# Force and Motion

JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 2. 2.

1 1. FORCE AND MOTION ANALYSING LINEAR MOTION Types of physical quantity: has only a magnitude (i) Scalar quantity: …………………………………………………………………. has both magnitude and direction (ii) Vector quantity: ………………………………………………………………… The difference between distance and displacement: length of the path taken (i) Distance: ………………………………………………………………………… distance of an object from a point in a certain direction (ii) Displacement: …………………………………………………………………… Distance always longer than displacement.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 traveled by: a.

The car b. The plane Kota Tinggi 41 km 53 km Distance and displacement 2. 3. 4. Solution: a. b. by car = 41 + 53 = 94 km Johor Bahru by plane = 60 km Desaru The path traveled by the plane is shorter than travelled by the car. So, Distance = 94 km Displacement = 60 km 60 km Hands-on Activity 2.

2 pg 10 of the practical book.Idea of distance and displacement, speed and velocity. Speed and velocity 1. 2. 3. 4. the distance traveled per unit time or rate of change of distance Speed is ..

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………………………………………………………………………………… the speed in a given direction or rate of change of displacement Velocity is: .. …………………………………………………………………………….

.. total distance traveled, s (m) , v = s m s-1 Average of speed: ……………………………………………………………………… time taken, t (s) t -1 Average of velocity:displacement, s (m) , v = s ms ………………………………………………………………………

Time taken, t (s) t 1 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 5.Example: An aero plane 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 aero plane ii. The velocity of the aero plane Solution: i. Speed = Distance Time = 300 + 400 4 = 175 km h-1 ii. velocity = displacement time (Determine the displacement denoted by AC and its direction) = . 500 .

4 C 400 km A 300 km B C 400 km A 300 km Acceleration and deceleration 1. B = 125 km h-1 (in the direction of 0530) Study the phenomenon below; 0 m s-1 20 m s-1 40 m s-1 . 3. The velocity of the car increases. Observation: ……………………………………………………………………………… the rate of change of velocity Acceleration is, ………………………………………………………………………. Final velocity – initial velocity Or, a = v – u Then, a = Time of change t Example of acceleration; t=2s t=2s A B C 0 m s-1 20 m s-1 2 40 m s-1 JPN Pahang 20 – 0 2 = 10 m s-2 Physics Module Form 4 Chapter 2 : Force and Motion Calculate the acceleration of car; i) from A to B aAB = 20 – 0 2 ii) 4.

5. From B to C aBC = 40 – 20 2 = 10 m s-2 = 10 m s-2 when the velocity of an object decreases, In calculations, a will Deceleration happens … ……………………………………………………………… be negative ……………………………………………………………………………………………… 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. Answer : v = 0 m s-1, u = 30 m s-1, t = 5 s Then , a = 0 – 30 5 = -6 m s-2 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: .

. . . . . .

. the frequency of the ticker timer = 50 Hz ( 50 ticks in 1 second) so, 1 tick = 1 second = 0. 2 seconds 50 (ii) Determination of displacement as the length of ticker tape over a period of time.

x y (iii) . . . . . . .

. xy = displacement over time t measure by ruler s . . .

. . . . . Uniform velocity ……………………………………………………………………………………….

. . . .

. . . . . Acceleration …

………. …………………………………………………………………………….. . .

. . . . . .

Acceleration, then deceleration 3 Determine the type of motion; JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion .……………………………………………………………………………………..

(iv) . Determination of velocity . . . . . .

. 12. 6 cm displacement = ……………………… -1 Velocity, v = 12. 6 = 90. 0 cm s 0. 4 (v) Determine the acceleration Length/cm v 8 7 6 5 4 3 2 1 u 7 x 0. 02 = 0. 14 s time = ……………………………….

. a= v–u t = 40. 0 – 15. 0 .. 5(0.

2) 25. 0 = 1. 0 -2 = 25. 0 cm s The equation of 0 motion 1. 2. ticks s : displacement, v : final velocity The important symbols : ……………………………………………………………….

. u : initial velocity, t : time, a : acceleration ……………………………………………………………………………………………… a= v? u t 1 2 at 2 The list of important formula; 1 1. s = (u + v)t 2. 2 3. 5. v = u + at v 2 = u 2 + 2as 4. s = ut + 3.

Example 1 : A car traveling 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. given : u = 10 m s-1 , a = 3 m s-2 , t = 20 s. s=? s = ut + ? at2 s = (10)(20) + ? (3)(20)2 = 800 m s = 800 m 4 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Example 2 : A van that is traveling with velocity 16 m s-1 decelerates until it comes to rest.

If the distance traveled is 8 m, calculate the deceleration of the van. 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 Exercise 2. 1 1. Length / cm Figure 2. 1 shows a tape chart consisting of 5-tick strip.Describe 16 the motion represented by AB and BC. In each case, determine the ; 12 A to B acceleration, BC uniform velocity (a) displacement 8 s = 4 + 8 + 12 + 16 + 16 + 16 = 72.

0 cm 4 (b) average velocity 72. 0 0 Figure 2. 1 vaverage = A B C Time/s 6(0.

1) -1 = 120. 0 cm s 16. 0 (c) acceleration Note : v = = 160 cm s-1 0. 1 v ? u 160 ? 40 4. 0 a= = u= = 40.

0 cm s-1 t 0. 5 0. 1 = 240 cm s-2 t = 5 (0. 1) = 0. 5 s 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 the car has stopped. u = 40 ms-1 v=0 t=8s s initial = 180 m (from car to obstacle when the driver start to step on the brake) sfinal = ? ( from car to abstacle when the stopped) obstacle sinitial s sfinal 1 1 s = ( u + v ) t = ( 40 + 0 )8 = 160m 2 2 sfinal = sinitial – s = 180 – 160 = 20 m 5 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 2. 2 ANALYSING MOTION GRAPHS 0m 0s 100m 10s 200m 20s 300m 400m 500m displacement The object moves with uniform velocity for time t seconds. 0s 40s 50s After t seconds, the object returns to origin (reverse) with in the form of graph called a motion graphs The data of the motion of the car can be presented…………………………………. uniform velocity Total displacement is zero The displacement-time Graph a) displacement (m) Graph analysis: Uniformis quadratic form the time Graph displacement all ……………………………………………………………… .

Graph gradientincreases with 0 Displacement = velocity = time. ……………………………………………………………… time (s) b) displacement (m) The object is stationary or is not moving Graph gradient increases uniformly ……………… …………………………………………… The object moves with increasing velocity with uniform Graph analysis: acceleration. Displacement increases uniformly …….

. ………………………………………………………… Graph gradient is fixed ………………………………………………………………… time (s) c) displacement (m) The object move with uniform velocity ………. ………………………………………………………… Graph analysis: …….

…………………………………………………………… ………………………………………………………………… time (s) d) Displacement (m) .. ………………………………………………………………… Graph analysis: ……………………………. ……………………………………… …………………………………………….

. ……………………… time (s) ……………………………………………………………………… ………………………………………………………………… 6 JPN PahangPhysics Module Form 4 Chapter 2 : Force and Motion e) displacement (m) Graph analysis: Graph is quadratic form. …………………………………………………………..

Displacement increases with time. …………………………………………………………..

Graph gradient decreases uniformly ………………………………………………………….. time (s) f) displacement (m) A B The object moves with decreasing velocity, with uniform ………………………………………………………….. deceleration. Graph analysis: OA ………………………………………………………….

. = uniform velocity (positive – move ahead) AB = velocity is zero (rest) …………………………………………………………..

BC = uniform velocity (negative – reverse) ……………………………………………………………O The velocity-time Graph a) v/ m s-1 ….. ….. b) -1 …v/ m s C time (s) Graph analysis: 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=vxt Graph analysis: 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 = ? v x t) v (m s-1) Graph analysis: 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 : 7 s = ? vt2 t ….. … t /s c) t1 t2 t (s) JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion d) v (m s-1) …….. t (s) Graph analysis: 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 Graph analysis: 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.

…………………………………………………… ……………………………………………………… Calculate:Given : QR and m (i) Velocity over OP,SOP = 20 RS SOQ = 20 m SOR = 0 m (ii) Displacement SOS = – 10 m tOP = 2 s tPQ = 3 s tQR = 2 s tRS = 1 s Solution : 20 0 ? 20 1 1 =10ms? VQR = = ? 10ms? (i) VOP = 2 2 -10 ? 0 1 = ? 10ms? R VRS = 1 8 t/s (ii) S = -10m S e) v (m s-1) ……….. t (s) Examples 1. s/m 20 10 O P Q 0 -10 2.

2 4 6 v/m s-1 10 5 P 0 2 4 6 Calculate:(i) acceleration,a over OP, PQ and QR (ii) Displacement Given : VO = 0 m s-1, VP = 10 m s-1 , Q Solution : VQ = 10 m s-1 VR = 0 m s-1 tOP = 4 s tPQ = 4 s tQR = 2 s 10 ? 0 10 ? 10 = 2. ms? 2 aPQ= = 0 ms ? 2 (i) aOP = 4 4 R 0 ? 10 8 10 t/s = ? 5. 0 ms ? 2 aQR = 2 1 8 (ii) S = (4 +10)(10) = 70. 0m 2 JPN Pahang O Physics Module Form 4 Chapter 2 : Force and Motion Excercise 2.

2 1. (a) s/m (b) s/m 10 t/s -5 0 -10 Figure 2. 21 Describe and interprete the motion of a body which is represented by the displacement time graphs in Figure 2. 21 a) The body remains in rest 5 m at the back of initial point b) The body start move at 10 m infront of the initial point, then back to initial point in 2 s. The body continue it motion backward 10 m..

The body move with uniform velocity. ) The body move with inceresing it velocity. 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 -5 0 -10 Figure 2.

22 The body move with uniform velocity , 5 m s-1 backward. The body start it motion with 10 m s-1 backward and stop at initial point in 2 s, then continue it motion forward with increasing the velocity until 10 m s-1 in 2 s. 2 4 t/s 2 4 t/s t/s (c) s/m (a) (b) 9 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion . 3 UNDERSTANDING INERTIA Idea of inertia A pillion rider is hurled backwards when the motorcycle starts to move.

1. ……………………………………………………………………………………………… Bus passengers are thrust forward when the bus stop immediately. 2. ……………………………………………………………………………………………… Large vehicle are made to move or stopped with greater difficulty. 3.

……………………………………………………………………………………………… Hand-on activity 2. 5 in page 18 of the practical book to gain an idea of inertia 4. Meaning of inertia : The inertia of an object is the tendency of the object to remain at rest or, if moving, to ………….. …………………………………………………………………………………. ontinue its uniform motion in a straight line ……………………………………………………………………………………………… Refer to figure 2.

14 of the text book, the child and an adult are given a push to swing. An adult (i) which one of them will be more difficult to be moved ……………………… An adult (ii) which one of them will be more difficult to stop? ……………………………. The relationship between mass and inertia : ………………………………. ……………… The larger the mass, the larger its inertia. …………………………………………….

. have the tendency to remain its situation either at rest or in The larger mass …………………………………………………………………………. moving. ………………………………………………………………………………………………Mass and inertia 1. 2.

3. Effects of inertia 1. Application of inertia Positive effect : ………………………………………………………………………… Drying off an umbrella by moving and stopping it quickly. (i) ……………………………………………………………………………………… Building a floating drilling rig that has a big mass in order to be stable and safe. (ii) ……………………………………………………………………………………… To tight the loose hammer (iii) ……………………………………………………………………………………… We should take a precaution to ovoid the effect. 2.

Negative effect : …………………………………………………………………………. During a road accident, passengers are thrust forward when their (i) ……………………………………………………………………………………… ar is suddenly stopped. …………………………………………………………………………… Passengers are hurled backwards when the vehicle starts to move and are hurled forward ………..

when it stops immediately. (ii) ……………………………………………………………………………………… A person with a heavier/larger body will find it move difficult to stop his movement. 10 A heavier vehicle will take a long time to stop. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion ……………………………………………………………………………………… (iii) (iv) Execise 2. 3 1. What is inertia? Does 2 kg rock have twice the inertia of 1 kg rock?Inetia 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.

……………………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… 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, A wooden block move up of a wooden dowel.

…………………………………………………………………………………… A wooden block has inertia to remains at rest. ……………………………………………………………………………………… hit the end of the dowel on the floor. The wooden block move downward of a wooden dowel. ……………………………………………………………………………………… A wooden block has inertia to continue it motion. …………………………………………………………………………………… (b) 2. 4 1.

2. ANALYSING MOMENTUM it has momentum. When an object is moving, …… ………………………………………………………… depends on its mass and velocity. The amount of momentum .

.. …………………………………………………………… as the product of its mass and its velocity, that is Momentum, p = m x v 11 Unit= kg m s-1Idea of momentum JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 3. Momentum is defined……………………………………………………………………. ……………………………………………………………………………………………… Conservation of momentum mb mg vb vg = 0 Momentum = mbvb Starting position before she catches the ball Receiving a massive ball (mb + mg) vb&g Momentum = (mb+mg)vb&g mb vb Starting position before she throws the ball vg mg Momentum = mgvg Momentum = mbvb Throwing a massive ball The principle of conservation of momentum : ………………………………………………… In the absence of an external force, the total momentum of a system remains …………………………………………………… unchanged. …………………………………………………………………………………………………… The colliding objects move separately after collision.

1. Elastic collision . ………………………………………………………………………….. u1 m1 u2 m2 12 Momentum : m1u1 + m2u2 = m1v1 + m2v2 m1 v2 m2 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Before collision 2. after collision The colliding objects move together after collision.

Inelastic collision :………………………………………………………………………… u1 m1 u2 = 0 m2 v m1 + m2 Before collision after collision Momentum : m1u1 + m2u2 = (m1 + m2) v 3. explosion : The objects involved are in contact with each other before explosion and are ……….

. ………………………………………………………………… separated after the explosion. v1 v2 (m1 + m2), u = 0 m2 Before + m )u Momentum : (mexplosion = m1 vv – m2 v2 1 2 Example 1 : after explosion Car A Car B 20 25 m Car A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at m s-1 in front of it. Car A and B move separately after collision.

If Car A is still moving at s-1 after collision,m = 100 kg velocity = 30 mB -1, v = 25 m s-1, m = 90 kg, Given : determine the , u of Car s after collision. A A A B Solution : uB = 20 m s-1 , vB = ? mAuA + mBuB = mAvA + mBvB (100)(30) + (90)(20) = (100)(25) + (90)(vB) vB = 25. 6 m s-1 Example 2 : 13 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Car A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at 20 -1 m s 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 : Given : mA = 100 kg , uA = 30 m s-1, mB = 90 kg, mAuA + mBuB = (mA + mB ) v (B+A) (100)(30) + (90)(20) = (100 + 90) v (B+A) v(A + B) = 25. 26 m s-1 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 : Given ; mb = 2 g = 0.

002 kg, vg = ? 0 = mgvg – mb vb, 0 = (1)(vg) – (0. 002)(150), mg = 1 kg, u(g+b) = 0 , vb = 150 m s-1 uB = 20 m s-1 , v(A+B) = ? vg = 0. 3 m s-1 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. ma = 150 g mwb = 450 g m (a+wb) = 600 g va = 15 m s-1 vwb = 0 v(a+ wb) = ? (0. 15 x 15) + (0. 450 x 0) = 0. 6 v(a+ wb) v(a+ wb) = 3.

75 m s-1 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 riflle is much less than the velocity of the bullet. mr = 5. 0 kg vr = ? ( 5. 0 ) vr = ( 0. 05)(80) vr = 0. 8 m s-1 mb = 50 g vb = 80 m s-1 mava + mwbvwb = m(a+wb)v(a+wb) , 2.

mr vr = mb vb , 2. 5 UNDERSTANDING THE EFFECT OF A FORCE 14 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Idea of force 1. What will happen when force act to an object? Force can make an object; ……………………………………………………………………………………………… 1.

Move 2. Stop the moving ……………………………………………………………………………………………… 3. Change the shape of the object 4.Hold the object at rest ……………………………………………………………………………………………… An object is said to be in balance when it is: 1. In a stationary state ……………………………………………………………………………………………… 2. Moving at uniform velocity ……………………………………………………………………………………………… Stationary object Normal reaction, N ……………………………… Stationary object explanation : Magnitude R = W but R acts in an opposite ……………………………………………… direction to the weight. ……………………………………………… ( object is in equilibrium ) ………. …………………………………….

. Idea of balanced forces 1. 2. weight, w = mg ………………………………………… 3. An object moving with uniform velocity Normal reaction, N …………………………….. xplanation : 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. …………………………………… ……….

. ……………….. 1. Resultant force Idea of unbalanced forces …………………………… when it is moving in acceleration. The ball move in acceleration A body is said to be in unbalanced.

. …………………………………………………… because the forces act are not balanced. F 2.

………………………..Explanation; > F’ …………………………………… F F’ So, the ball move in F direction ………… 15 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion ……………………………………………… ……………………………………………… ……… …….. ……………………………………………… 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. 1. Refer to the result of experiment 2. 2 and 2.

3, it is found that; a ? F when m is constant and a ? /m when F is constant. …………………………………………………………………………………………… Therefore, a ? F/m …………………………………………………………………………………………… From …… a ? F/m, F ? ma …………………………………………………………………………………………… …………………………………………………………………………………………… Therefore, F = kma … k =constant =1 …… ……………………………………………………………………………………………… 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 its acting on an object of mass 1 kg ( m = 1 kg) F = ma So, ………………………………………………………………………………………… Example 1 : Calculate F, when a = 3 m s-2 dan m = 1000 kg F = ma F = (1000)(3) F = 3000 N Example 2 : m = 25 kg F = 200 N . Calculate the acceleration, a of an object.

F = ma 200 = 25 a a = 8. 0 ms-2 Exercise 2. 5 16 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 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.

m = 30 kg , F – Ff = ma , F = 50 N , Ff = 20 N , a =? 50 – 20 = 30 a a = 1. 0 m s2 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? = 1000 kg , u = 72 km h-1, v = 0, s = 40 m, F = ? F = ma, = 1000 x 5.

0 = 5000. 0 N Note : u = 72 km h-1 =20 m s-1 v2 = u2 + 2as 0 = 202 + 2a(40) a = 5. 0 m s2 2. 6 ANALYSING IMPULSE AND IMPULSIVE FORCE Impulse and impulsive force The change of momentum 1. Impulse is ………………………………………………………………………………. The large force that acts over a short period of time during collision 2. Impulsive force is ……………………………………………………………………… and explosion.

……………………………………………………………………………………………… 3. Formula of impulse and impulsive force: It is known that a= (v–u)/t Refer, F = ma Therefore, So, F = m( v – u) t Ft = mv – mu , Unit = N sFt 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 5(10) – (100 = 100 N short period of time 5(10)) Example 1; = 100 Ns v u 1 wall If ; u = 10 m s-1 , v = – 10 m s-1 , m = 5 kg Impulse, Ft = Example 2; 5(10) – (- 5(10)) v = 100 Ns and t = 1 s and impulsive force, F = u 17 100 = 50 N 2 Impulsive force , F ? 1 / t Therefore, F decreases when the time of collision increases ( refer to examples ) JPN PahangPhysics Module Form 4 Chapter 2 : Force and Motion Wall with a soft surface If ; u = 10 m s-1 , v = – 10 m s-1 , m = 5 kg Impulse, Ft = 4. and t = 2 s and impulsive force, F = The relationship between time of collision and impulsive force. ……………………………………………………………………………………………… ……………………………………………………………………………………………… Exercise 2. 6 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? Fimpulse = Ft = 20 x 0. 8 = 16. 0 Ns 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? 1. 2. A large canvas bag will increase the time of collision. When the time of collision increase the impulsive force will decrease. 2. 7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES 18 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Safety features in vehicles Reinforced passenger compartment Head rest Windscreen Crumple zones Crash resistant door pillars Anti-lock brake system (ABS) Traction control Air bags bumpersImportance of safety features in vehicles Safety features Padded dashboard Rubber bumper Shatter-proof windscreen Air bag Importance Increases the time interval of collision so the impulsive force produced during an impact is thereby reduced Absorb impact in minor accidents, thus prevents damage to the car.

Prevents the windscreen from shattering Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers. Prevents the passengers from being thrown out of the car. Slows Safety seat belt down the forward movement of the passengers when the car stops abruptly /suddenly. The absorber made by the elastic material Prevents the collapse of the front (hentaman) during itinto the : To absorb the effect of impact and back of the car moving Side bar in doors the soft material of bumper passenger compartment. Also gives good protection from a side-on – Made by collision.

: To increase the time during collision, then the impulsive force will be decreased. – The passenger’s space made by the strength materials. Exercise 2. 7 : To decrease the risk trap to the passenger during accident.

– Keep an air bag at the in explain the board and in front bus that help 1.By using physics concepts, front of dashmidifications to the of passengersto improve that : will be more comfortable. safety of passengers andActs 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.

19 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 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 .

m where, F : gravitational force m : mass of an object g = 9. 8 N kg-1 2. 8 UNDERSTANDING GRAVITY that an object of mass 1 kg will experience a gravitational force of 9. 8 N. Carry out hands-on activity 2. 8 on page 35 of the practical book. Acceleration due to gravity.

1. 2. 3. 4. 5. It pulled by the force of gravity. An object will fall to the surface of the earth because……………………………….

.. Solution : F = mg = (60) (9. 8) gravitational force. as earth’s The force of gravity also known ……………………………………………………….

.. = 588. 0 N When an object falls under the force of gravity only, ………………………………… …………………………………………………………………………………………… The acceleration of objects falling freely ……………………………………………… The magnitude of the acceleration due to gravity depends ……………………….

.. Given : m = 600 kg. F = 4800 N, g = ? ……………………………………………………………………………………………… g = F = 4800 .

= 8 N kg-1 Gravitational field m 600 1. The region around the earth is …………………………………………………………. 2. 3. 4. 5. The object in gravitational field ………………………………………………………… The gravitational field strength is defined ……………………………………………. 