Pre-U STPM Physics Term 1 CC039242a

44 2 Physics Term 1 STPM Chapter 2 Kinematics 5. The figure below represents the motion of a ball rebounding from the floor after being released from a point above the floor. What is the quantity represented on the y-axis? y Time 0 A Displacement C Acceleration B Velocity D Kinetic energy 6. A falling stone strikes a soft ground at speed u and decelerates uniformly until it stops. Which one of the following graphs best represents the variation of the stone’s speed v with distance s moved into the ground? A v s 0 u C v 0 s u B v s 0 u D v s 0 u 7. The graph below shows the variation with time t of the velocity v of a bouncing ball, released from rest. Downwards velocities are taken as positive. At which time does the ball start to leave the floor on the rebounce. A B C D v 0 Time 8. A puck slides across an icy surface and travels a distance x in time t under uniform retardation. Which of the following when plotted to represent the motion of the puck would give a straight-line graph? A x against t C x t against t 2 B x t against t D x against t 2 9. A car starts from rest and accelerates at 2.0 m s–2 when a lorry moving at 16 m s–1 in the same direction passes it. What is the distance travelled by the car when it overtakes the lorry? A 16 m C 64 m B 32 m D 128 m 10. The velocity-time graph of an object is as shown in the figure. t1 t2 Time Velocity 0 Sketch (a) the displacement-time graph, (b) the acceleration-time graph. 11. A particle moves in a straight line with uniform retardation. At time t = 0, its speed is u and its displacement from the origin is zero. Sketch labelled graphs to show how (a) the velocity v of the particle, (b) the displacement s of the particle from the origin, vary with time t. Explain the relation between the graphs. 12. A ball is released from rest from a height of 1.00 m. It falls freely to the floor and rebounces to a height of 0.80 m. Neglecting the time of impact with the floor, sketch the velocity-time graph of the motion, taking upwards velocity as positive.

45 2 Physics Term 1 STPM Chapter 2 Kinematics 2.2 Projectile Students should be able to: • solve problems on projectile motion without air resistance • explain the effects of air resistance on the motion of bodies in air Learning Outcomes 1. A projectile is a body that travels under the action of gravity after being projected at an angle to the horizontal. 2. The motion of a projectile consists of two components: (a) a vertical component which is motion with uniform acceleration, g the acceleration due to gravity, (b) a horizontal component which is motion with constant velocity. 3. The path of a projectile is a curve known as parabola. 40 m s–1 30 m s–1 20 m s–1 10 m s–1 30 m s–1 30 m s–1 30 m s–1 30 m s–1 30 m s –30 m s –1 –1 –40 m s–1 –20 m s–1 –10 m s–1 30 m s–1 30 m s–1 30 m s–1 30 m s–1 50 m s–1 50 m s–1 53° a = –g Figure 2.6 Projectile 4. Figure 2.6 shows the motion of a tennis ball with an initial velocity of 50 m s–1 at an angle of projection 53° to the horizontal. An analysis of the motion shows that: (a) The horizontal component of velocity is constant at 30 m s–1, because there is no horizontal force on the ball. (b) The horizontal displacements are the same for equal time intervals. (c) The vertical component of velocity continuously changes during the motion. (d) At the highest point of the trajectory, the vertical component of velocity is zero. (e) The acceleration a is constant and vertically downwards, a = –g. 5. To study the motion of a projectile, we consider the horizontal component and vertical components separately. u cos x R θ u sin u O a = –g θ θ y H Figure 2.7 2008/P2/Q1, 2013/P1/Q2, 2014/P1/Q16

46 2 Physics Term 1 STPM Chapter 2 Kinematics 6. Figure 2.7 shows a body projected with a velocity u at an angle θ to the horizontal. 7. Consider the vertical component of motion: (a) Initial velocity = u sin θ Acceleration = a = –g Suppose H = maximum height reached At the maximum height, vertical component of velocity = 0 Using v 2 = u2 + 2as 0 = (u sin θ)2 – 2gH Maximum height, H = u2 sin2 θ 2g (b) If t1 = Time taken by the body to reach the maximum height, using v = u + at 0 = u sin θ – gt1 t1 = u sin θ g (c) The instantaneous height y, at any time t is given by s = ut + 1 2 at2 y = (u sin θ) t – 1 2 gt2 (d) Let T = Total time of flight, the time taken by the body to travel up and fall back to the ground. When the body lands on the ground, the vertical displacement s = 0. Using s = ut + 1 2 at2 0 = (u sin θ) T – 1 2 gT2 Time of flight, T = 2u sin θ g = 2t1 This means that the time taken by the object to go to its maximum height is the same as the time it takes to move from the maximum height to the ground. 8. Consider the horizontal component of motion: (a) Horizontal component of velocity = u cos θ = constant Instantaneous horizontal displacement at any time t, is x = (u cos θ) t (b) The range R of the projectile, R = (u cos θ) T = (u cos θ) ( 2u sin θ g ) R = u2 sin 2θ g

47 2 Physics Term 1 STPM Chapter 2 Kinematics (c) The maximum range is u2 g and it occurs when sin 2θ = 1 2θ = 90° θ = 45° To obtain the maximum range for a particular speed of projection, the body must be projected at an angle of 45° to the horizontal. 9. Figure 2.8 shows three possible ranges for a body projected with the same speed but at different angles of projection. 30° 45° 60° Maximum range The range is maximum when the angle of projection is 45° Figure 2.8 Example 6 A motorcycle stunt-rider moving horizontally takes off from a point 5.0 m above the ground with a speed of 30 m s–1. How far away does the motorcycle land? 5.0 m x Solution: Consider the vertical component of motion: Initial velocity = 0 Acceleration a = +g (Consider downwards direction as positive) Displacement s = 5.0 m Using s = 1 2 ut + at2 5.0 = 0 + 1 2 (9.81)t 2 Time of flight, t =  2 × 5.0 9.81 = 1.01 s Consider horizontal component of motion: Horizontal displacement x = 30 × t = 30 × 1.01 = 30.3 m Exam Tips At the highest point 1. velocity of projectile is u cos θ horizontally and not zero. 2. kinetic energy = 1 2 m (u cos θ)2

48 2 Physics Term 1 STPM Chapter 2 Kinematics Example 7 An aircraft flies at a height h with a constant horizontal velocity u so as to fly over a cannon. When the aircraft is directly over the cannon, a shot is fired to hit the aircraft. Neglecting air resistance, find in terms of u, h and g, the acceleration due to gravity the minimum speed of the shell in order to hit the aircraft. Solution: If v = minimum speed θ = angle of projection With this minimum speed, the shell hits the aircraft at the maximum height reached by the shell, and since the shell is fired when the aircraft is above the cannon, horizontal component of shell velocity = speed of aircraft, u v cos θ = u .............................. ① Using maximum height, h = v2 sin2 θ 2g v2 sin2 θ = 2gh .............................. ② ① + ②: v2 sin2 θ + v2 cos2 θ = 2gh + u2 v = u2 + 2gh Example 8 The diagram shows the path of a bullet fired horizontally with a velocity of 20.0 m s–1 from a height of 2.0 m. Calculate (a) the speed of the bullet v, (b) the angle θ when the bullet hits the ground. Solution: The horizontal component of velocity vx = 20.0 m s–1 (constant) To find the vertical component of velocity when the bullet hit the ground, consider vertical component of motion: Initial vertical component of velocity = 0 Acceleration, a = g Vertical displacement, s = 2.0 m Using s= ut + 1 2 at2 2.0 = 0 + 1 2 × 9.81 × t 2 t = 0.639 s v u θ h θ 2.0 m v 20.0 m s–1

49 2 Physics Term 1 STPM Chapter 2 Kinematics Vertical component of velocity when bullet hits the ground vy = u + at = 0 + 9.81 × 0.639 = 6.27 m s–1 Speed of bullet v = vx 2 + vy 2 = 202 + 6.272 = 21.0 m s–1 tan θ = vy vx = 6.27 20 θ = 17.4° Quick Check 4 θ vy vx v 1. A stone is thrown from O and follows a parabolic path. The highest point reached is P. Which of the following is zero at the highest point P? A Acceleration B Vertical component of velocity C Kinetic energy D Momentum 2. A projectile is fired with an initial velocity u at an angle θ to the horizontal. Neglecting air resistance, the horizontal distance x it has travelled, and the height y reached after a time t are A x = ut sin θ, y = ut cos θ – 1 2 gt2 B x = ut sin θ – 1 2 gt2 , y = ut cos θ C x = ut cos θ, y = ut sin θ + 1 2 gt2 D x = ut cos θ, y = ut sin θ – 1 2 gt2 3. A projectile is fired with a horizontal velocity v from a height h as shown in the figure. It strikes the ground at an angle of θ to the horizontal. Which of the following values of v and h will give the greatest value of the angle θ? θ h v v/m s–1 h/m A 20 15 B 20 25 C 15 25 D 25 20 4. An aeroplane flying in a straight line at constant height of 2 000 m with a speed of 200 m s–1 drops an object. The object takes a time t to reach the ground and travels a horizontal distance s. Taking g as 10 m s–2 and ignoring air resistance, which of the following gives the values of t and s? t s A 200 s 10 km B 100 s 05 km C 020 s 04 km D 010 s 02 km 5. A bullet is fired horizontally from the top of a cliff on the surface of the earth with a speed of 40 m s–1. Assuming no air resistance, what is the speed of the bullet 3 s later? (g = 10 m s–2) A 30 m s–1 C 50 m s–1 B 40 m s–1 D 70 m s–1 6. A projectile leaves the ground at an angle of 60° to the horizontal. Its initial kinetic energy is K. Neglecting air resistance, find in terms of K its kinetic energy at the highest point of the motion.

50 2 Physics Term 1 STPM Chapter 2 Kinematics 7. A ball is thrown with a velocity of 8.0 m s–1 at 60° to the horizontal. (a) Draw on the same axes, a graph to represent the variation with time of (i) vH, the horizontal component of velocity, (ii) vV, the vertical component of velocity. Identity your graphs and show suitable values of velocity and time. (b) Use your graph to find the maximum height reached by the ball. 8. v R θ A missile is fired with a speed of 500 m s–1 from the ground. (a) What is the angle of projection θ for the missile to achieve the maximum range? (b) Calculate the maximum range, R. (c) What is its maximum height above the ground when its range is maximum? (Neglect air resistance) 9. An aeroplane is flying at a constant horizontal velocity of 50 m s–1 at a height of 1 000 m. What is its horizontal distance from a target on the ground, so that a parcel released from the plane will hit the target. 10. A coin is pushed off from the smooth horizontal surface of a table of height 2.0 m. It falls and strikes the floor at a horizontal distance of 3.2 m from the edge of the table. Calculate (a) the time taken by the coin to fall through the air, (b) the speed at which the coin leaves the table, (c) the velocity of the coin when it strikes the floor. Effects of Air Resistance 1. When a body moves through the air, the air resistance against the motion of the body is known as the viscous drag. 2. The viscous drag on a body depends on (a) shape of the body Objects which are streamlined experience less drag. (b) velocity of the body The viscous drag is proportional to the square of the velocity. 3. When a body is released from rest and falls through the air, its velocity initially increases. As its velocity increases, the viscous drag increases. The acceleration of the body decreases. Finally when the acceleration is zero, the velocity is constant. This maximum constant velocity is known as terminal velocity. 4. Figure 2.9 shows how the velocity of the body varies with time. Figure 2.10 shows the variation of acceleration with time. Time 0 Velocity Terminal velocity Time 0 Acceleration g Figure 2.9 Figure 2.10

51 2 Physics Term 1 STPM Chapter 2 Kinematics Figure 2.11 5. To slow down his fall, a skydiver spreads himself wide to increase the viscous drag (Figure 2.11). What happens to the skydiver when his parachute opens? Example 9 A ball bearing is released from rest below the surface of a viscous liquid in a tall and wide container. Sketch a graph to show how the height h of the ball bearing from the base of the container varies with time. Explain the shape of the graph. Solution: If z= Instantaneous displacement from the liquid surface. H = Height of liquid z = H – h h = height from base of container h = H – z dh dt = – dz dt dH dt = 0 = –v v = velocity of ball bearing The initial velocity of the ball bearing is zero. Hence, when t = 0, gradient of graph = 0 The velocity of the ball bearing increases until the terminal velocity is achieved. H h 0 l Since the gradient of h - t graph, dh dt = –v, the gradient becomes more negative until a constant negative value is obtained. When the terminal velocity is achieved, the h-t graph is straight. Hence, the h-t graph is as shown. H z h Info Physics Highest free-fall October 14, 2012, Austria’s Felix Baumgartner jumped off a helium-filled ballon at an altitude of 39,045 m and free-fall for 4 minutes, He reached a speed of 1342.0 km h–1 (1.24 times the speed of sound)

52 2 Physics Term 1 STPM Chapter 2 Kinematics 1. A small metal sphere is held just below the surface of a viscous liquid in a tall vessel. It is then released and its displacement s with time t is plotted. Which of the following is the s-t graph? A t s 0 C t s 0 B t s 0 D t s 0 2. A steel sphere is held a short distance below the surface of a liquid in a deep vessel and then released. The terminal velocity of the sphere in the liquid is independent of A the height of the liquid in the vessel B the density of the liquid C the diameter of the sphere D the temperature of the liquid 3. Which one of the following graphs represents the acceleration of a small rain drop falling through the air. Acceleration due to gravity g = 10 m s–2. A 10 Acceleration / m s–2 Time 0 B 10 Acceleration / m s–2 Time 0 C 10 Acceleration / m s–2 Time 0 D 10 Acceleration / m s–2 Time 0 4. A ball is thrown vertically upwards in a viscous medium. The time of flight for the upwards motion is tu and for the downwards motion is td. Which of the following is correct? A td > tu because the ball moves faster on its downwards motion and the viscous drag is greater. B td > tu because the magnitude of the acceleration when the ball is moving down is smaller than the magnitude of the retardation during the upwards motion. C td < tu because the viscous drag is the greatest at the moment of projection. D td < tu because at a given speed the net accelerating force when the ball is moving downwards is greater that the net retarding force when it is moving upwards. 5. Four spheres are released from the top floor of a skyscraper and fall through air to the ground. The spheres attain the terminal velocity before reaching the ground. The sphere with the greatest terminal velocity is A steel sphere of radius 1 cm B steel sphere of radius 2 cm C polystyrene sphere of radius 1 cm D polystyrene sphere of radius 2 cm Quick Check 5

53 2 Physics Term 1 STPM Chapter 2 Kinematics 1. An object is projected vertically upwards from the top of a building of height 12.0 m with a speed of 6.0 m s–1. What is the time taken by the object to reach the ground? [Acceleration of free fall is 10.0 m s–2.] A 1.06 s B 2.00 s C 2.26 s D 4.90 s 2. The displacement-time graph of a body is shown below. Displacement Time S P Q R 0 Which statement about the motion of the body is true? A At P, the body is accelerating B At Q, the speed of the body is zero C The velocities at P and R are the same D At S, the speed is constant 3. An object starts from rest and its accelerationtime graph is as shown below. 2 4 8 12 16 4 –2 0 Acceleration / ms –2 Time / s What is the maximum velocity of the object? A 16 m s–1 B 22 m s–1 C 24 m s–1 D 26 m s–1 4. A vehicle decelerates uniformly from a speed u. It takes a time t, and travels a distance d before stopping. Its speed, and distance travelled after a time t 2 is given by Speed Distance travelled A u 2 less than d 2 B u 2 more than d 2 C Less than u 2 less than d 2 D More than u 2 more than d 2 Important Formulae 1. Equations for motion with constant accelaration v = u + at s = 1 2 (u + v)t s = ut + 1 2 at2 v2 = u2 + 2as 2. Projectile Maximum height, H = u2 sin2 θ 2g Range, R = u2 sin 2θ g Range is maximum when = 45° At time = t, horizontal displacement, x = (u cos q)t vertical displacement, y = (u sin q)t – 1 2 gt2 STPM PRACTICE 2

54 2 Physics Term 1 STPM Chapter 2 Kinematics 5. An object is projected vertically from the top of a building and it falls to the ground below. Neglecting air resistance, which is the graph of displacement-y against time t? A y 0 t C y 0 t B y 0 t D y 0 t 6. A stone is thrown vertically upwards with an intial speed of 24.0 m s-1 from the top of a tower 60.0 m above the ground. (a) Determine the speed of the stone when it strikes the ground. (b) Calculate the time taken to reach the ground from the instant it is thrown. 7. A stone is thrown from a cliff which is at a height h above the water surface with a velocity v at an angle θ to the horizontal as shown in the figure below. After a time t, the stone is at P(x,y). y x R P(x, y) v θ Cliff h (a) Write the expressions for the x- and y-coordinates of P. (b) The height h = 10.0 m, v = 12.0 m s–1 and θ = 60o . Calculate the horizontal distance R. 8. (a) An object is projected with a speed of v at an angle θ to the horizontal. Derive the Cartesian equation for the trajectory of the object. (b) The figure below shows a boat approaching a cliff of height 15.0 m at a constant speed of 12.0 m s-1. A bullet is fired horizontally with a speed u when the boat is 800 m from the cliff. The bullet strikes the boat. 800 m Cliff Bullet 15.0 m v (i) The bullet strikes the boat how many seconds after been fired? (ii) What is the distance of the boat from the cliff when it is struck by the bullet? (iii) Find the initial speed v of the bullet. (iv) Determine the magnitude and direction of the velocity of the bullet when it strikes the boat. 9. A ball is kicked horizontally from a cliff of height H with a velocity of 8.0 m s–1. The ball hits the water at an angle of 60° to the horizontal as shown in the diagram. Cliff H 60º 8.0 m s–1 (a) Neglecting air resistance, calculate (i) speed of the ball when it hits the water, (ii) H, the height of the cliff. (b) If air resistance is not negligible, discuss the change in the angle that the ball hits the water.

55 2 Physics Term 1 STPM Chapter 2 Kinematics 10.(a) An object in linear motion has an initial velocity, u. The object has an uniform acceleration of a. When the displacement is s, the velocity is v. Define uniform acceleration and deduce the equation v2 = u2 + 2as (b) A car is travelling at 54 km h–1, and when the car is 25 m from the stop line, the traffic light turns red. The reaction time of the driver is 0.60 s. (i) Find the minimum deceleration of the car for the car to stop behind the stop line. (ii) Sketch a graph to show how the displacement of the car from the stop line varies with time. (c) A ball is kicked from the ground and it follows the path shown in the figure below. The greatest height reached is 2.0 m and the range in 22.0 m. 22.0 m 2.0 m Determine (i) the velocity of projection of the ball, (ii) the time of flight. 11. A car starts from rest and travels along a straight road. The acceleration-time graph of the car is as shown below. 0 Time / s 2 2 4 6 8 10 4 –2 Acceleration / m s–2 (a) What is the maximum speed of the car? (b) What is the speed after 10 s? 12. A missile is to be launched from a point P, 500 m from the peak of a hill of height 2000 m into enemy territory. If the missile is to just pass over the hill and strike a target 1200 m from P, calculate (a) the velocity at which the missile should be launched, (b) the time of flight of the missile. 13. (a) Define acceleration. Explain how it is possible for a body to undergo acceleration when its speed remains constant. (b) A ball falls 2.00 m from rest and a further distance of 0.10 m as shown in the figure. 2.00 m 0.10 m Neglect air resistance, calculate (i) the speed of the ball after falling 2.00 m, (ii) the time to fall a distance of 0.10 m. 14.(a) Define velocity. Explain how the displacement and the acceleration of an object may be deduced from a velocity-time graph. (b) A trolley is placed at the top of a slope as shown in the figure. A block is fixed rigidly to the lower end of the slope. At time t = 0, the trolley is released from the top of the incline and its velocity v varies with time t as shown below. v O A C B

56 2 Physics Term 1 STPM Chapter 2 Kinematics ANSWERS (i) Describe qualitatively the motion of the trolley during the periods OA, AB and BC. (ii) Calculate the acceleration of the trolley down the incline. (iii) Find the length of the incline. (iv) Calculate the distance the trolley moved up the incline on the rebounce. 15. The table below shows the total stopping distance x for different velocities v of a vehicle. v/m s–1 10.0 15.0 20.0 25.0 30.0 x/m 20.0 37.5 60.0 87.5 120.0 The total stopping distance is the sum of the distance travelled by the vehicle during the reaction time of the driver and the distance travelled when the brakes are applied. Draw a graph of x v against v. Use the graph to find (a) the total stopping distance if the velocity of the vehicle is 35 m s–1. (b) the reaction time of the driver. 1 1. (a) Use v = u + at (i) 2.0 m s–1 (ii) 4.0 m s–1 (iii) 10.0 m s–1 (b) Use average speed = 1 2 (u + v) (i) 1.0 m s–1 (ii) 2.0 m s–1 (ii) 5.0 m s–1 (c) Use s = average speed × time (i) 1.0 m (ii) 4.0 m (iii) 25.0 m 2. 25 m 3. 48 m s–1 2 1. (a) Use v = u + at, t = 0.815 s (b) Use v2 = u2 – 2gH H = 3.26 m above the building. = 12.26 m from the ground (c) Use s = uT – 1 2 gT2 , s = –9.0 m T = 2.40 s 2. 22 m s–1 3. g = 2h t2 2 – t1 2 3 1. C 2. B 3. C 4. C 5. A 6. C 7. C 8. B 9. C 10. (a) Displacement Time t1 t2 0 (b) 0 Acceleration Time t1 t2

57 2 Physics Term 1 STPM Chapter 2 Kinematics 11. (a) t v u 0 (b) t s 0 v = ds dt = gradient of (s-t) graph or s =v dt = area under (v-t) graph. 12. t v 0 4 1. B 2. D 3. C 4. C 5. C 6. 1 4 K 7. (a) v /m s–1 t /s vH 6.9 –6.9 1.41 4.0 0 vV (b) Maximum height = 2.43 m 8. (a) 45° (b) 2.55 × 104 m (c) 6.37 × 103 m 9. 714 m 10. (a) 0.639 s (Use s = ut + 1 2 gt2 for vertical motion) (b) 5.01 m s–1 (Use 3.2 m = vt) (c) vx = 5.01 m s–1, vy = 0 + (9.81)(0.639) = 6.27 m s–1 V = vx 2 + vy 2 = 8.03 m s–1 At an angle to the horizontal, tan–1 ( vy vx ) = 51.4° 5 1. D 2. A 3. C 4. B 5. B STPM Practice 2 1. C : s = ut + 1 2 at2 –12.0 m = (6.0)t – 1 2 (10.0)t2 t = 2.26 s 2. B : Velocity = gradient of graph At Q, gradient = 0. 3. C : Increase in velocity = area between graph and time-axis = 1 2 (12)(4) m s–1 = 24 m s–1 4. B : Shaded area under graph > total area under graph Speed t/2 Time u t u/2 0 Area = distance travelled in time t/2 5. C : y = ut – 1 2 gt2 Gradient of y-t graph decrease until zero at the highest point. 6. (a) v2 = u2 + 2as = 24.02 + 2(–9.81)(–60.0) v = 41.9 m s–1 (b) s = 1 2 (u + v)t –60.0 m = 1 2 (24.0 + (–41.9)t t = 6.70 s Alternative: s = ut + 1 2 at2 –60.0 m = (24.0)t + 1 2 (–9.81)t2 Solving t = 6.70 s 7. (a) x = (v cos θ)t s = ut + 1 2 at2 = (v sin θ)t – 1 2 gt2 (b) Vertical motion, –10.0 = (12.0 sin 60o )t – 1 2 (9.81)t2 9.81t2 – 20.8t – 20 = 0 t = 20.8 ± 20.82 – 4(9.81)(–20) 2(9.81) = 2.84 s Horizontal motion, R = (12.0 cos 60o )(2.84) m = 17.0 m 8. (a) v sin θ v cos θ x x θ v y P(x, y)

58 2 Physics Term 1 STPM Chapter 2 Kinematics x = (v cos θ)t, t = x v cos θ y = (v sin θ)t – 1 2 gt2 = (v sin θ) ( x v cos θ ) – 1 2 g( x v cos θ )2 = x tan θ – gx2 2v2 cos2 θ (b) (i) Vertical motion s = ut + 1 2 at2 15.0 = 0 + 1 2 (9.81)t2 t = 2(15.0) 9.81 s = 1.75 s (ii) Distance of boat from the cliff = 800m – (12.0)(1.75) m = 779 m (iii) v(1.75) = 779 m v = 445 m s–1 (iv) Vx = 445 m s–1 Vy = 0 + (9.81)(1.75) m s–1 = 17.2 m s–1 V = 4452 + 17.22 m s–1 = 445.3 m s–1. . At an angle tan-1 ( Vy Vx ) to the horizontal = 2.2° 9. (a) (i) vx = 8.0 m s–1 60º v vy vx = 8.0 ms –1 Cos 60° = vx v v = 8.0 cos 60° m s–1 = 16.0 m s–1 (ii) vy = vx tan 60° = 13.9 m s–1 Vertical motion: vy 2 = 02 + 2(9.81)H H = 13.92 2(9.81) m = 9.85 m (b) If there is air resistance, path of the ball is shown in the diagram. Cliff θ New path The angle θ  60° 10. (a) Refer to page 34 (b) (i) 54 km h–1 = 54 × 103 60 × 60 m s–1 = 15 m s–1 Distance travelled under deceleration = (25 – 15 × 0.60) m = 16 m v2 = u2 + 2as a = 0 – 152 2(16) m s-2 = –7.0 m s-2 (ii) s (m) 0 t 25 16 (c) (i) H = 2.0 m = u2 sin2 q 2g R = 22.0 m = u2 sin2q g Solving, q = 10.3o and u = 35 m s–1 (ii) T = 22.0 ucos q = 0.64 s 11. (a) vmax = 1 2 (8)(4) m s–1 = 16 m s–1 (b) Speed = 16 – 1 2 (2)(2) m s–1 = 14 m s–1 12. (a) 140 m s–1 at 19° to the horizontal (b) 9.1 s 13. (a) Acceleration is the rate of change of velocity. Eg. A body accelerates from rest. (b) (i) 6.26 m s–1 (ii) 0.0157 s 14. (a) Refer to page 33 (b) (i) OA – constant acceleration AB – constant deceleration BC – constant acceleration (ii) 1.67 m s–2 (iii) 1.20 m (iv) 0.65 m 15. (a) 157.5 m (b) 1.0 s

CHAPTER Concept Map DYNAMICS 3 Newton’s Laws of Motion Conservation of Momentum Frictional Force Elastic and Inelastic collision Centre of Mass Dynamics 59 INTRODUCTION 1. Dynamics is the study of forces that cause motion. Classical dynamics are based on Newton’s laws of motion. 2. Newton’s laws of motion are only applicable for bodies moving with velocities which are small compared to the speed of light, c = 3.0 × 108 m s–1. Action: Tindakan Coefficient: Pekali Collision: Perlanggaran Elastic: Kenyal Force: Daya Friction: Geseran Impulse: Impuls Inertia: Inersia Reaction: Tindak balas Rest / stationary: Pegun Bilingual Keywords

60 3 Physics Term 1 STPM Chapter 3 Dynamics 3.1 Newton’s Laws of Motion Students should be able to: • state Newton’s laws of motion • use the formula F = m dv dt + v dm dt for constant m or constant v only Learning Outcomes 1. First Law (a) A body will remain at rest or move along a straight line with constant speed unless it is acted upon by a force. (b) This means that a body resists changes to its state of rest or motion. This property of a body is known as inertia. The inertia of a body is measured by its mass. 2. Second Law The rate of change of momentum of a body is directly proportional to the resultant force and it takes place along the direction of the resultant force. Rate of momentum change ∝ F, resultant force. Resultant force F ∝ mv – mu t where m = Mass of body u = Initial velocity v = Final velocity t = Time taken for the change of momentum Resultant force, F = k ( mv – mu t ) = km ( v – u t ) = kma (a = v – u t ) where k = constant of proportionality, the value of which depends on the system of units being used. 3. The SI unit for force is newton (1 N). The newton (1 N) is defined as the force, which causes a mass of 1 kg to have an acceleration of 1 m s–2. That is, if F = 1 N, m = 1 kg, a = 1 m s–2, substituting into the equation F = kma 1 N = k (1 kg)(1 m s–2) = k kg m s–2 Hence, k= 1 and N = kg m s–2 Resultant force, F = ma 4. Impulse From the equation F = ma = mv – mu t Ft = mv – mu = change in momentum The product of force and its duration of action t, Ft is known as impulse of a force F. Hence, impulse = Ft = Change in momentum Exam Tips The 2 main points in Newton’s second law: • Rate of change of momentum directly proportional to the resultant force. • Direction of rate of change of momentum is the same as the direction of resultant force. Exam Tips In the equation F = ma F is the resultant force acting on the mass m in the direction of the acceleration a. 2008/P1/Q3, 2011/P1/Q2, 2013/P1/Q3, Q16, 2014/P1/Q3,Q4, 2015/P1/Q2, Q3, 2017/P1/Q4

61 3 Physics Term 1 STPM Chapter 3 Dynamics F 0 t 1 t 2 t Figure 3.1 5. In general, resultant force, F = d(mv) dt Impulse = t1 t2 F dt = Change of momentum = Shaded area under the (F – t) graph (Figure 3.1) 6. Third Law Action and reaction are of the same magnitude but in opposite directions. F1 a1 a2 F2 m M Exam Tips Action and reaction must be on two different bodies. Figure 3.2 7. When a bat of mass M hits a ball of mass m, the ball experiences a force of F1. At the same instant, the ball exerts a force F2 on the bat. 8. The force F2 has the same magnitude as F1 but in the opposite direction. Hence,magnitude of F1 = magnitude of F2 = F 9. Using F = ma, the acceleration of the ball, a1 = F m which is large since the mass of the ball is small. The acceleration of the bat, a2 = F M . Since M >> m, then a2 << a1. 10. Another example of action-reaction pair is the gravitational force on a body of mass m by the Earth of mass M. 11. Magnitude of gravitational force of the Earth on the body = magnitude of the gravitational force of the body on the Earth = F. Acceleration of the body of mass m towards the Earth a1 = F m . Acceleration of the Earth of mass M towards the body a2 = F M . Since M >> m, a2 << a1 and is negligible. Exam Tips When applying the equation F = ma, draw a free-body diagram for each of the bodies in the system. A free-body diagram shows only the body alone with all the external forces on it. This is illustrated in Example 2. Figure 3.3 a a F m M Body Earth Info Physics Sir Isaac Newton (1642 –1727) He was an English physicist, mathematician, astronomer and alchemist who laid the foundations of classical mechanics, universal gravitation. Newton buit the first practical reflecting telescope and used a prism to disprese white light into the various colours of the visible spectrum. He also formulated Newton’s law of cooling.

62 3 Physics Term 1 STPM Chapter 3 Dynamics Example 1 A body of mass 3 kg experiences a force F which varies with time t as shown in the figure. What is the momentum of the object after 8.0 s? Solution: Momentum of object = Area under (F – t) graph = 1 2 (8 + 6) × 5 = 35 N s Example 2 Three blocks X, Y and Z of masses m1, m2 and m3 respectively are on a smooth table. The masses are connected by inextensible threads and dragged by a force F. m3 m2 m1 F Z Y X Find the tension in each of the threads connecting the blocks in terms of F, m1, m2 and m3. Solution: m2 m1 T2 T1 m3 F Z Y X m1+m2+m3 F a m1 T1 F X a m2 T2 T1 Y a (i) m1 + m2 + m3 (ii) m1 (iii) m2 Free-body diagrams The free-body diagrams for (i) the whole system, (ii) block X and (iii) block Y are as shown above. Apply F = ma for the whole system F =(m1 + m2 + m3) a Acceleration, a = F m1 + m2 + m3 For X, F – T1 = m1a Tension, T1 = F – m1( F m1 + m2 + m3 ) = ( m2 + m3 m1 + m2 + m3 )F

63 3 Physics Term 1 STPM Chapter 3 Dynamics For Y, T1 – T2 = m2a Tension, T2 = T1 – m2( F m1 + m2 + m3 ) = ( m2 + m3 m1 + m2 + m3 ) F – ( m2 m1 + m2 + m3 ) F = ( m3 m1 + m2 + m3 ) F Example 3 A plumb-line is hung from the roof of a vehicle. When the vehicle is stationary, the bob is above a mark on the floor. Draw figures to show the position of the bob relative to the mark on the floor when (a) the vehicle accelerates uniformly, (b) the vehicle travels with uniform velocity, (c) the vehicle retards uniformly. Explain your figures. Solution: (a) Uniform acceleration = a. The net force on the bob is the horizontal component of the tension in the string, T cos θ = ma (b) Since the velocity is uniform, the acceleration, a = 0. There is no resultant force on the bob. Hence, the plumb-line hangs vertically. (c) When the vehicle retards, acceleration = –a Using F = ma –T cos θ = m (–a) T cos θ = ma Mark Uniform acceleration, a T θ T Mark mg Uniform velocity, v T Mark Acceleration, –a T θ T mg Mark

64 3 Physics Term 1 STPM Chapter 3 Dynamics Example 4 A rocket stands vertically on its launching pad. The total mass of the rocket and its fuel before launch is 2.0 × 103 kg. On ignition, gas is ejected from the rocket at a speed of 2.5 × 103 m s–1 relative to the rocket and fuel is consumed at 7.6 kg s–1. Find the thrust of the rocket. Explain why there is an interval between ignition and lift-off. Solution: Using F = d dt (mv) = v dm dt Thrust of rocket, F= (2.5 × 103 )(7.6) = 1.90 × 104 N Initial weight of rocket = mg = (2.0 × 103 )(9.81) = 1.96 × 104 N Since the weight is greater than the thrust, the rocket does not lift off from the launching pad. The ejection of gas from the rocket results in the total weight of the rocket to continually decrease. When the total weight of the rocket is less than the thrust, the rocket lifts off. Hence, there is an interval between ignition and lift-off. Example 5 A water rocket is partially filled with water. The space above the water contains compressed air as shown in the figure. During the flight of the rocket upwards, water of density ρ is forced out through the nozzle of diameter d at a speed v relative to the rocket. (a) Deduce in terms of ρ, d and v, (i) the mass of water ejected per unit time from the nozzle. (ii) the rate of change of momentum of the water. (b) Show that the accelerating force F on the rocket is given by F = πd2 v 2 ρ 4 – mg where m is the instantaneous mass of the rocket and content during its upwards motion. (c) The recommended volume of water in the rocket before take off is 600 cm3 . If the initial air pressure in the rocket is 1.6 × 105 Pa, all the water in the rocket will be expelled and the pressure is just reduced to atmospheric pressure. A child filled the rocket with 750 cm3 water, but the initial pressure in the rocket was still 1.6 × 105 Pa. Without further calculation, explain the effect of the increased volume of water on (i) the initial thrust, (ii) the initial accelerating force, (iii) the initial acceleration, (iv) the final mass of the rocket and water, (v) the maximum height reached. Compressed air Water Nozzle

65 3 Physics Term 1 STPM Chapter 3 Dynamics Solution: Mass of water per unit time = Volume of water per unit time × Density = (Cross-sectional area × Length per unit time) × Density = πd2 4 vρ (a) Rate of change of momentum = d dt (mv) = v dm dt = v ( πd2 4 vρ) = πd2 v 2 ρ 4 (b) Thrust on rocket = Rate of change of momentum Accelerating force = Thrust – Weight F = πd2 v 2 ρ 4 – mg (c) (i) Initial thrust remains unchanged as speed v of water ejected remains unchanged. (ii) Using acceleration force, F = thrust – weight, the accelerating force decreases because the weight of the water has increased. (iii) Using a = F m , initial acceleration decreases since the accelerating force F decreases and the mass m has increased. (iv) The final total mass is bigger because there is still water in the rocket when water stopped being forced out. (v) The maximum height reached decreases since the acceleration is smaller. Quick Check 1 1. The graph shows how the force F on a body in collision with another body varies with time t. The area under the graph represents F 0 t A the acceleration B the work done C the change in momentum D the velocity 2. A body of mass 6 kg is acted by a force F which varies with time t as shown in the figure. If the initial velocity of the body is 2.0 m s–1, what is the velocity of the mass at time t = 8.0 s? F / N t / s 12 2 4 6 8 0 A 8.0 m s–1 B 12.0 m s–1 C 48.0 m s–1 D 60 m s–1

66 3 Physics Term 1 STPM Chapter 3 Dynamics 3. When a force F, which varies with time is applied to a mass of 2 kg, the gain in momentum after a time t is 40 kg m s–1. What is the time t? A 4.0 s C 16.0 s B 8.0 s D 32. 0 s 4. Two blocks, X and Y of mass 2.0 kg and 1.5 kg on a smooth horizontal surface are applied a force of 7.0 N as shown in the figure. What is the magnitude of the force exerted by block Y on block X during the motion? X Y A 1.0 N B 3.0 N C 4.0 N D 7.0 N 5. A body of mass 5 kg is moving on a horizontal smooth surface with a speed of 1.41 m s–1 in a north-easterly direction. A force of 0.5 N acts on the body in a westerly direction for 10 s. What is the final velocity of the body? A 0.41 m s–1 in a north-easterly direction B 1.00 m s–1 in a northerly direction C 2.00 m s–1 in a easterly direction D 2.41 m s–1 in a westerly direction 6. A block of mass 1.5 kg on a smooth surface is connected to a block of mass 0.5 kg by a light string passing over a smooth pulley. T T 1.5 kg 0.5 kg What is the acceleration of the 1.5 kg block and the tension T in the string? [Assume g = 10 m s–2] Acceleration Tension T A 2.0 m s–2 3.00 N B 2.5 m s–2 3.75 N C 5.0 m s–2 7.50 N D 10 m s–2 15.0 N 7. A worker of mass 80 kg stands in a platform of mass 40 kg pulls the platform up with an acceleration of 2.5 m s–2 using a smooth pulley as shown in the figure. What is the tension T in each side of the rope? (g = 10 m s–2) A 300 N C 750 N B 500 N D 900 N 8. A plumb-line hangs form the ceiling of a car which accelerates uniformly at 0.5 m s–2. What is the angle between the plumb-line and the vertical? A 3° B 10° C 30° D 60° 9. Two bodies, X and Y of mass m and M respectively, exert forces on each other and no other forces act on them. The force on X is F and the acceleration is a. Which of the following pairs is correct? Magnitude of Magnitude of force on Y acceleration of Y A m M F a B M m F a C F Ma m D F ma M

67 3 Physics Term 1 STPM Chapter 3 Dynamics 10. Which of the following statements relating to Newton’s third law of motion is not correct? A The two forces must be the same type. B The two forces must act on two different bodies. C The two forces must always opposite in direction. D The bodies are in equilibrium. 11. A pendulum bob of mass m hangs from the roof of a train that moves with an acceleration a. The tension in the string is T as shown in the figure. θ T a m Which equation is the correct motion of the pendulum bob? A T sin θ = ma C T tan θ = ma B T cos θ = ma D T tan θ = ma 12. A bullet of mass 50 g hits a sand bag with a speed of 250 m s–1. It moves through a distance of 0.75 m before stopping. What is the time taken for the bullet to stop and what is the average resistive force? Time taken Average resistive force A 12.0 ms 2.1 kN B 12.0 ms 1.0 kN C 6.0 ms 2.1 kN D 6.0 ms 1.0 kN 13. The velocity of a car varies with time as shown in the graph below. 5 5 10 15 20 10 15 0 Velocity / ms–1 Time, t / s Which statement about the force on the car is not correct? A For 0  t  5 s, the resultant force is not zero B For 5 s  t  10 s, the resultant force is zero C Resultant force during 10 s  t  15 s is greater than during 0  t  5 s D For 10 s  t  15 s, the thrust of the car’s engine is greater that the resistive force 14. Which of the following statement is not consistent with Newton’s first law of motion? A A stationary body experiences no resultant force. B No resultant force acts on a body which travels with uniform velocity C A force can cause the direction of motion of a body to change. D No resultant force acts on a body that moves with uniform acceleration. 15. A toy helicopter of mass 5.0 kg rises with uniform acceleration from rest until a height of 60 m in 10 s. What is the upthrust produced by the helicopter rotor? 16. A conveyor belt used to transfer luggage consists of endless belt running over rollers, moving at 1.5 m s–1. To keep the belt moving at a constant speed when transporting luggage, a greater driving force is required than for an empty belt. Luggage is placed at a rate of 40 kg per second on one end and removed at the other end. Why is an additional force required and what is its value? 17. A passenger is firmly restrained in his seat by means of a safety belt as shown in the figure. During a head-on collision, the car in which the passenger is in is brought to rest from a speed of 14 m s–1 in 0.25 s.

68 3 Physics Term 1 STPM Chapter 3 Dynamics 2009/P1/Q7, 2009/P2/Q2, 2010/P1/Q7, 2015/P1/Q18, 2016/P1/Q3 (a) What is the average extra force on the neck of the passenger if the mass of his head is 3.0 kg and remains firmly attached to the body. (b) Comment whether the force is big enough to cause serious injury to the passenger. 18. In order to stop a car of mass 2 000 kg travelling at 20 m s–1, the driver applies his brakes so that the stopping force F increases linearly to a maximum and then decreases to zero as shown in the figure. (a) Calculate (i) the change in momentum between t = 0 and t = 10 s. (ii) the value of Fmax. (iii) the maximum retardation of the car. (b) Sketch graphs to show the variation with time of (i) the velocity, (ii) the distance travelled. (c) Explain why this would seem to be a more gentle stop to a passenger compared to a constant force of Fmax is applied throughout the 10 second. 19. (a) State the relation between force and momentum. (b) A stream of sand falls at a constant rate from a height h on the pan of a balance marked in kg. After a time T, a total mass M has fallen and the sand stops falling. (i) Sketch a graph to show how the reading m of the balance changes with time t. Give the values of m at t = 0, t = T, and t = 2T. Explain the shape of the graph. (ii) How would the graph differ if the sand was allowed to fall from a height 2h at the same rate? 3.2 Friction Students should be able to: • explain the variation of frictional force with sliding force • define and use coefficient of static function and coefficient of kinetic friction Learning Outcomes 1. Friction between two solid surfaces is the force which opposes relative motion between the surfaces. 2. Figure 3.4 shows a block on a rough horizontal surface. When the block is pulled by a small force F, it does not move because static friction between the surfaces is in contact opposes the force F. Magnitude of static friction = magnitude of the force F. 3. When the pulling force is increased, and the block remains stationary, then the static friction too has increased. 4. The static friction between the surfaces in contact is maximum just before the block starts to move. 5. This maximum static friction is known as limiting static friction. mg R F Static friction Figure 3.4 Friction opposes motion INFO Normal Force and Friction

69 3 Physics Term 1 STPM Chapter 3 Dynamics 6. Limiting static friction depends on • the nature of the solid surfaces in contact • the normal reaction R but it does not depend on the surface area in contact. Limiting static friction = R where  = coefficient of static friction between the two surfaces. 7. The value of  is greater for rough surfaces. 8. Once the block is set into motion, the force F’ required to keep it moving with uniform velocity is less than the force required to start the block moving. 9. When the velocity of the block is uniform, forces on the block are in equilibrium. 10. The force F’ is balanced by the kinetic friction. Magnitude of kinetic friction = Magnitude of force F’. 11. Kinetic friction is the friction between two solid surfaces when there is a relative motion between the surfaces. 12. Kinetic friction depends on • the nature of the surfaces in contact • the normal reaction, R but it does not depend on the surface area in contact and the relative velocity between the surfaces. Dynamic friction = ’ R where ’ = coefficient of dynamic friction 13. Kinetic friction < limiting static friction. Example 6 A wooden block of mass m is placed on a rough plane. The plane is slowly tilted. The maximum angle between the plane and the horizontal before the block slides down the plane is θ. Find the coefficient of static friction . m θ Solution: The component of the weight mg (i) along the inclined plane = mg sin θ (ii) perpendicular to the inclined plane = mg cos θ. Since the block is in equilibrium, R = mg cos θ Limiting static friction = mg sin θ Also, limiting friction = R = mg sin θ Coefficient of static friction,  = mg sin θ R = mg sin θ mg cos θ = tan θ Limiting static friction mg R mg cos θ θ mg sin θ θ mg R F ' Kinetic friction Figure 3.5 Kinetic friction is less than the limiting static friction

70 3 Physics Term 1 STPM Chapter 3 Dynamics Quick Check 2 1. The figure shows a system which consists of a 10 kg and a 5 kg mass tied to a string in equilibrium. The 10 kg mass on the table is on the point of sliding. 30° 5 kg 10 kg Table Calculate the coefficient of static friction between the 10 kg mass and the table. 2. A block of mass 4.0 kg on a plane inclined at an angle of 30° to the horizontal, slides from rest through a distance of 1.0 m in 2.0 s. θ = 30° Calculate the coefficient of kinetic friction between the block and the inclined plane. 3.3 Conservation of Linear Momentum Students should be able to: • state the principle of conservation of momentum, and verity the principle using Newton’s laws of motion • apply the principle of conservation of momentum • define impulse as ∫F dt • solve problems involving impulse Learning Outcomes Momentum 1. The momentum p of a mass m moving with a velocity v is p = mv Momentum is a vector quantity. Its direction follows the direction of the velocity, v. 2. Principle of conservation of linear momentum states that the total linear momentum of a system of bodies is constant, if no external forces act on the system. 3. Collision: A collision is a phenomenon which (a) occurs in a short time interval. (b) what happens after collision differs from what happens before collision. (c) the colliding bodies constitute a closed system. (d) momentum and energy are conserved during collision. 4. The principle of conservation of linear momentum for a system of colliding bodies can be deduced using Newton’s laws of motion. Figure 3.6

71 3 Physics Term 1 STPM Chapter 3 Dynamics 5. Figure 3.6 shows two bodies of masses m1 and m2 moving with velocities u1 and u2 respectively before collision. During collision m1 exerts a force F1 on m2, and m2 exerts a force F2 on m1. 6. According to Newton’s third law of motion: Action = – Reaction F1 = – F2 Using Newton’s second law of motion, F = m(v – u) t Hence, m2 (v2 – u2) t = –m1(v1 – u1) t where v1, v2 are the velocities of m1 and m2 after collision and t is the time of collision. m2 (v2 – u2) = – m1 (v1 – u1) m1 v1 + m2 v2 = m1 u1 + m2 u2 Total linear momentum after collision = Total linear momentum before collision 3.4 Elastic and Inelastic Collisions Students should be able to: • distinguish between elastic collisions and inelastic collisions (knowledge of coefficient of restitution is not required) • solve problems involving collisions between particles in one dimension Learning Outcomes 1. Newton’s Law of Restitution. When two bodies collide: Relative velocity after collision = –e (Relative velocity before collision) where e = coefficient of restitution. (0  e  1) For a perfectly elastic collision, kinetic energy is conserved and e = 1. For an inelastic collision, e = 0. 2. Elastic collision A B A B Figure 3.7 Figure 3.7 shows two bodies of masses m1 and m2 moving with velocities u1 and u2 respectively before collision. After collision, the velocities are v1 and v2 respectively. Since momentum is conserved, m1 v1 + m2 v2 = m1 u1 + m2 u2 .............................. a 2017/P1/Q3

72 3 Physics Term 1 STPM Chapter 3 Dynamics Using Newton’s law of restitution: Relative velocity of A to B after collision = – 1 (Relative velocity of A to B before collision) v1 – v2 = –1 (u1 – u2) v1 – v2 = –u1 + u2 .............................. b b × m2 : m2 v1 – m2 v2 = –m2 u1 + m2 u2 .............................. c From a : m1 v1 + m2 v2 = m1 u1 + m2 u2 .............................. d c + d : (m1 + m2) v1 = (m1 – m2) u1 + 2m2 u2 v1 = (m1 – m2) (m1 + m2) u1 + 2m2 (m1 + m2) u2 .............................. e b × m1 : m1 v1 – m1 v2 = –m1 u1 + m1 u2 .............................. f From a : m1 v1 + m2 v2 = m1 u1 + m2 u2 .............................. g g – f : (m1 + m2) v2 = (m2 – m1) u2 + 2m1 u1 v2 = (m2 – m1) (m1 + m2) u2 + 2m1 (m1 + m2) u1 .............................. h The equation b v1 – v2 = –u1 + u2 can also be obtained using the fact that kinetic energy is conserved in a perfectly elastic collision. 1 2 m1 v1 2 + 1 2 m2 v2 2 = 1 2 m1u1 2 + 1 2 m2 u2 2 m1(v1 2 – u1 2 ) = m2 (u2 2 – v2 2 ) .............................. (a) Since linear momentum is conserved. m1v1 + m2 v2 = m1 u1 + m2 u2 m1 (v1 – u1) = m2 (u2 – v2) ............................. (b) (a) (b) : m1 (v1 2 – u1 2 ) m1 (v1 – u1) = m2 (u2 2 – v2 2 ) m2 (u2 – v2) v1 + u1 = u2 + v2 v1 – v2 = u2 – u1 3. Case 1 : When m1 = m2 From equation e : v1 = u2 From equation h : v2 = u1 After elastic collision, the two bodies of equal mass interchange velocities. For example, if B was initially at rest, u2 = 0, after collision v1 = 0, A stopped v2 = u1, B moved off with velocity of A before collision. 4. Case 2 : When m1 << m2 When a small mass collides into a very large mass which is stationary, m1 << m2 and u2 = 0. From equation e : v1  –u1 and v2 = 0. After collision, the small mass rebounces with the same speed, and the large mass remains stationary. Example, a ball hitting a wall.

73 3 Physics Term 1 STPM Chapter 3 Dynamics 5. Case 3 : when m1 >> m2 A very large mass m1 colliding into a small mass m2 which initially is stationary, m1 >> m2, u2 = 0. From equation e : v1  u1 From equation h : v2  2u1 After collision, the large mass continue its motion with the same velocity, and the small mass moves off with a velocity which is twice the initial velocity of the large mass. 6. Inelastic collision During an inelastic collision, (a) momentum is conserved, (b) energy is conserved, (c) kinetic energy is not conserved. 7. When the collision is perfectly inelastic, the two bodies coalesce and move with a common velocity. Example 7 Two particles A and B of mass 7 kg and 5 kg move with velocities 6 m s–1 and 8 m s–1 on a smooth horizontal surface. The particles collide and coalesce after collision. Calculate (a) the velocity after collision, (b) the fractional loss of kinetic energy. Solution: (a) Using the principle of conservation of linear momentum: Total momentum of A and B after collision = Vector sum of the momentum of A and B before collision p = p1 + p2 which is represented by the triangle of vectors (12 v)2 = (40)2 + (42)2 12v = 58 v = 4.83 m s–1 tan θ = 40 42 θ = 43° (b) Total kinetic energy before collision = 1 2 × 7 × 62+ 1 2 × 5 × 82 = 286 J Total kinetic energy after collision = 1 2 × 12 × 4.832 = 140 J Loss in kinetic energy = (286 – 140) J = 146 J Fractional loss of kinetic energy = 146 286 = 0.510 A 6 m s–1 8 m s–1 5 kg 7 kg B y x O p2 = 40 kg m s–1 p1 = 42 kg m s–1 p = 12v θ

74 3 Physics Term 1 STPM Chapter 3 Dynamics Example 8 A particle A moves with a velocity u, along the x-axis makes an elastic oblique collision with another particle of the same mass, B initially at rest. After collision, A moves with a velocity v1 at an angle θ1 to the x-axis, and particle B moves with velocity v2 at an angle θ2 to the x-axis, as shown in the figure. (a) Draw a labelled vector diagram to show v, the vector sum of the velocities v1 and v2 after collision. (b) Show that the principle of conservation of linear momentum requires the v should be equal in magnitude and direction as u1. (c) Write down an equation to represent the elastic nature of the collision. Use this equation and the vector diagram in (a) to show that θ1 + θ2 = 90°. Solution: (a) B A C (b) Suppose mass of A = mass of B = m Using the principle of conservation of linear momentum: mu = mv1 + mv2 = m (v1 + v2) = mv v = v1 + v2 u = v Hence, v has the same magnitude and direction as u. (c) Since the collision is elastic, kinetic energy is conserved, 1 2 mu2 = 1 2 mv1 2 + 1 2 mv2 2 u2 = v1 2 + v2 2 v2 = v1 2 + v2 2 v = u From the vector diagram in (a) if v2 = v1 2 + v2 2 then, ABC = 90° θ1 + θ2 = 90° The above example shows an important fact about oblique elastic collision between two bodies of the same mass. After collision, the directions of the velocities of the two bodies are perpendicular to each other. Example 9 A trolley filled with sand moves on a horizontal surface of negligible friction. At time t = 0, sand starts to drop vertically at a constant rate from a hole at the base of the trolley. (a) Sketch graphs to show (i) how the velocity, v, and (ii) how the momentum, p of the trolley varies with time. A B B A

75 3 Physics Term 1 STPM Chapter 3 Dynamics (b) A trolley P of mass 2.0 kg moves with a velocity of 1.10 m s–1 collides elastically with another trolley Q of mass 4.0 kg moving in the opposite direction with a speed of 0.50 m s–1. Calculate (i) the speeds and directions of motion of the trolleys after collision. (ii) the impulse on each trolley during the collision. (c) The trolley P has a spring rod fitted at the front. Describe what happens to the spring during and after collision. What is the maximum energy stored in the spring. Explain why the total kinetic energy of the trolley before and after collision are the same. Solution: (a) (i) The velocity of the trolley remains constant because the sand falls perpendicularly from the trolley. No force acts on the trolley in the horizontal direction. (ii) Since sand falls at a constant rate from the trolley. The mass and momentum of the trolley decrease uniformly with time until all the sand has fallen. The mass and momentum of the trolley then remain constant. (b) P Q P Q Exam Tips In solving problems using conservation of linear momentum, draw diagrams to show system (a) before collision, (b) after collision. Mark directions of velocities in both diagrams. (i) Suppose the velocities of the trolleys P and Q after collision are v1 and v2 respectively (figure). Using the principle of conservation of linear momentum, 2.0 v1 + 4.0 v2 = 2.0 × 1.10 – 4.0 × (0.50) = 0.20 v1 + 2.0 v2 = 0.1 .............................. a Using Newton’s law of restitution for elastic collision, relative velocity of P to Q after collision = –1 (relative velocity of P to Q before collision) v1 – v2 = –(1.10 – (–0.50)) = –1.60 ........................ b From a : v1 + 2.0 v2 = 0.10 a – b : 3v2 = 1.70 v2 = 0.567 m s–1 From a : v1 = 0.10 – 2v2 = 0.10 – 1.134 = –1.034 m s–1 Velocity, v 0 Time Momentum, p 0 Time

76 3 Physics Term 1 STPM Chapter 3 Dynamics After collision the velocity of P is 1.034 m s–1 in the direction opposite to its initial velocity, the velocity of Q is 0.567 m s–1 in the opposite direction of its initial velocity. (ii) Using Newton’s third law of motion, Impulse on trolley P = Impulse on trolley Q but in the opposite direction Impulse on Q = Change of momentum of Q = m (v – u) = 4.0 (v2 – u2) = 4.0 (0.567 – (–0.50)) = 4.268 N s (c) During collision, the velocity of trolley P decreases, the spring is compressed. When the compression of the spring is maximum, both the trolleys move with the same velocity. As the length of the spring increases, the trolleys move apart from each other. After collision, the trolleys move in opposite directions. If v = velocity of the 2 trolleys when the compression is maximum. Using the principle of conservation of linear momentum, (2.0 + 4.0) v = 2 × 1.10 – 4.0 × 0.5 v = 0.20 6.0 = 0.0333 m s–1 Maximum energy stored in the spring = (Total initial kinetic energy) – (Kinetic energy of trolleys when compression of spring is maximum) = 1 2 × 2.0 × (1.10)2 + 1 2 × 4.0 × (0.50)2 – 1 2 (2.0 + 4.0)(0.0333)2 = 1.707 J Example 10 A body P of mass m moving with a velocity u collides inelastically with a body Q of mass 2m which initially is stationary. (a) Find, in terms of m and u, the total kinetic energy of P and Q after collision. Explain why there is a decrease in kinetic energy. (b) Sketch labelled graphs, on the same axes, to show how the momentum of P and momentum of Q varies with time. (c) Explain how the forces on the body P and body Q can be deduced from the graphs in (b). Then, sketch graphs to show how the instantaneous force on P and on Q varies with time. Solution: (a) After collision 3m P v Q Before collision m 2m P u u2 = 0 Q Exam Tips Method of checking answers to v1and v2:  v1 – v2 should be equal to  u1 – u2 From answers to (b) (i),  v1 – v2 = [0.567 – (–1.034)] m s–1 = 1.60 m s–1  u1 – u2 = [1.10 – (–0.50)] m s–1 = 1.60 m s–1

77 3 Physics Term 1 STPM Chapter 3 Dynamics Since the collision is inelastic, P and Q move together with the same velocity v after collision. Using the principle of conservation of linear momentum, 3mv = mu v = 1 3 u Total kinetic energy before collision = 1 2 mu2 Total kinetic energy after collision = 1 2 (3m) ( u 3 ) 2 = 1 6 mu2 The kinetic energy decreases because part of it is converted into heat during collision. (b) P P Q Q Momentum 2 3 1 3 mu mu mu Before collision During collision After collision Time (c) The force on each body = Gradient of the momentum-time graph P Q 0 Force Before collision During collision After collision Time Quick Check 3 1. The figure shows two trolleys, X and Y approaching each other. The momentum of each trolley before impact is as shown. X Y After collision, the trolleys travel in opposite directions and the momentum of trolley X is 2 kg m s–1. What is the magnitude of the momentum of trolley Y? A 2 kg m s–1 C 6 kg m s–1 B 4 kg m s–1 D 14 kg m s–1 2. A particle X moves with momentum p and kinetic energy E collides head-on with an identical particle Y which is initially stationary. The two particles coalesce and move together after collision. Which of the following represents the momentum of the particle X and the system, and the kinetic energy of X and the system after the collision? Momentum of Kinetic energy of X System X System A 0 p 0 E B p 4 p 2 E 4 E 2 C p 2 p E 4 E 2 D p 2 p E E 3. A gas molecule hits the wall of the container normally at a speed u. Which of the following is true if the collision is perfectly elastic? A The particle returns along its original path with speed u, kinetic energy is conserved. B The particle returns along its original path with speed < u, kinetic energy is not conserved. C The particle does not rebound, kinetic energy is not conserved. D The particle returns along its original path with speed u, kinetic energy is not conserved.

78 3 Physics Term 1 STPM Chapter 3 Dynamics 8. Two masses m1 and m2 have the same kinetic energy. What is the ratio of the momentum of m1 to the momentum of m2? A m1 m2 C m1 2 m2 2 B m2 m2 D m1 m2 9. A trolley P of mass m travelling at a speed of 6u towards another trolley Q of mass 4m which is at rest. The trolleys collide elastically. What is the speed of trolley Q after collision? A 1.2u C 2.4u B 1.5u D 6.0u 10. An alpha particle travelling at a velocity u makes an oblique elastic collision with a helium atom moving with a velocity which negligible compared to u. After collision, the alpha particle moves in the direction inclined at 30° to its initial direction of motion. What is the angle between the directions of motion of the helium atom and the initial direction of motion of alpha particle? A 0° C 60° B 30° D 150° 11. State Newton’s first law of motion. Explain how Newton’s first law of motion is consistent with the principle of conservation of linear momentum. 12. A particle of mass m travelling with a velocity u makes a head-on elastic collision with another particle of mass M which is at rest. If m < M, show that the particles with mass m moves in the direction opposite to its initial velocity after collision. 13. A particle of mass m travelling at a speed u collides head-on with an identical particle which is at rest. The two particles coalesce after collision and move with the same velocity. (a) Calculate the common velocity of the particles after collision. (b) Find the ratio of the kinetic energy of the system after collision to the kinetic energy before collision. (c) What happens to the loss in kinetic energy? 4. Two particles P and Q of equal mass travelling towards one another with speeds u and v respectively, make an elastic collision. At some instant during the collision, P is brought momentarily to rest. What is the speed of Q at that instant? A (v – u) C 1 2 (v – u) B 2(v – u) D Zero 5. A neutron of mass m makes a head-on elastic collision with a stationary oxygen nucleus of mass 16m. The velocity of the neutron after collision is A equal to its initial velocity but in the opposite direction. B equal in magnitude as the final velocity of the oxygen nucleus. C less in magnitude than the final velocity of the oxygen nucleus. D less in magnitude that its initial velocity. 6. Two gas molecules, X of mass 4m and Y of mass 9m move directly towards each other, collide and then separate. If ΔvX and ΔvY are the changes in the velocities of X and Y respectively, the magnitude of the ratio ΔvX ΔvY is A 9 4 B 3 2 C 2 3 D 4 9 7. An apple of mass 200 g is on top of a pole of height 5.0 m. An arrow of mass 50 g travelling horizontally with a velocity u hits the apple. The apple and arrow then fall together towards the ground as shown in the figure. If air resistance is negligible and the value of g is 10 m s–2, what is the velocity u of the arrow? A 8.0 m s–1 B 10.0 m s–1 C 16.0 m s–1 D 40.0 m s–1

79 3 Physics Term 1 STPM Chapter 3 Dynamics 14. (a) State the principle of conservation of linear momentum. (b) State two laws of motion which are used to prove that momentum is conserved during a collision. (c) A B After collision A B Before collision The figure shows a particle A of mass m moving with a velocity u towards a particle B of mass M initially at rest. After colliding elastically, A and B move off with velocities v and V respectively as shown. (i) Write equations that summarise the conservation of linear momentum and energy during the collision. (ii) From these two equations, it can be deduce that v = V – u. Using this result, find the fractional loss of kinetic energy of A in terms of m and M only. (iii) Evaluate this fractional loss for M = 50m. 15. (a) A body A of mass m moves with a velocity u collides elastically with an identical body B which is stationary. Describe the motion of the bodies after collision. (b) During such elastic collision as in (a) above, the bodies are compressed and coalesce for a short duration during collision. (i) On the same axes, sketch graphs to show the variation of the velocities of A and B with time. Show on your graph the time duration before, during and after collision. Mark on the time axis, t0 when the compression is maximum. (ii) Explain why the velocities of the two bodies are the same at time = t0. Find this velocity. Hence, find in terms of m and u, the total kinetic energy of the two bodies. (iii) at time = t0, and (iv) after collision. Comment on the difference in the kinetic energies in (iii) and (iv). 3.5 Centre of Mass Students should be able to: • define centre of mass for a system of particles in a plane • predict the path of the centre of mass of a two particle system Learning Outcomes 1. The centre of mass of a body is the point where the whole mass of the body is assumed to be concentrated. 2. Suppose a system consists of particles of mass m1, m2, m3, …, mn in the x-y plane at the points (x1, y1), (x2, y2), (x3, y3), …, (xn, yn). The coordinates of the centre of mass is (– x, – y) where – x = m1x1 + m2x2 + m3x3 + … + mnxn m1+ m2 + m3 +… + mn = ∑ i = n i = 1 (mixi) ∑ i = n i = 1 mi – y = m1y1 + m2y2 + m3y3 + … + mnyn m1+ m2 + m3 +… + mn = ∑ i = n i = 1 (miyi) ∑ i = n i = 1 mi

80 3 Physics Term 1 STPM Chapter 3 Dynamics 3. The centre of gravity (CG) of a body is the point where the whole weight of the body can be assumed to act. 4. The position of the centre of gravity depends on the gravitational field strength, g. In a uniform gravitational field, the centre of gravity and the centre of mass are at the same point. 5. When a body is freely suspended from a point O, the centre of gravity (CG) is directly below the point of suspension (Figure 3.8) CG O Plumb-line Lamina Figure 3.8 Example 11 A system consists of four particles at the points as shown in the figure. 5 kg 3 kg 2 kg 4 kg 0 5 5 x / m y / m Find the coordinate of the centre of mass. Solution: The particles and their coordinates are: 5 kg(0,2),3 kg(2,5), 4 kg(3, 0) and 2 kg(4,3) – x = 5(0) + 3(2) + 4(3) + 2(4) 5 + 3 + 4 + 2 m = 1.86 m – y = 5(2) + 3(5) + 4(0) + 2(3) 5 + 3 + 4 + 2 m = 2.21 m

81 3 Physics Term 1 STPM Chapter 3 Dynamics Example 12 A uniform lamina consists of a square ABCE of sides 12.0 cm and an isosceles triangle DEC of height 9.0 cm. Find the position of the centre of mass of the lamina. (The centre of mass of a triangular lamina = 2 3 along the median from the apex). Solution: The centre of mass of the square lamina ABCE is at its centre G1, 6.0 cm from the point O. The centre of mass of the triangular lamina is at G2, Distance of G2 from D = 2 3 × 9.0 cm = 6.0 cm Distance OG2 = (12.0 + 3.0) cm = 15.0 cm The mass for each section of the lamina is directly proportional to its area Mass of square ABCE = (12 × 12) = 144 unit Mass of the triangle DEC = 1 2 × 12 × 9 = 54 unit Total mass of lamina = (144 + 54) = 198 unit Distance of centre of mass of lamina from the point O, x – = ∑ (mi xi) (∑ mi ) = 144 × 6 + 54 × 15 (144 + 54) = 8.45 cm Quick Check 4 BO C A G1 G2 D E 1. Find the position of the centre of mass of the system shown in the figure. 6 kg 2 kg 4 kg 8 kg 0 5 5 x / m y / m . 2. A system consists of three particles along a straight line as shown in the figure. 2 kg 5 kg 3 kg 2.0 m 5.0 m Where is the centre of mass for the system? 3. Differentiate between centre of mass and centre of gravity. What is the necessary condition for the centre of mass and the centre of gravity of a body to be on the same point?

82 3 Physics Term 1 STPM Chapter 3 Dynamics 4. The length of each wooden block shown in the above figure is L. How should the wooden blocks be arranged so that the total lengths of the two blocks that overhang the bottom block is maximum? STPM PRACTICE 3 1. A machine gun fires 90 bullets per minute, each at a speed of 250 m s–1 relative to the gun. The mass of each bullet is 20.0 g. What is the magnitude of the recoil force on the gun? A 4.5 N C 45 N B 7.5 N D 75 N 2. An object is placed on a rough horizontal surface. The static friction force that prevents an object from sliding increases when A a load is placed on top of the object. B rollers are placed between the object and the surface. C there is a layer of liquid between the object and the surface. D the surface area of the object in contact with the surface is increased. 3. The limiting static friction between a body and a surface is 7 N. In the figures below, the applied force is F and the frictional force is f. In which of the figures is the magnitude of the frictional force between the body and the surface incorrect? A C f = 4 N F = 4 N Body is stationary f = 6 N F = 9 N Body is moving B D f = 5 N F = 5 N Body is stationary f = 7 N F = 10 N Body is moving Important Formulae 1. Force, F = d dt (mv) = ma 2. Impulse = ∫t1 t2 Fdt = area under F-t graph = (mv – mu), change of momentum 3. Newton’s law of restitution for elastic collision Relative velocity after collision = –(relative velocity before collision) (v1 – v2) = –(u1 – u2) 4. Limiting static friction, F = µR 5. Kinetic friction, F’ = µ’R 6. Coordinates of centre of mass – x = m1x1 + m2x2 + m3x3 + … + mnxn m1+ m2 + m3 +… + mn = ∑ i = n i = 1 (mixi) ∑ i = n i = 1 mi – y = m1y1 + m2y2 + m3y3 + … + mnyn m1+ m2 + m3 +… + mn = ∑ i = n i = 1 (miyi) ∑ i = n i = 1 mi

83 3 Physics Term 1 STPM Chapter 3 Dynamics 4. A block of mass m1 on a smooth surface which is connected to another block of mass m2 by a string passing over a smooth pully is shown in the diagram below. m1 m2 What is the acceleration of the block of mass m1? A m1g m2 C m1g m2 + m1 B m2g m1 D m2g m2 + m1 5. The velocity-time graph of a body is as shown below. Velocity Time x y z Which statement about the force on the body is not correct? A In the interval x, the resultant force on the body is in the direction of the velocity. B In the interval y, the resultant force on the body is maximum. C In the interval z, the resultant force on the body opposes the motion of the body. D The magnitude of the resultant force in the interval x is greater than the magnitude of the resultant force in the interval z. 6. A force causes a body on a frictionless horizontal surface to move in a straight line. After the force is removed, the body A continues its motion with uniform velocity. B decelerates and finally stops. C continues to accelerate. D stops immediately. 7. A block A is on top of another block B which is on a horizontal surface. The coefficient of static friction between block A and block B is 0.30. Acceleration A B What is the maximum acceleration of the blocks before block A moves relative to block B? A 0.30 m s–2 B 2.9 m s–2 C 9.8 m s–2 D 33 m s–2 8. A truck of mass 4 000 kg travelling at 72 km h–1 collides into a car of mass 1 000 kg which is at rest. The two vehicles move together after collision. What is the kinetic energy of truck after collision? A 5.12 × 105 J B 6.40 × 105 J C 6.40 × 106 J D 6.64 × 106 J 9. A jet of water hits a wall normally with a speed v. The pressure on the wall is p. If the speed of the jet of water is 4v, what is the pressure on the wall? A 2p B 4p C 8p D 16p 10. Two bodies P and Q with different masses on a smooth horizontal surface are connected by an elastic string. P and Q are displaced so that the string is stretched and then released. Which of the following quantities have the same magnitude for both P and Q before the string becomes slack? A Speed B Acceleration C Momentum D Kinetic energy

84 3 Physics Term 1 STPM Chapter 3 Dynamics 11. A coin slides along a smooth horizontal surface until it falls off the edge at time T. X Y Coin Which graph represents how the resultant force F on the coin varies with time t as the coin moves from the position X to position Y? A F t T 0 C F t T 0 B F t T 0 D F t T 0 12. During a head-on collision between two vehicles of different masses, which of the following has the same magnitude for both vehicles? A Change of speed B Change of velocity C Change of kinetic energy D Change of momentum 13. The graph shows how the force F on a body varies with time. Assuming that the body is moving in a straight line, by how much does its momentum change in 6.0 seconds? F / N t / s 6 4 2 –2 2 4 6 0 A 1.5 N s C 12 N s B 10 N s D 14 N s 14. A block of mass 15.0 kg is on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.75. F (a) What is the minimum force required to move the block across the surface? (b) If the force F = 50.0 N, what is the frictional force on the block? (c) If F = 200 N, what is the acceleration of the block? 15. (a) State Newton’s second law of motion. Hence deduce the relation: force = mass × acceleration (b) A rocket of mass 5400 kg lifts off vertically from the launching pad with an acceleration of 12.0 m s–2. Gas is ejected from the rocket at a speed of 2.50 × 103 m s–1. Calculate the rate at which gas is ejected. (c) A block of mass m on a rough horizontal surface is connected to another block of mass 2.40 kg by a light string that passes over a smooth pulley as shown in the figure. The coefficient of static friction, and kinetic friction between the block and the rough surface is 0.45 and 0.38 respectively. 2.40 kg m (i) Define coefficient of static friction, and coefficient of kinetic friction. (ii) Find the minimum value of m that would prevent the block from sliding. (iii) If the value of m is 1.50 kg, what is the acceleration of the blocks? 16. (a) State the principle of conservation of momentum. (b) A particle A of mass m moving with speed v makes a head-on collision with an identical particle B which is initially at rest. How would you conclude from the subsequent motion of the particles whether the collision is (i) perfectly elastic, (ii) completely inelastic? (c) Calculate the fractional loss of kinetic energy of the system as a whole.

85 3 Physics Term 1 STPM Chapter 3 Dynamics (ii) From the two equations, obtain an equation relating v1 to u and v2. (iii) Hence, deduce the condition necessary for the mass m1 to stop after collision. Explain your answer. (iv) A trolley of mass 400 g moving with a velocity of 0.80 m s–1 collides elastically with another trolley of mass 600 g initially at rest. Calculate the velocities of the trolleys after collision. 17. Explain the terms force and linear momentum. (a) State how force is related to linear momentum. (b) In an elastic collision, kinetic energy is conserved. Discuss whether in a completely inelastic collision kinetic energy is lost completely. Use an example to illustrate your reasoning. (c) A mass m1 moving with a velocity u collides elastically with a mass m2 initially stationary. After collision, m1 and m2 move with velocity v1 and v2 respectively. (i) Write equations to represent the use of the principle of conservation of linear momentum and the principle of conservation of energy in the above collision. ANSWERS 1 1. C 2. B 3. B 4. B 5. B 6. B 7. C 8. A 9. D 10. D 11. B: Resultant force in the direction of a is T cos θ. Apply F = ma gives T cos θ = ma 12. C:s = (u + v 2 )t gives t = 2(0.75) 250 + 0 s = 6.0 ms F = m( v – u t ) = (0.050)( 0 – 250 6.0 × 10–3 ) N = 2.1 kN 13. C: Acceleration, a = gradient of v-t graph Resultant force F = ma 14. D 15. 55 N F – mg = ma s = ut + 1 2 at2 16. F = d dt (mv) F = 60 N 17. (a) 168 N F = ma, a = v – u t (b) Yes 18. (a) (i) 4.00 × 104 kg m s–1 Δp = m (v – u) (ii) Fmax = 8 000 N Δp = area under graph (iii) 4.00 m s–2 amax = Fmax m (b) (i) (ii) (c) Greater change of momentum

86 3 Physics Term 1 STPM Chapter 3 Dynamics 19. (a) Force = rate of change of linear momentum. (b) (i) t = 0, F = d dt (mv) = 2gh Reading = F g = M T 2g g t = T, Reading = M + F g t = T to 2T, Reading = M (ii) t = 0, F = M T 2g(2h) Reading = F g = 2M T h g > M T 2h g 2 1.  = 0.866 2. s = ut + 1 2 at2 gives a = 2(1.0) 2.02 = 0.50 m s–2 mg sin 30o – µmg cos 30o = ma µ = g sin 30° – 0.50 g cos 30° = 0.518 3 1. C 2. C 3. A 4. A 5. D 6. A 7. D 8. D 9. C 10. C 11. If no external forces act on a body, (i) body at rest remains at rest – momentum zero and stays zero. Hence momentum is conserved. (ii) moving body continues to move with constant velocity – since velocity is constant, momentum is constant, and is conserved. 12. Show that v1 = (m – M) (m + M) u < 0 because m < M 13. (a) v = u 2 (b) 1 2 (c) Energy ‘loss’ is converted into heat. 14. (c) (i) 1 2 mu2 = 1 2 mv2 + 1 2 MV2 mu = mv + MV (ii) 4mM (m + M)2 (iii) 0.0769 15. (a) A stops, velocity of B = u (b) (i) A u A B B (ii) No relative motion between the two bodies at t = t0, v0 = u 2 (iii) 1 4 mu2 (iv) 1 2 mu2 4 1. –x = 6(0) + 8(2.0) + 2(3.0) + 4(5.0) 6 + 8 + 2 + 4 m = 2.1 m –y = 6(3.0) + 8(0) + 2(4.0) + 4(2.0) 6 + 8 + 2 + 4 m = 1.7 m 2. Distance from 2 kg mass, –x = 2(0) + 5(2.0) + 3(7.0) 2 + 5 + 3 m = 3.1 m 3. Condition: Object is small or gravitational field is uniform. 4. 4 mg mg _L _ 2 L STPM Practice 3 1. B : Recoil force = rate of change of momentum = (number of bullet s-1)(mass) (velocity) = ( 90 60 )(0.020)(250) N = 7.5 N

87 3 Physics Term 1 STPM Chapter 3 Dynamics 2. A : When a load is added, the normal force of the object on the surface increases. Static friction is directly proportional to the normal force. 3. D : Dynamic friction is less than limiting static friction. 4. D : Use F = ma for (m1 + m2) m2g = (m1 + m2)a a = m2g m1 + m2 5. B 6. A 7. B Limiting friction = a µmg A B F = ma ma < mg a < (0.30)(9.81) m s–2 < 2.9 m s–2 8. A : 72 km h-1 = 72 × 103 m 3600 s = 20 m s–1 Momentum is conserved (4000 + 1000)v = (4000)(20) v = 16.0 m s-1 Kinetic energy = 1 2 (4000)(16.0)2 J = 5.12 × 105 J 9. D : p = 1 A d(mv) dt = 1 A (Avρ)v ∝ v2 When speed = 4v, pressure = 42 p = 16p 10. C : Momentum is conserved MV + mv = 0, MV = - mv 11. C : Coin on the table: resultant force F = 0 Off the table: resultant force F = mg = constant 12. D : Impulse = change of momentum Impulses on the vehicles are of the same magnitude but in opposite directions. 13. B : Change of momentum = area under graph = 1 2 (4)(6) + 1 2 (2)(–2) N s = 10 Ns 14. (a) Minimum force = mg = (0.75)(15.0)(9.81) N = 110 N (b) If F = 50.0 N < 110 N, the block does not move. Friction = 50.0 N (c) F – 110 N = (15.0)a a = 200 – 110 15.0 m s–2 = 6.00 m s–2 15. (a) Refer to page 60 (b) a F mg F – mg = ma F = mg + ma = d(mv) dt m(g + a) = vdm dt dm dt = m(g + a) v = 47.1 kg s–1 (c) (ii) Tension in string, T = (2.40)g µsmg > T = (2.40)g m > 2.40 (0.45) kg = 5.33 kg (iii) (2.40)g – T = (2.40)a T – µk(1.50)g = (4.00)a a= [2.40 – (0.38)(4.00)](9.81) (2.40 + 4.00) m s–2 =1.35 m s–2 16. (a) Refer to page 70 (b) (i) A stops, velocity of B = v. (ii) A and B move together. (c) 1 2 17. (a) Force = Rate of change of the linear momentum. (b) Refer to Quick Check 3, no 2 (c) (i) m1u = m1v1 + m2v2 1 2 m1 u2 = 1 2 m1v1 2 + 1 2 m2 v2 2 (ii) v1 = v2 – u (iii) m1 = m2 (iv) v1 = –0.16 m s–1 v2 = 0.64 m s–1

CHAPTER Concept Map WORK, ENERGY AND 4 POWER Bilingual Keywords Efficiency: Kecekapan Energy: Tenaga Potential energy: Tenaga keupayaan Power: Kuasa Work: Kerja Work Work done = Fs cos θ Work done = ∫ Fds = area under F-s graph Energy • Potential energy change = mgh • Kinetic energy = 1 2 mv2 • Work-energy theorem • Principle of conservation of energy Power • Power = Work done Time taken • Efficiency Work, Energy and Power 88

89 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 4.1 Work Students should be able to: • define the work done by a force dW = F • ds • calculate the work done using a force-displacement graph • calculate the work done in certain situations, including the work done in a spring Learning Outcomes 1. The work done by a constant force F when the displacement of its point of application in s is given by the scalar product of F and s, Work done W = F.s = Fs cos θ where θ = angle between F and s 2. When the directions of the force F and the displacement s are the same (Figure 4.2), θ = 0°. Work done, W = Fs 3. If the force F is perpendicular to the displacement s (Figure 4.3), θ = 90° Work done W = Fs cos 90° = 0 4. Unit for work is the joule (J). One joule (1 J) is the work done by a force of 1 N when its point of application is displaced 1 m in the direction of the force. 1 J = 1 N × 1 m = 1 N m Work Done by a Variable Force 1. Figure 4.4 shows a force F that varies with the displacement x. The work done by the force F when x varies from x = 0 to x = s is W =  s 0 F dx = Shaded area under the (F-x) graph 2. Figure 4.5 shows how the force F stretching a wire varies with the extension x. Work done to produce an extension e is W =  s 0 F dx = Shaded area under the (F-x) graph = 1 2 Fe Figure 4.1 Figure 4.2 Figure 4.3 F 0 s x Work done Figure 4.4 Force, F 0 e F x Work done B A Figure 4.5 2013/P1/Q18, 2014/P1/Q5

90 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Exam Tips Note the difference Force, F 0 Time, t Force, F 0 Displacement, x (a) Area below force-time graph (b) Area below force-displacement graph = change of momentum = work done Work Done by a Gas During Expansion 1. Figure 4.6 shows a fixed mass of gas in a cylinder fitted with a smooth light piston. Suppose p = Pressure of the gas A = Cross-sectional area of the piston Force on the piston, F = pA When the gas expands and pushes the piston through a small distance dx, Work done, dW = F dx = (pA) dx A dx = dV = p dV Hence, work done by the gas when its volume increases from V1 to V2 is W =  s 0 dW =  v2 v1 p dV = Shaded area under the (p-V) graph Work done 0 p V V1 V2 Work done 0 p V V1 V2 (a) Pressure of gas constant (b) Temperature of gas constant Figure 4.7 2. When a gas expands under constant pressure, the work done is represented by the area under the (p-V) graph as shown in Figure 4.7(a). 3. Figure 4.7(b) shows the work done by a gas when the temperature of the gas remains constant when it expands. Piston A F = pA Figure 4.6

91 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Example 1 A mass of 2.0 kg initially stationary is acted by a force F which varies with the displacement x of the mass as shown in the figure. What is the speed of the mass when the displacement x = 8.0 m? Solution: Work done by the force = Area under (F-x) graph = 1 2 (6 + 2) 10 – 1 2 × 2 × 5 = 35 J Gain in kinetic energy, 1 2 mv2 = Work done 1 2 × (2.0)v 2 = 35 v = 5.92 m s–1 Example 2 A catapult consists of two rubber bands. The tension T in each rubber band is given by T = (50x – 100x2 ) where x = extension. Find the total work done in stretching the rubber bands until the extension of each is 0.20 m. Solution: Work done in stretching one rubber band W=  0 0.20 T dx =  0 0.20 (50x – 100x2 ) dx = 50 2 x2 – 100 x3 3 0 0.20 = 25 (0.20)2 – 100 3 (0.20)3 = 0.733 J Work done in stretching both rubber bands = 2 × 0.733 J = 1.466 J

92 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Quick Check 1 1. Two springs A and B are separately stretched by a force F that varies with the extension x as shown in the graph below. 8 10 12 Spring A Spring B 642 12 10 8 6 4 2 0 x / cm F / N Calculate the work done to stretch (i) spring A (ii) spring B by 12 cm. 2. The force F on an object varies with displacement s as shown in the graph. 8642 10 12 –2 –4 8 6 4 2 0 s / m F / N Find the work done by the force when the displacement is (i) 8 m (ii) 12 m 4.2 Energy Students should be able to: • derive and use the formula: potential energy change = mgh near the surface of the Earth • derive and use the formula: kinetic energy = 1 2 mv2 • state and use the work-energy theorem • apply the principle of conservation of energy in situations involving kinetic energy and potenial energy Learning Outcomes 1. Energy enables a body to do work. There are various forms of energy: mechanical energy, heat, sound, electrical energy, nuclear energy and chemical energy. 2. In this chapter, we will discuss only mechanical energy. There are two types of mechanical energy, namely kinetic energy and potential energy. 3. A body that moves has kinetic energy. A body of mass m moving with a speed v has translational kinetic energy, K = 1 2 mv2 . 4. Potential energy is the energy of a body due to its state or its position. When a body of mass m is raised through a height h near the Earth’s surface, its gravitational potential energy increases. A stretched spring has an elastic potential energy. Info Physics Hydrogen seems to be our future source of enegry. Hydrogen as a fuel is high in energy. Hydrogen fuel cells are used to power space shuttle’s electrical system. There is plentiful supply of hydrogen, combined with oxygen in water (H2 O). 2016/P1/Q5, 2016/P1/Q7, 2017/P1/Q5

93 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Work-energy Theorem 2008/P1/Q4, 2009/P1/Q4, 2010/P1/Q4, 2011/P1/Q3 1. The work-energy theorem states that work done on a system increases the mechanical energy of the system. Conversely, work done by a system equals the decrease in its mechanical energy. F m s v F m u Figure 4.8 2. Figure 4.8 shows that when a force F acts on a mass m, the velocity of the mass increases from u to v in a distance s. Work done by the force = Fs = (ma)s (F = ma) = 1 2 m(v 2 – u2 ) (From v 2 – u2 = 2as) = 1 2 mv 2 – 1 2 mu2 = Gain in kinetic energy of the mass 3. If initially the mass was stationary, thus u = 0, Gain in kinetic energy= 1 2 mv 2 – 1 2 mu2 = 1 2 mv2 – 0 = 1 2 mv2 Hence kinetic energy of a mass m moving with a speed v is 1 2 mv2 . 4. To lift a mass m at constant speed, a force F = mg is required. When the mass is raised through a height h, work done by the force = Fh = (mg)h = Gain in potential energy of the mass 5. Principle of conservation of energy states that energy cannot be created nor destroyed, but energy can be transformed from one form to another. 6. For a system that has mechanical energy only, Total energy, E = Kinetic energy + Potential energy = constant E = K + U = constant Therefore ∆ K + ∆U = 0 ∆K = –∆U Increase in kinetic energy = decrease in potential energy or vice versa. 7. For example when a mass falls from rest and its speed after falling through a height h is v, Gain in kinetic energy, ∆K = 1 2 mv 2 (v 2 = u2 + 2as = 0 + 2gh) = 1 2 m(2gh) = mgh, decrease in potential energy INFO Work – Energy Theorem