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 .. ………………………………………………………………………………… 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 Pahang

Physics 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-1

Idea 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 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 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 Pahang

Physics 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 bumpers

Importance 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 …………………………………………….. The gravitational field strength, g can be calculate as; At the surface of the earth, ……………. ………………………………………………… …………………………….. 20 JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion 6. 7. This means ………………………………………………………………………………… ………….. 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. 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. Given : u = 0 ms-1, t = 1. 2 s, a = g = 9. 8 ms-2 (a) v = ? = u + at = 0 + (9. 8)(1. 2) = 11. 76 ms-1 (b) Depth = s = ? s = ut + ? at2 = (0)(1. 2) + ? (9. 8)(1. 2)2 = 7. 0561 m Weight 1. 2. as the gravitational force acting on the object. The weight of an object is defined …………………………………………………….. For an object of mass m, the weight can be calculate as : weight, W = mg where, g = acceleration due to gravity. 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 = mg 9. 78 N kg-1? W = 600 x 9. 78 = 58 68 N 21 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion

Exercise 2. 8 1. Sketch the following graphs for an object that falling freely. (a) (b) (c) Displacement-time graph, Velocity-time graph Acceleration-time graph (a) s / m (b) v / m s-1 (c) a / m s2 t/s t/s t/s 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. It = 200 g m is in a stationary = 3. 0 m s state t = 0. 9 s u=0 g=? = moving with uniform velocity 0. 2 kg It is s = ut +Normal reaction, R ? g t2 3. 0 = 0 (0. 79) + ? g (0. 792) g = 9. 6 m s-2 Normal reaction, R The answer less than the constant because of the airW=mg Weight, W=mg weight, frictional force. 2. 9 IDEA OF EQUILIBRIUM FORCES Magnitude of R = W Magnitude of R = mg cos ? An object is inacts in opposite direction. And acts in opposite direction. equilibrium when : R and W Resultant force = W – R = 0 So ,Resultant force = mg cos ? – R = 0 1. So,……………………………………………………………………………………………… ( object in equilibrium ) ( object in equilibrium ) 2. …………………………………………………………………………………………… normal reaction, R friction force Weight, W Force , F = Frictional force Resultant force = F – Frictional force =0 (object 22 equilibrium) in force, F JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion stationary object An object moving with uniform velocity Addition of Force 1. a resultant force is a single force the Addition of force is defined as … …………………………………………………….. 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) …………………………………………………………………………………………… Examples : the forces are acting in one direction F1 = 10 N F2 = 5 N Resultant force, F = F1 + F2 = 10 + 5 = 15 N Example : the forces are acting in opposite directions F1 = 10 N F2 = 5 N Resultant force, F = F1 – F2 = 10 – 5 = 5 N Example : the forces are acting in different directions F2 = 5 N 23 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 500 F F1 = 10 N Parallelogram method: 1. 2. 3. 4. Draw to scale. Draw the line parallel with F1 to the edge of F2, and the line parallel with F2 to the edge of F1 Connect the diagonal of the parallelogram starting from the initial point.

Measure the length of the diagonal from the initial point as the value of the resultant force. F2 F F1 Triangle method Solution : Resultant force, F = 6000 – 5300 1. Draw to scale. =700 N They forcesnot theequilibrium were to in edge of another force. 2. Displace one of the 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? Resultant force, F = 10. 5 x 50 = 525 N 24 JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion 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 Fx force of 300 N. The ropes used make an angle 250 with each other. Draw a Cos ? = , therefore Fx = F cos ? F parallelogram and label the resultant force using scale of 1 cm : 50 N. Fy Determine the magnitude= resultant force. Fy = F sin ? Sin ? of , therefore F 250 10. 5 cm Fx Fx = F cos ? = 50 cos 60 = 50 (0. 5) = 25 N Fy = F Sin ? = 50 sin 600 = 50 (0. 8660) Resolution of a force = 43. N reverse process of finding the resultant force 1. Resolution of a force is ………………………………………………………………… Fy Fy F Vertical Component ? Fx mg = 800 N Refer to trigonometric formula: F is the resultant force of Fx and Fy Therefore, F can be resolved into FxF = mg sin 400 + 200 and Fy = 800(0. 6427) + 200 = 514. 2 + 200 = 714. 2 N horizontal component 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 600 25 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Example of resolution and combination of forces F=? 200 N 400 400 Problem solving 1. 2. 3. he resultant force is equal to zero. When a system is in equilibrium, ………………………………………………………. If all forces acting at one point are resolved into horizontal and vertical the sum of each component is equal to zero. components, …………………………………………………………………………… Example 1; Show on a figure; a) the direction of tension force, T of string b) the resultant force act to lamp 700 700 c) calculate the magnitude of tension force, T Fmaximum when both of forces act in same direction; T b) T’ T (c ) T’ = 2T sin 700 a) Fmaximum = 18 + 6 18 N 24 N Therefore, mlampg = 2T sin 700 = 24 N 6N mlamp g T= m forces act in Fminimum when the lamp = 1. kg opposite direction ; 2sin70 0 Fminimum = 18 – 6 lamp = 14. 7 N 18 N 12 N W = 12 N 6 N= 1. 5(9. 8) = 7. 82 N 2sin70 0 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 maximun possible value and the minimum possible value of the resultant force. F = Resultant of Force F2 = 2202 + 2002 F = 297. 32 N F 26 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 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 . 10 UNDERSTANDING WORK, ENERGY AND EFFICIENCY When a force that acts on an object moves the object through a Work is done, …………………………………………………………………………….. distance in the direction of the force. ……………………………………………………………………………………………… of a force and the distance traveled in the direction of WORK is the product. ……………………………………………………………………. the force. ……………………………………………………………………………………………… WORK = FORCE X DISPLACEMENT The formulae of work; W =Fxs W : work in Joule/J F : force in Newton/N s : displacement in meter/m Work 1. 2. 3. 4. Example 1; Force, F s W = Fs If, F = 40 N and s = 2 m Hence, W = 40 x 2 = 80 J Example 2; 27

JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 80 N 600 W = Fs = 80 cos 600 (5) = 80 (0. 5) (5) = 200 J s= 5m Example 3; T F = 30 N h = 1. 5 m W=Fs=Fh = 30 (1. 5) = 45. 0 J T Example 4; F = 600 N W=Fs = 600 x 0. 8 = 480 J S = 0. 8 m Energy 1. It is the potential to do work. Energy is ………………………………………………………………………………………………….. created nor be destroyed. potential energy, kinetic energy, electrical 28 energy, sound energy, nuclear energy, heat and chemical energy. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 2. . 4. Energy cannot be ………………………………………………………………………………………. Exist in various forms such as …………………… …………………………………… ……………………………………………………………………………………………… Example of the energy transformation; When we are running up a staircase the work done consists of energy change from ……………………………………………………………………………………………… Chemical Energy a Kinetic Energy a 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.

Example : ………………………………………………………………………………… …………… … 5. Work done and the change in kinetic energy Force, F s 1. 2. energy of an object due to its motion. Kinetic energy is ………………………………………………………………………… Refer to the figure above, Work = Fs = mas = m ( ? v2) Through, v2 = u2 +2as u=0 and, as = ? v2 The formulae of Kinetic energy, Ek = ? mv2 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? Given : m = 100 kg , F = 200 N a. Kinetic energy, Ek = Fs = 200 x 10= 2000 J b.

Velocity, v a ? mv2 = 2000 v = 6. 32 m s-1 29 Solution : JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Work done and gravitational potential energy 1. …2. created or destroyed but can be changed from one form to h = 1. 5 m another form. energy of an object due to its position. Gravitational potential energy is………………………………………………………… (possessed by an object due to its position in a gravitational field) Maximum Potential energy …………………………………………………………………………………………… Refer to the figure above; W = Fs = mg h where, F = mg So, Gravitational energy,energy decrease potential Ep = mgh W = 10 (10) 1. and kinetic energy increase = 150 J Kinetic energy decrease and potential energy 3. Increase Example; If m = 10 kg Therefore Work done = 150J Ep = 150 J Principle of conservation of energy And, Maximum kinetic energy Carry out hands-on activity 2. 10 on page 38 of the practical book. To show the principlehof conservation of energy. -2 , v = ? Given : = 20 m, u = 0 , g = 9. 8 ms 1. 2. Energy cannot be ……………………………………………………………………… Ep = Ek …………………………………………………………………………………………… mgh = ? mv2 Example : a thrown ball upwards will achieve a maximum height before changing its m(9. 8)(20) = ? v2 direction and falls v2 = 392, v = 19. 8 m s-1 the rate of doing work. Therefore, power, P = workdone , so, timetaken P= W t Where, P : power in watt/W W : work in joule/J t : time to do work in seconds/s 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? 30 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 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? -2 g = 9. 8 ms-2) Solution : Given : m = 180 kg, h = 2 m, t = 0. 8 s and g = 9. 8 ms . P = ? W mgh P= = t t 180 ? 9. 8 ? 2 = = 4 410 W Efficiency 0. 8 as the percentage of the energy input that is transformed into useful energy. 1. Defined…….. ……………………………………………………………………………. 2. Formulae of efficiency : Efficiency = 3. Useful energy output ? 100% Energy input Analogy of efficiency; unwanted energy Energy input, Einput Device/ mechine Useful energy, Eoutput Energy transformation Solution : Given : m = 0. 12 kg, s= 0. m, t = 5 s, 4. Einput = 0. 8 J Example;(a) Eoutput = ? 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 E of output s (a) The useful output = F xof the motor. = (0. 12 x 10) (b) The efficiency of the motor x 0. 4 = 0. 48 J (b) Efficiency = ? Efficiency = = Eoutput Einput x 100% 0. 48 x 100% = 60% 0. 80 31 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Carry out hands-on activity 2. 11 on page 39 of the practical book to measure the power.

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. W=Fs The energy transferred to the force = 900 J = 90 x 10 = 900 J 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. m = 3. 0 x 50 = 150 kg h = 1. 5 m W = mhg = 150 x 9. 8 x 1. 5 = 2205 J (b) P= = W t 2205 = 8. 82 W 250 2.

What is his power if he completes this work in 250 s? 32 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 2. 11 APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY OF DEVICES During the process of transformation the input energy to the useful output energy,……… some of 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. ……………………………………………………………………………………………… Example of wasting the energy; Kinetic energy ……….. ……………… Input enegy from the petrol output energy 1. 2. 3. 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 .. ………………….. ……………. …………………. ……………………. 4. 5. The world we are living in face acute shortage of energy. …………………………………………… It is very important that a device makes the best possible use of the input energy. …………………… Ways of increasing the efficiency of devices Engine must be designed with the capability to produce greater amount of 1.

Heat engines …………………….. ……………………………………………………… mechanical work. ……………………………………………………………………………………………… 2. Electrical devices. … ……………………………………………………………………… 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. ……………………………………………………………………………………………… ……………………………………………………………………………………………… 33 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 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 defrost 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. ……………………………………………………………………………………………… Operation of electrical devices 1. 2. 3. 2. 12 hen they are in good operating The electrical devices increase the efficiency…………………………………. …… condition. will increase the life span of device. Proper management …….. ……………………………………………………………… Example : -the filter in an air-conditioner and fins of the cooling coil of a refrigerator ………….. ……………………………………………………………………………… must be periodically cleaned. ……………………………………………………………………………………………… UNDERSTANDING ELASTICITY Carry out Hands-on activity 2. 12 page 40 of the practical book. the property of an object that enables it to return its original shape and 1. Elasticity is ……………………………………………………………………………… imensions after an applied external force is removed. ……………………………………………………………………………………………… The property of elasticity is caused by the existence of forces of 2. Forces between atoms ………………………………………………………………….. repulsion and attraction between molecules in the solid material. ……………………………………………………………………………………………… 3. Forces between atoms in equilibrium condition Force of attraction Force of repulsion Force of repulsion Explanation : 1. The atoms are separated by a distance called the equilibrium distance and vibrate at ……………………………………………………………………………………………… it position. …………………………………………………………………………………………… 2. Force of repulsion = Force of attraction ……………………………………………………………………………………………… 4. Forces between atoms in compression compressive force compressive force 34 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Force of repulsion Force of repulsion 5. Explanation ; 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.

Forces between atoms in tension force of attraction stretching force stretching force Explanation ; 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. ……………………………………………………………………………………………… 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 that the extension of a spring is directly proportional to the applied force 1.

Hooke’s Law states ……………………………………………………………………… provided that the elastic limit is not exceeded. ……………………………………………………………………………………………… as the maximum force that can be applied to 2. Elastic limit of a spring is defined………………………………………………………. spring such that the spring will return to its original length when the force released. ……………………………………………………………………………………………… when the length of the 3. The spring is said to have a permanent extension,… ………………………………… spring longer than the original length even though the force acts was released and the ……………………………………………………………………………………………… elastic limit is exceeded. …………………………………………………………………………………………… When the spring obey Hooke’s Law. 4. The elastic limit is not exceeded,……………………………………………. ………… The mathematical expression for Hooke’s Law is : ……………………………………………………………………………………………… F ? x ……………………………………………………………………………………………… F = kx, k = Force constant of the spring Force constant, k = 5. Graf F against x F/ N E Q P F x with unit N m-1, N cm-1 or N mm-1 35 0 R x (cm) JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion F = kx Spring obeying Hooke’s Law Spring not obeying Hooke’s law (exceeded the elastic limit) F x with unit N m-1, N cm-1 or N mm-1

Force constant, k = 6. Spring Constant, k F/N 0. 8 k is the gradient of the F – x graph k= F x 0. 8 = 8 = 0. 1 N cm-1 0 Example 1; 8 x/cm A spring has an original length of 15 cm. With a load of mass 200 g attached, the length of the spring is extend 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] Given : lo = 15 cm, a. b. k = ? , m = 200 g , F = 2. 0 N, l = 20 cm x = 5 cm 2. 0 ? k = F = 5 = 0. 4Ncm 1 x l = ? , when m = 150 g, F = 1. 5 N From a, k = 0. 4 N cm-1 x= Example 2; F 3. 5 = = 8. 5cm k 0. 4 l = 15 + 8. 75 = 23. 75 cm The graph shows the relationship between the stretching force, F and the spring extension, x. Graph F against x of (a) Calculate the spring constant of P and Q. F (N) spring P and spring Q (b) Using the graph, determine the stretching force acts to spring P and 8 P spring Q, when their extension are 0. 5 cm 7 Solution 6 a. Spring constant, k = gradient of graph Q 5 6 1 =15. 79N cm? kP = 4 0. 38 3 3 1 = 6. 0 N cm? kQ = 0. 5 2 36 When x = 0. 5, FP = 8. 0 N b. 1 ( extrapolation of graph P) FQ = 3. 0 N 0 0. 1 0. 2 0. 3 0. 4 0. 5 x (cm) JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion Elastic potential energy 1. the energy stored in a spring when it is extended or compressed Elastic potential energy ……………………………………………………………….. spring with the original length F compression x x spring compressed F x = compression F spring extended x = extension F, extension x x Other situation where the spring extended and compressed Relationship between work and elastic potential energy F/N Graph F against x Area under the graph F x x / cm = work done = ? Fx So, Elastic potential energy = ? Fx Example ; 5 kg 15 cm 8 cm x = 15 – 8 = 7 cm = 0. 7 m Force act to the spring, F = 5 x 10 = 50 N Elastic potential energy = ? Fx = ? 50 (0. 07) = 1. 75 J Factors that effect elasticity Hands-on activity 2. 13 on page 42 the practical book to investigate the factors that affect elasticity. 37 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Type of material different Diameter of spring wire same Diameter of spring same Length of spring same Summarise the four factors that affect elasticity Factor Length Diameter of spring Diameter of spring wire Type of material Exercise 2. 12 1. same different same Same same same different same ame same same different Change in factor Effect on elasticity Shorter spring Less elastic Longer spring More elastic Smaller diameter Less elastic Larger diameter More elastic Smaller diameter More elastic Larger diameter Less elastic the elasticity changes with the type of materials 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. 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 = ? = k (2) 18 = 3 x -1 k = 3 N cm x = 6 cm If a 20 N force extends a spring from 5 cm to 9 cm, (a) what is the force constant of the spring? F = 20 N, x = 9 – 5 = 4 cm, k = ? F = kx 20 = k (4) k = 5 N cm-1 (b) Calculate the elastic potential energy stored in the spring. E = ? Fx = ? (20)(0. 04) = 0. 4 J 2. Reinforcement Chapter 2 Part A : Objective Questions 1. When a coconut is falling to the ground, which of the following quantities is constant? 38 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion A. B. C. D. 2. Velocity Momentum Acceleration Kinetic energy Velocity / ms-1 4

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 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. B. C. D. E. 0. 1 N 60 g = 0. 06 kg W = 0. 06 (1/6)(10) 0. 2 N = 0. 1 N 0. 4 N 0. 6 N 0. 8 N 0 2 4 6 Time / s Calculate the momentum of the trolley from t = 2s to t = 4s. A. B. C. D. E. 7. 1. 5 kg m s-1 P = mv 3. 0 kg m s-1 = 1. 5 x 4 4. 0 kg m s-1 = 6. 0 kg ms-1 -1 6. 0 kg m s 7. kg m s-1 3. This figure shows an aircraft flying Lift in the air. Thrust Air friction 4. The momentum of a particle is depend on A. mass and acceleration B. weight and force C. mass and velocity Which of the following diagrams shows a body moving at constant velocity? A. B. C. D. 2N 12 N 12 N 20 N 2N 7N 14 N 17 N Weight The aircraft above accelerates if A. B. C. D. 8. Lift > Weight Thrust > Lift Lift > Air friction Thrust > Air friction m = 0. 3 kg 5m What is the momentum of the stone just before it hits the ground? A. B. C. D. E. 0. 15 kg m s-1 0. 3 kg m s-1 1. 5 kg m s-1 3. 0 kg m s-1 15. 0 kg m s-1 . 6. The graph below shows the motion of a trolley with mass 1. 5 kg. 39 Solution : P = mv (find v first to calculate the P) Ep = Ek ? mgh = ? mv2 (0. 3)(10)(5) = ? (0. 3) v2 v = 10 m s-1 P = (0. 3)(10) = 3. 0 kg m s-1 JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion 10. 9. A big ship will keep moving for some distance when its engine is turned off. This situation happens because the ship has A. B. C. D. great inertia great acceleration great momentum great kinetic energy 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 B C D E 2 0. 6 s s = ut + ? gt = (0)t + ? (9. 8/6)t2 1. 4 s 1. 7 s t