/s
Solution : Given : m = 180 kg, h = 2 m, t = 0.8 s and g = 9.8 ms-2. P = ?
P = t
W = t
mgh
= 0.8
29.8180 ×× = 4 410 W
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Efficiency
1. Defined……..…………………………………………………………………………….
2. Formulae of efficiency :
3. Analogy of efficiency;
Energy transformation
4. Example; An electric motor in a toy crane can lift a 0.12 kg weight through a height of
0.4 m in 5 s. During this time, the batteries supply 0.8 J of energy to the motor. Calculate (a) The useful of output of the motor.
(b) The efficiency of the motor
Carry out hands-on activity 2.11 on page 39 of the practical book to measure the power.
Device/ mechineDevice/ mechine
as the percentage of the energy input that is transformed into useful energy.
%100×=inputEnergy
outputenergyUsefulEfficiency
unwanted energy
Energy input, Einput Useful energy, Eoutput
Solution : Given : m = 0.12 kg, s= 0.4 m, t = 5 s, Einput = 0.8 J
(a) Eoutput = ?
Eoutput = F x s
= (0.12 x 10) x 0.4
= 0.48 J
(b) Efficiency = ?
Efficiency %100xEE
input
output=
100%x0.800.48
= %60=
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Exercise 2.10
1. What is the work done by a man when he pushes a box with a force of 90 N through a distance of 10 m? State the amount of energy transferred from the man to the force.
2. A sales assistant at a shop transfers 50 tins of milk powder from the floor to the top shelf.
Each tin has a mass of 3.0 kg and the height of thee top shelf is 1.5 m.
(a) Calculate the total work done by the sales assistant.
(b) What is his power if he completes this work in 250 s? 2.11 APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY
OF DEVICES
1. During the process of transformation the input energy to the useful output
energy,……………………………………………………………………………………..
2. .……………………………………………………………………………………………..
3. ………………………………………………………………………………………………
Example of wasting the energy;
………..…………………
Input energy output from the petrol energy
…………………… ……………. ……………… …………………….
..………………….. …………….. ………………….. …………………….
..………………….. ……………. …………………. …………………….
some of the energy transformed into unwanted forms of energy.
The efficiency of energy converters is always less than 100%.
The unwanted energy produced in the device goes to waste.
Kinetic energy
Energy loss due to Energy loss Energy loss Energy loss due to friction at
friction in as heat as sound other parts in the
moving parts engine
W = F s The energy transferred to the force = 900 J = 90 x 10 = 900 J m = 3.0 x 50 = 150 kg h = 1.5 m W = mhg = 150 x 9.8 x 1.5 = 2205 J
P = t
W
= 2502205 = 8.82 W
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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4. The world we are living in face acute shortage of energy.
5. It is very important that a device makes
…………………………………………………………………
Ways of increasing the efficiency of devices
1. Heat engines ……………………..………………………………………………………
………………………………………………………………………………………………
2. Electrical devices. ...……………………………………………………………………...
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
Operation of electrical devices
1. The electrical devices increase the efficiency………………………………….……
2. Proper management ….....………………………………………………………………
3. …………..………………………………………………………………………………
………………………………………………………………………………………………
when they are in good operating
condition. will increase the life span of device.
Example : -the filter in an air-conditioner and fins of the cooling coil of a
refrigerator must be periodically cleaned.
Refrigerators
- choose the capacity according to the size of the family.
- installed away from source of heat and direct sunlight.
- the door must always be shut tight.
- more economical use a large capacity refrigerator.
- use manual defrosts consumption.
Washing machines
- use a front loading as such more economical on water and electricity
- front loading use less detergent as compared to a top loading machine.
the best possible use of the input energy.
Engine must be designed with the capability to produce greater amount
of mechanical work.
Light Fittings
- replace filament light bulb with fluorescent lamps which have higher efficiency.
- use a lamp with a reflector so that the illumination can be directed to specific areas
of the user.
Air-conditioners.
- choose a model with a high efficiency.
- accommodate the power of air-conditioner and the size of the room
- Ensure that the room totally close so that the temperature in the room can be
maintained.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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2.12 UNDERSTANDING ELASTICITY
Carry out Hands-on activity 2.12 page 40 of the practical book.
1. Elasticity is ……………………………………………………………………………...
………………………………………………………………………………………………
2. Forces between atoms …………………………………………………………………..
………………………………………………………………………………………………
3. Forces between atoms in equilibrium condition
Explanation :
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
4. Forces between atoms in compression
Explanation ;
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
5. Forces between atoms in tension
force of attraction
stretching force stretching force
Explanation ;
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
Force of repulsion
Force of attraction
Force of repulsion
compressive force compressive force
Force of repulsion Force of repulsion
1. Force of attraction takes effect.
2. When the compressive force is removed, force of repulsion between the
atoms pushes the atom back to their equilibrium positions.
1. Force of repulsion takes effect.
2. When the compressive force is removed, force of repulsion between the atoms
pushes
the atom back to their equilibrium positions.
the property of an object that enables it to return its original shape and
dimensions after an applied external force is removed.
The property of elasticity is caused by the existence of forces of
repulsion and attraction between molecules in the solid material.
1. The atoms are separated by a distance called the equilibrium distance and vibrate
at it position.
2. Force of repulsion = Force of attraction
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Carry out Experiment 2.4 on page 41 of the practical book To investigate the relationship between force and extension of a spring Hooke’s Law
1. Hooke’s Law states ………………………………………………………………………
………………………………………………………………………………………………
2. Elastic limit of a spring is defined……………………………………………………….
………………………………………………………………………………………………
3. The spring is said to have a permanent extension,...…………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
4. The elastic limit is not exceeded,…………………………………………….…………
………………………………………………………………………………………………
………………………………………………………………………………………………
5. GGrraaff FF aaggaaiinnsstt xx
F/ N
6. Spring Constant, k
F/N
0.8
0 8 x/cm
x (cm) 0
F = kx Spring obeying Hooke’s Law
E Q P
Spring not obeying Hooke’s law (exceeded the elastic limit)
R
Force constant, k = F with unit N m-1, N cm-1 or N mm-1 x
k is the gradient of the F - x graph
k = xF
= 80.8
= 0.1 N cm-1
When the spring obey Hooke’s Law.
The mathematical expression for Hooke’s Law is :
F ∝ x
F = kx, k = Force constant of the spring
Force constant, k = F with unit N m-1, N cm-1 or N mm-1 x
that the extension of a spring is directly proportional to the applied
force provided that the elastic limit is not exceeded.
as the maximum force that can be applied to
spring such that the spring will return to its original length when the force released.
when the length of the
spring longer than the original length even though the force acts was released and the
elastic limit is exceeded.
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Example 1; A spring has an original length of 15 cm. With a load of mass 200 g attached, the length of the spring is extended to 20 cm.
a. Calculate the spring constant. b. What is the length of the spring when the load is in increased
by 150 g? [assume that g = 10 N kg-1]
Example 2;
Elastic potential energy
1. Elastic potential energy ………………………………………………………………..
spring with the original length
F compression
x spring compressed x
F x = compression x
x F spring extended
x = extension F, extension
Other situation where the spring extended and compressed
Given : lo = 15 cm, m = 200 g , F = 2.0 N, l = 20 cm x = 5 cm
a. k = ?, k = 10.4Ncm52.0
xF−==
The graph shows the relationship between the stretching force, F and the spring extension, x. (a) Calculate the spring constant of P and Q. (b) Using the graph, determine the stretching force acts to spring P and spring Q, when their extensions are 0.5 cm Solution a. Spring constant, k = gradient of graph
kP = 1cm N 15.790.38
6 −=
kQ = 1cm N 6.00.53 −=
b. When x = 0.5, FP = 8.0 N ( extrapolation of graph P) FQ = 3.0 N
the energy stored in a spring when it is extended or compressed
b. l = ? , when m = 150 g, F = 1.5 N From a, k = 0.4 N cm-1
x = cm75.8==0.43.5
kF
l = 15 + 8.75 = 23.75 cm
F (N)
x (cm)
P
Q
8
7
6
5
4
3
2
1 0 0.1 0.2 0.3 0.4 0.5
Graph F against x of spring P and spring Q
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Relationship between work and elastic potential energy
Graph F against x
Example ;
Factors that effect elasticity
Hands-on activity 2.13 on page 42 the practical book to investigate the factors that affect elasticity.
Type of material different same same same
Diameter of spring wire same different same same
Diameter of spring same same different same
Length of spring same Same same different
Summarise the four factors that affect elasticity
Factor Change in factor Effect on elasticity
Shorter spring Less elastic Length
Longer spring More elastic
Smaller diameter Less elastic Diameter of spring
Larger diameter More elastic
Smaller diameter More elastic Diameter of spring wire
Larger diameter Less elastic
Type of material the elasticity changes with the type of materials
x / cm
F/N
F x
Area under the graph = work done = ½ Fx So, Elastic potential energy = ½ Fx
15 cm
5 kg
8 cm
x = 15 – 8 = 7 cm = 0.07 m Force act to the spring, F = 5 x 10 = 50 N Elastic potential energy = ½ Fx
= ½ 50 (0.07) = 1.75 J
JPN Pahang Physics Module Form 4 Teachers’ Guide Chapter 2 : Force and Motion
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Exercise 2.12
1. A 6 N force on a spring produces an extension of 2 cm. What is the extension when the force is increased to 18 N? State any assumption you made in calculating your answer.
2. If a 20 N force extends a spring from 5 cm to 9 cm,
(a) what is the force constant of the spring?
(b) Calculate the elastic potential energy stored in the spring. Reinforcement Chapter 2 Part A : Objective Questions 1. When a coconut is falling to the
ground, which of the following quantities is constant?
A. Velocity B. Momentum C. Acceleration D. Kinetic energy 2. In an inelastic collision, which of the
following quantities remains constant before and after the collision?
A. Total acceleration B. Total velocity C. Total momentum D. Total kinetic energy
3. Calculate the weight of a stone with mass 60 g on the surface of the moon. (The gravitational acceleration of the moon is 1/6 that of the Earth.)
A. 0.1 N B. 0.2 N C. 0.4 N D. 0.6 N E. 0.8 N 4. The momentum of a particle is
depend on A. mass and acceleration B. weight and force C. mass and velocity
60 g = 0.06 kg W = 0.06 (1/6)(10) = 0.1 N
To solve the problem, determine the spring constant to use the formula F = k x F = 6 N , x = 2 cm F = kx When, F = 18 N, x = ? 6 = k (2) 18 = 3 x k = 3 N cm-1 x = 6 cm F = 20 N, x = 9 – 5 = 4 cm, k = ? F = kx 20 = k (4) k = 5 N cm-1 E = ½ Fx = ½ (20)(0.04) = 0.4 J
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5. Which of the following diagrams shows a body moving at constant velocity?
A. 2 N 2N B. 12 N 7 N C. 12 N 14 N D. 20 N 17 N
6. The graph below shows the motion of a trolley with mass 1.5 kg.
Velocity / ms-1 4 0 2 4 6 Time / s
Calculate the momentum of the trolley from t = 2s to t = 4s.
A. 1.5 kg m s-1 B. 3.0 kg m s-1 C. 4.0 kg m s-1 D. 6.0 kg m s-1 E. 7.5 kg m s-1
7. This figure shows an aircraft flying in the air.
8. m = 0.3 kg
5 m
What is the momentum of the stone just before it hits the ground?
A. 0.15 kg m s-1 B. 0.3 kg m s-1 C. 1.5 kg m s-1 D. 3.0 kg m s-1 E. 15.0 kg m s-1
Solution :
9. A big ship will keep moving for some distance when its engine is turned off. This situation happens because the ship has A. great inertia B. great acceleration C. great momentum D. great kinetic energy 10. An iron ball is dropped at a height of
10 m from the surface of the moon. Calculate the time needed for the iron ball to land. (Gravitational acceleration of the moon is 1/6 that of the Earth and g = 9.8 N kg-2)
A 0.6 s B 1.4 s C 1.7 s D 3.5 s E 12.0 s
P = mv = 1.5 x 4 = 6.0 kg ms-1
Lift Thrust Air friction Weight
The aircraft above accelerates if A. Lift > Weight B. Thrust > Lift C. Lift > Air friction D. Thrust > Air friction
P = mv (find v first to calculate the P) Ep = Ek è mgh = ½ mv2 (0.3)(10)(5) = ½ (0.3) v2 v = 10 m s-1 P = (0.3)(10) = 3.0 kg m s-1
s = ut + ½ gt2 = (0)t + ½ (9.8/6)t2 t = 3.5 s
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Part B : Structure Questions 1.
(i) Car A (ii) Car B
Diagram 1.1
Diagram 1.1(i) and (ii) show two methods used by the mechanic to move a breakdown car. A constant force, F = 500 N is used to push and pull the car in method A and B. (a) (i) Which method is easier to move the car?
………………………………………………………………………………
(ii) State a reason for your answer in (a)(i).
………………………………………………………………………………
………………………………………………………………………………
(b) The frictional force acting between the car and track surface in both methods is 200 N. Calculate, the
(i) horizontal resultant force in method A.
(ii) horizontal resultant force in method B.
(iii) acceleration of the car in method B.
( c ) Suggest a method to move Car B so that the acceleration produced is equal to that of method A. ……………………………………………………………………………..………..
………………………………………………………………………………………
Method (a)
The forces given parallel with the surface of motion,
So, all the forces given used to move the car.
F = Fgiven - Ffriction
= 500 – 200
= 300 N
F = Fgiven Cos 500 – Ffriction
= 500 cos 600 – 200
= 50.0 N
F = m a
50.0 = 1000 a
a = 0.05 m s-2
The acceleration of Car A = 0.3 m s-2
To move Car B with the same acceleration of Car A, increase the force given
to 1000 N
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2. ceiling Tin water M N hand P Q R (i) Diagram 2.1 (ii)
a) Diagram 2.1(i) shows tin P that is empty and tin Q that is filled with water. A student find difficult to pushed tin Q. Write the inference about the observation.
………………………………………………………………………………………
b) Diagram 2.1(ii) shows a tin being released from the different positions M and N. The hand of a student at position R needs greater force to stop the motion of the tin falling from position M. Explain this observation. ………………………………………………………………………………………
……………………………………………………………………………………… c) Based on the observation (i) and (ii), state two factors that affect the magnitude of
the momentum of the object.
……………………………………………………………………………………… d) If water flows out from a hole at the bottom of the tin Q, how would the inertia of
Tin Q depends on time ?
……………………………………………………………………………………
3. 2 ms-1
P iron ball ( 2 kg ) S T 3.0 m smooth surface 1.0 m 2.0 m
Q R
Diagram 3 Rough surface
The figure shows a iron ball that is rolled through PQRST. The rough surface of QR has frictional force of 4 N. a) Calculate
(i) the kinetic energy of the iron ball at P.
F
The difficulty to move the tin depends to its mass. From position M the velocity of tin is more than the velocity compare when it is
from N. Ek increase then the force to stop it will be increased. mass and velocity inertia of tin Q will decrease because the mass of tin decreased.
Ek = ½ mv2 = ½ (2)(22) = 4.0 J
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(ii) the potential energy of the iron ball at P.
(iii) the total of energy of the iron ball at P.
b) c) (i) Calculate the total of energy of the iron ball when it reaches at Q ?
(ii) Calculate the work done against friction along QR.
d) Calculate the total kinetic energy of the iron ball at S.
e) Calculate the speed of the ball at position T. Part C : Essay Questions 1.
(i) (ii) Diagram 1.1
Diagram 1.1(i) shows the condition of a car moving at high velocity when it suddenly crashes into a wall. Diagram 1.1(ii) shows a tennis ball hit with racquet by a player.
a) (i) What is the meaning of momentum?
(ii) Based on the observations of Diagram (i) and (ii), compare the characteristics of car when it crashes into the wall and the tennis ball when it is hit with a racquet. Hence, relate these characteristics to clarify a physics concept, and name this concept.
Ep = mgh = (2) (10) (3.0) = 60.0 J E = Ek + Ep = 4.0 + 60.0 = 64.0 J 64.0 J ( the conservation of energy )
W = Ff x s = 4 x 1.0 = 4.0 J
Es = E – Ef Ek at S = Es - Ep at s = 64.0 – 4.0 = 60.0 – (2)(10)(2.0) = 60.0 J = 20.0 J Ek at T = 20.0 J v2 = 20 = ½ m v2 v = 4.5 m s-1 = ½ (2)(v2)
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b) Explain why a tennis player uses a taut racquet when playing.
c) In launching a rocket, a few technical problems have to be overcome before the rocket can move upright to the sky. By using appropriate physics concepts, describe the design of a rocket and the launch techniques that can launch the rocket upright.
Answer
a) (i) momentum is product of mass and velocity (ii) - The shape of car changed but the shape of wall remained.
- The shape of ball remained but the shape of the racquet string was changed. (The racquet string is elastic but the wall is harder) - The time taken of collision between the ball and racquet string more than the time taken when the car hit the wall.
- The impulsive force will decrease when the time of collision increased. - The concept is the impulsive force.
b) - To decrease the time of collision between the ball and the racquet string. - Impulsive force will be increased. - The force act to the ball will be increased. - The velocity of ball will be increased.
c) - Make a gradually narrower at the front shape (tapering) : To decrease air friction
- Made by the high strength and high rigidity of materials : To decrease the probability to become dented (kemik). - Made by the low density of material. : To reduce the mass/weight - The structure is fractional engine : The mass will be decreased and the velocity will increase. - Made by the high of heat capacity of materials : It will be high heat resistance.
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2. Brand Reaction time / s Mass / kg Engine thrust
force / N Resistance force
/ N A 0.3 1.5 10.0 4.0 B 0.5 1.8 12.5 2.4 C 0.2 0.9 6.5 2.2 D 0.6 2.5 16.0 6.5
In a radio-controlled car racing competition, 4 mini-cars branded A, B, C and D took part. The information of the 4 cars is given in the table above. Details of the above information are given as below;
Reaction time - Duration between the moment the radio-controlled is switched on and the moment the car starts moving. Resistance - Average value of opposing forces includes the friction between
wheels and track, and air resistance.
(a) What is the meaning of acceleration?
(b) Draw a graph of velocity against time that shows a car moving initially with constant acceleration, then moving with constant velocity and followed by constant deceleration until it stops.
(c) Explain the suitability of the properties in the above table in constructing a radio-controlled car racing purpose. Hence, determine which brand of car will win the 50-metre race.
(c) If Car B in the above table is moved up the plane at the angle of 30o to the horizon, (i) Show that the car is able to move up the plane. (ii) Determine the acceleration of the car.
Answer : (a) Increase the velocity (b) v / ms-1
displacement = area under the graph t / s
(c) - time reaction mast be short : fast to detect the signal to start its move - has a small of mass : to decrease the inertia, then easier to start move and to
stop its moving. - thrust force is high : has more power during its moving / increase the
acceleration - friction force is low : decrease the lost of force - the best car is A : because it has short of time reaction, small of mass, high of
thrust force and low friction of force. (d) (i) EB = (12.5 – 2.4 ) (50) = 505.0 J 50 m 50Sin300 E (suitable to move up) = 1.8 (10)(50Sin300) = 450 .0 J EB> E ( car B can move up the plane) (ii) F = ma , 12.5 – 2.4 = 1.8 a, a = 5.61 ms-1
Properties
300