#### Transcript M1.4 Dynamics

AS-Level Maths: Mechanics 1 for Edexcel M1.4 Dynamics 1 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 48 © Boardworks Ltd 2005 Momentum Contents Newton’s laws of motion Applying Newton’s laws Momentum Impulse The principle of conservation of momentum Examination-style questions 2 of 48 © Boardworks Ltd 2005 Newton’s First Law of Motion Newton’s First Law of Motion states that: An object will remain at rest or move in a straight line with a constant speed unless acted on by an external force. For example, the body in the following diagram is moving with a constant velocity of 2 ms–1. R 2 ms–1 F D W There is no acceleration and so the forces are in equilibrium. So, F=D and R=W 3 of 48 © Boardworks Ltd 2005 Newton’s Second Law of Motion Newton’s Second Law of Motion states that: The rate of change of momentum of a body is proportional to the force applied to that body and in the direction of the force. Hence for a constant mass: F = d mv dt F = m dv dt F = ma 4 of 48 © Boardworks Ltd 2005 Newton’s Second Law example 1 Find the acceleration produced by a resultant force of 6 N acting on a particle of mass 1.5 kg. Applying Newton’s Second Law: F = ma 6 = 1.5a a=4 Therefore the acceleration produced is 4 ms–2. 5 of 48 © Boardworks Ltd 2005 Newton’s Second Law example 2 A particle of mass 3 kg is accelerating at 4 ms–2. Find the resultant force that produces this acceleration. Applying Newton’s Second Law: F = ma F=3×4 F = 12 Therefore the resultant force is 12 N. 6 of 48 © Boardworks Ltd 2005 Newton’s Second Law example 3 Find the magnitude of the resultant force acting on a particle of mass 2 kg needed to produce an acceleration of (3i + 4j)ms–2. Applying Newton’s Second Law: F = ma F = 2 × (3i + 4j) = 6i + 8j The magnitude of the force is 62 + 82 = 10 N. Therefore the magnitude of the resultant force is 10 N. 7 of 48 © Boardworks Ltd 2005 Newton’s Third Law of Motion Newton’s Third Law of Motion states that: For every action there is an equal and opposite reaction. For example when two surfaces are pressed against one another then a normal reaction force acts between the two surfaces. This force acts perpendicular to the area of the surfaces in contact. Normal reaction force Weight 8 of 48 © Boardworks Ltd 2005 Momentum Contents Newton’s laws of motion Applying Newton’s laws Momentum Impulse The principle of conservation of momentum Examination-style questions 9 of 48 © Boardworks Ltd 2005 Question 1 A car is travelling on a straight horizontal road. It passes point A with a speed of 15 ms–1, and 5 seconds later passes point B with a speed of 20 ms–1. The car has a mass of 1200 kg including the driver. Assume acceleration is constant. Find the driving force, D, if the resistance to motion is 500 N. 500 N 10 of 48 1200 kg D © Boardworks Ltd 2005 Question 1 Solution Acceleration = 20 15 = 1 ms–2 5 - Applying Newton’s Second Law: 5000 – R = 2050 × 2 D = 1200 + 500 D = 1700 Therefore the driving force of the car is 1700 N. 11 of 48 © Boardworks Ltd 2005 Question 2 A truck is travelling on a straight horizontal road. It passes point A with speed of 10 ms–1, and 5 seconds later passes point B with a speed of 20 ms–1. The truck has a mass of 2050 kg including the driver. Assume acceleration is constant. Find the resistive force, R, to the motion if the truck produces a driving force of 5000 N. R 12 of 48 2050 kg 5000 N © Boardworks Ltd 2005 Question 2 Solution Acceleration = 20 10 = 2 ms–2 5 Applying Newton’s Second Law: 5000 – R = 2050 × 2 R = 5000 – 4100 R = 900 Therefore the resistive force is 900 N. 13 of 48 © Boardworks Ltd 2005 Question 3 A car of mass 1450 kg including the driver is travelling along a straight road which is inclined at an angle of 4° to the horizontal. The car is travelling at a constant speed uphill and is subject to a resistive force of 850 N. a) Draw a diagram to show all the forces acting on the car. b) Calculate the driving force of the car. a) a=0 R D 850 N 14 of 48 4° 1450g © Boardworks Ltd 2005 Question 3 Solution b) As the car is travelling at a constant speed, acceleration = 0. The component of the car’s weight acting along the plane is 1450g cos86°. So the net force driving the car is D – 1450g cos86°. Applying Newton’s Second Law along the plane, D – 850 – 1450g cos86° = 0 D = 850 + 1450g cos86° D = 1841 (to 4 s.f.) Therefore the driving force of the car is 1841 N. 15 of 48 © Boardworks Ltd 2005 Question 4 A truck of mass 2000 kg is travelling along a straight road inclined at an angle θ to the horizontal. The truck is travelling at a constant speed and is experiencing a resistive force of 1000 N. The driving force of the truck is 4750 N. Calculate the value of θ. R a=0 4750 N 1000 N θ 16 of 48 2000g © Boardworks Ltd 2005 Question 4 Solution Applying Newton’s Second Law up the plane: 4750 – 1000 – 2000g cos(90 – θ)° = 0 3750 = 19600 sinθ sinθ = 3750 ÷ 19600 θ = 11.0° (to 3 s.f.) Therefore the road is inclined at an angle of 11.0° to the horizontal. 17 of 48 © Boardworks Ltd 2005 Question 5 A lift is accelerating upwards at 2 ms–2. A woman of mass 60 kg is standing in the lift. Find the normal contact force between the woman and the floor of the lift. Applying Newton’s Second Law: R – 60g = 60 × 2 R 60g 18 of 48 a = 2 ms–2 R = 60g + 120 R = 708 Therefore the normal contact force is 708 N. © Boardworks Ltd 2005 Question 6 A block of mass 3 kg is being pulled up a rough plane inclined at an angle of 20° to the horizontal by a force of 20 N parallel to the plane. The block is accelerating at a rate of 2 ms–2. Friction acts as a resistive force. Find the frictional force. a = 2 ms–2 R 20 N F 20° 19 of 48 3g © Boardworks Ltd 2005 Question 6 Solution Applying Newton’s Second Law up the plane, 20 – F – 3 F g c o s 7 0 ° =3 = 20 20 of 48 © Boardworks Ltd 2005 × 2 – Momentum t n e p l t s p t o y n A w n i n ’ o e s g l N C N a w e s w o t o f n m ’ o s l t a i w o n s Momentum When an object is in motion it is said to have momentum. The momentum of an object is given by the product of its mass and its velocity. momentum = mass × velocity Momentum is a vector quantity since it depends upon the velocity of the particle. We can therefore write the above formula as p = mv A stationary particle has no velocity and so its momentum is zero. 22 of 48 © Boardworks Ltd 2005 Momentum Multiplying mass and velocity gives the units for momentum as kilogram metres per second (kg ms–1). Momentum can also be given in Newton seconds (Ns). Newtons, the basic units of force, are given by N = kg ms–2, so Newton seconds = N × s = kg ms–2 × s = kg ms–1 23 of 48 © Boardworks Ltd 2005 Calculating momentum 1 A truck of mass 5 kg is travelling at 10 ms–1. Calculate the momentum of the truck and the direction in which it is acting. 10 ms–1 Momentum = mass × velocity = 5 × 10 Ns 5 kg = 50 Ns The momentum of the truck is acting in the direction of motion. 24 of 48 © Boardworks Ltd 2005 Calculating momentum 2 Calculate the momentum of a bullet of mass 4 g travelling at 750 ms–1. Mass of the bullet = 4 g = 0.004 kg Momentum = mass × velocity = 0.004 × 750 Ns = 3 Ns Again, the momentum is acting in the direction of the motion. 25 of 48 © Boardworks Ltd 2005 Momentum Contents Newton’s laws of motion Applying Newton’s laws Momentum Impulse The principle of conservation of momentum Examination-style questions 26 of 48 © Boardworks Ltd 2005 Impulse The impulse on a body is defined as its change in momentum. Impulse = change in momentum = mv – mu where u is the initial velocity and v is the final velocity. Impulse is denoted by the vector I. Note that the impulse I that a body A exerts on a body B is equal to the magnitude of the impulse that B exerts on A but in the opposite direction. Since impulse is change in momentum it is also measured in Newton seconds. 27 of 48 © Boardworks Ltd 2005 Calculating impulse 1 A ball of mass 0.5 kg hits the floor with a speed of 8 ms–1. It rebounds with a speed of 6ms–1. Find the impulse exerted by the floor on the ball. Taking the upwards direction to be positive: Impulse (in Newton seconds) = (0.5 × 6) – (0.5 × –8) =3+4 =7 28 of 48 © Boardworks Ltd 2005 Calculating impulse 2 A ball of mass 0.2 kg falls into a lake. Its velocity before impact with the water was 15 ms–1 and after impact with the water 10 ms–1. Calculate the impulse exerted on the ball by the water. Taking the downwards direction to be positive: Impulse = 0.2 × 10 – 0.2 × 15 =2–3 = –1 Ns Therefore the impulse exerted on the ball by the water is 1 Ns in an upwards direction. 29 of 48 © Boardworks Ltd 2005 Momentum Contents Newton’s laws of motion Applying Newton’s laws Momentum Impulse The principle of conservation of momentum Examination-style questions 30 of 48 © Boardworks Ltd 2005 The principle of conservation of momentum The principle of conservation of linear momentum states that if two objects moving in a straight line collide, the total momentum before the collision is equal to the total momentum after the collision. We can prove this as follows: Let the forces acting on two particles of masses m1 and m2 in an isolated system be F1(t) and F2(t) respectively, where the forces are variable and functions of time. Note that in an isolated system there are no external forces such as air resistance and friction. 31 of 48 © Boardworks Ltd 2005 The principle of conservation of momentum By Newton’s Third Law: F1 = –F2 F1 + F2 = 0 From Newton’s second Law: m1a1 + m2a2= 0 Integrating with respect to time we get: m1v1 + m2v2= constant Therefore, the total momentum of the system is conserved. In general, for two particles of mass m1 and m2 we have, m1u1 + m2u2 = m1v1 + m2v2 32 of 48 © Boardworks Ltd 2005 Modelling collisions 33 of 48 © Boardworks Ltd 2005 Conservation of momentum 1 Two smooth spheres A and B are projected towards each other with speeds of 5 ms–1 and 7 ms–1 respectively. Sphere A has mass 0.3 kg and sphere B has mass 0.4 kg. After they collide sphere A rebounds with a speed of 6 ms–1. Calculate the speed with which sphere B rebounds. Taking the positive direction of motion to be from left to right: (5 × 0.3) + (–7 × 0.4) = (–6 × 0.3) + (v × 0.4) 1.5 – 2.8 = –1.8 + 4v 5 ms–1 7 ms–1 4v = 0.5 Before impact 0.3 kg 0.4 kg v = 0.125 A B Therefore, sphere B v ms–1 6 ms–1 moves to the right with a speed of 0.125 ms–1. After impact 0.3 kg 0.4 kg A 34 of 48 B © Boardworks Ltd 2005 Conservation of momentum 2 A particle A of mass 0.3 kg collides directly with a particle B of mass 0.1 kg. Immediately before the collision A has a velocity of 5 ms–1 and B is at rest. After the collision the particles coalesce into a single particle C and it continues to move in the same direction as A. Find the velocity of C immediately after the collision. Total momentum before collision = total momentum after 0.3 × 5 = 0.4 × v 0.4v = 1.5 v = 3.75 Therefore C moves with velocity 3.75 ms–1 immediately after the collision. 35 of 48 © Boardworks Ltd 2005 Impulsive tension (jerk) in a string 36 of 48 © Boardworks Ltd 2005 Jerk in a string Two smooth spheres A and B of mass 2 kg and 3 kg respectively are at rest on a smooth, horizontal surface. They are connected by a light, inextensible string which is slack. A is projected directly away from B with a speed of 5 ms–1. Find their speed when the string becomes taut. Before Jerk 5 ms–1 0 ms–1 2 kg 3 kg A Let the speed when the string is taut be v ms–1 (2 × 5) + (3 × 0) = 5v B 5v = 10 After Jerk A 37 of 48 v ms–1 v ms–1 2 kg 3 kg B v=2 The spheres move to the left with a speed of 2 ms–1. © Boardworks Ltd 2005 Examination-style questions Contents Newton’s laws of motion Applying Newton’s laws Momentum Impulse The principle of conservation of momentum Examination-style questions 38 of 48 © Boardworks Ltd 2005 Examination-style question 1 A car is travelling along a straight road. Initially it is travelling at a constant speed on a horizontal section of the road. The mass of the car and the driver is 850 kg and the car is exerting a driving force of 1100 N. During the second stage of the journey the road inclines at an angle of 3° to the horizontal. Find the driving force that the car must exert if it is to accelerate at a rate of 0.5 ms–2. Assume that the resistive force remains constant through the whole motion. 39 of 48 © Boardworks Ltd 2005 Solution 1 By Newton’s First Law the resistive force acting on the car must be 1100N. a = 0.5 ms–2 R D 1100 N 3° 850g Applying Newton’s Second Law up the plane, D – 1100 – 850g cos87° = 850 × 0.5 D = 1960 (to 3 s.f.) Therefore the driving force required to generate an acceleration of 0.5 ms–2 is 1960 N. 40 of 48 © Boardworks Ltd 2005 Examination-style question 2 A particle A collides with a particle B. A has a mass of 0.3 kg and is moving with a velocity of 1 3 –1 –1 ms immediately before the collision and ms 2 4 immediately after the collision. a) Find the change in the momentum of A due to this collision and state the change in the momentum of B. The mass of B is 0.2 kg and immediately before the collision 1 –1 it has a velocity of ms . 6 b) Find the velocity of B immediately after the collision. 41 of 48 © Boardworks Ltd 2005 Solution 2 1 3 a) Change in momentum of A = 0.3 0.3 2 4 0.6 = 1 . 8 0.6 Therefore the change in momentum of B is Ns. 1.8 b) For B, v1 1 0.6 0.2 0.2 = v 6 1 . 8 2 v1 = 4 and v2 = 3 Therefore the velocity of B immediately after the collision is 4 –1 3 ms . 42 of 48 © Boardworks Ltd 2005 Examination-style question 3 Two particles A and B are moving directly towards each other when they collide. A is of mass 0.4 kg and B is of mass 0.8 kg. Immediately before the collision A is moving with a speed of 5 ms–1 and B is moving with a speed of 4 ms –1. Immediately after the collision the particles move away from each other with speeds of a ms–1 and b ms–1 respectively. a) Show that a – 2b = 3 b) After the collision A moves 7 m in 2 seconds. Find the values of a and b. 43 of 48 © Boardworks Ltd 2005 Solution 3 A A 5 ms–1 4 ms–1 0.4 kg 0.8 kg B a ms–1 b ms–1 0.4 kg 0.8 kg B a) Taking left to right the be the positive direction, by conservation of momentum, 0.4 × 5 – 0.8 × 4 = –0.4 × a + 0.8 × b 2 – 3.2 = –0.4a + 0.8b 4a – 8b = 12 a – 2b = 3 b) A moves 7 m in 2 seconds a = 3.5 3.5 – 2b = 3 2b = 0.5 b = 0.25 44 of 48 © Boardworks Ltd 2005 Examination-style question 4 Two trucks A and B, moving in opposite directions on the same horizontal track, collide. The mass of A is 800 kg and the mass of B is m kg. Immediately before the collision the speed of A is 5 ms–1 and the speed of B is 4 ms–1. Immediately after the collision the trucks are joined together and move with the same speed, 1 ms–1. The direction of motion of A is unchanged by the collision. Find a) the mass of truck B b) the magnitude of the impulse exerted on A in the collision. c) State any assumptions that you have made. 45 of 48 Before A 5 ms–1 4 ms–1 800 kg m kg B 1 ms–1 After (800 + m) kg © Boardworks Ltd 2005 Solution 4 a) Using Conservation of Momentum, 5 × 800 – 4m = 800 + m 4000 – 4m = 800 + m 3200 = 5m m = 640 mass of truck B is 640 kg b) Impulse = change in momentum, I = 800 × 1 – 800 × 5 = 800 – 4000 = –3200 the magnitude of the impulse exerted on A is 3200 Ns. c) The trucks are modelled as particles and there are no external forces, such as friction, acting. 46 of 48 © Boardworks Ltd 2005 Examination-style question 5 Two particles A and B have masses m kg and 2 kg respectively. The particles are moving directly towards each other on a smooth horizontal surface when they collide. Immediately before the collision the speed of A is 6 ms–1 and the speed of B is 2 ms–1. Given that the impulse exerted on B by A is 10 Ns, a) find the speed of B immediately after the collision. b) If the speed of A immediately after the collision is 2 ms–1 in the same direction as before the collision, calculate the value of m. 47 of 48 © Boardworks Ltd 2005 Solution 5 6 ms–1 A m kg 2 ms–1 B 2 kg 2v + 4 = 10 v ms–1 B 6 ms–1 A A m kg 2 kg 2 ms–1 B 2 kg 2 ms–1 3 ms–1 m kg 2 kg B a) Taking left to right to be the positive direction, change in momentum of B is 2v = 60 v = 30 Therefore the speed of B after the collision is 3 ms–1. b) Taking left to right to be the positive direction, by Conservation of Momentum 6m – 4 = 2m + 6 4m = 10 Therefore the mass of A is 2.5kg. 48 of 48 © Boardworks Ltd 2005