Types of force Direction of force Friction Weight

Types of force Direction of force Friction Weight Tension in string Thrust in rod Friction

Resolving forces into components Why do some cricket bowls bounce higher than others? A C H means O S H means The velocity of the ball has a horizontal and a vertical component Resolving (separating) these components helps you understand the trajectory of the ball

Resolving forces into components Consider a block on a slope: The weight of the block has: a component acting perpendicular to the slope a component acting parallel to the slope Consider a block being pulled along a surface: The tension of the wire exerts: a component acting vertically on the block a component acting horizontally on the block When you intend to resolve forces, you can use the following notation: resolve parallel to the slope resolve horizontally resolve perpendicular to the slope resolve vertically

Resultant forces The combined effect of any number of forces are called the resultant force What direction would the resultant force act in? Now try Ex 3 C, Q 2

Newton’s laws of motion Newton’s 2 nd law states that the force needed to accelerate a particle is equal to the product of the particle’s mass and the acceleration produced. If the resultant force is zero, the object will not move Eg 6 kg block is being pulled by a cable with tension 12 N. The surface exerts a resistance of 4 N. Find the acceleration of the block Eg a 70 kg skydiver is falling at terminal velocity. Write down the force caused by air resistance Now try Ex 3 A, Q 7 -10

Particles on slopes One of the common scenarios you encounter is a particle on a slope. By resolving the weight into components acting parallel and perpendicular to the slope before applying Newton’s law F = ma, the acceleration of the particle can be determined. Eg a block released on a smooth slope Eg a block released on a rough slope with friction F N

Normal reaction The normal reaction is the force which acts perpendicular to a surface when an object is contact with the surface. This must be equal to the resultant force an object is applying to the surface, as the object is not accelerating. Eg a block at rest on a slope Eg a block at rest on a surface, despite being pulled by a string

Friction The coefficient of friction, μ measures the roughness of 2 surfaces in contact. The larger the value of μ, the greater the friction. As the applied force increases, so does the force of friction resisting movement, up to a limiting value. If the applied force exceeds this limit, the block will move. The limiting value depends on the normal reaction and coefficient of friction between 2 surfaces, so that Hence

Will it move? A block of mass 5 kg lies at rest on a rough surface with a coefficient of friction of 0. 4 A horizontal force of P N is applied to the block Find the size of the frictional force and where applicable, acceleration of the block when P is: a) 10 N b) 19. 6 N c) 30 N a) P is less than Fmax, so the block will not move b) P is equal to Fmax, so the block will not move but is in limiting equilibrium c) P is greater than Fmax, so the block will move Now try Ex 3 D

The typical scenario For a block that is about to slide down a slope: The weight of the block has components acting parallel & perpendicular to the slope The slope in turn applies a normal reaction equal to the perpendicular component Proportional to this reaction, friction acts up the slope up to a limit equal to the component of weight acting down the slope Weight Reaction Friction Acceleration

Eg A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane inclined at an angle of 30° to the horizontal. The coefficient of friction between P and the plane is 0. 4. The initial speed of P is 6 m s-1. Find (a) the frictional force acting on P as it moves up the plane, (b) the distance moved by P up the plane before P comes to instantaneous rest.

WB 10 The diagram above shows a boat B of mass 400 kg held at rest on a slipway by a rope. The boat is modelled as a particle and the slipway as a rough plane inclined at 15° to the horizontal. The coefficient of friction between B and the slipway is 0. 2. The rope is modelled as a light, inextensible string, parallel to a line of greatest slope of the plane. The boat is in equilibrium and on the point of sliding down the slipway. (a) Calculate the tension in the rope. The boat is 50 m from the bottom of the slipway. The rope is detached from the boat and the boat slides down the slipway. (b) Calculate the time taken for the boat to slide to the bottom of the slipway.

WB 11 A small parcel of mass 2 kg moves on a rough plane inclined at an angle of 30° to the horizontal. The parcel is pulled up a line of greatest slope of the plane by means of a light rope which is attached to it. The rope makes an angle of 30° with the plane, as shown in the diagram above. The coefficient of friction between the parcel and the plane is 0. 4. Given that the tension in the rope is 24 N, (a) find, to 2 significant figures, the acceleration of the parcel.

The rope now breaks. The parcel slows down and comes to rest. (b) Show that, when the parcel comes to this position of rest, it immediately starts to move down the plane again. (c) Find, to 2 significant figures, the acceleration of the parcel as it moves down the plane after it has come to this position of instantaneous rest. Now try Ex 3 E

Ex 3 E Q 1) Q 2) g ½k 2 kg

Q 5) g 2 k (1) + (2) Q 6) 4 kg (1) + (2)

Q 3) Q 4) g k 40 kg 20

Q 7) kg 10 (1) + (2)

Q 8) B A

Finding the value of μ If no other forces but friction and weight are acting on a particle, the coefficient of friction can be determined by raising a slope until the block slides Eg a block can be raised to an angle of 45 o before sliding, what is the coefficient of friction?

WB 12 A particle P of mass 0. 4 kg is moving under the action of a constant force F newtons. Initially the velocity of P is (6 i – 27 j) m s– 1 and 4 s later the velocity of P is (-14 i + 21 j) m s– 1. (a) Find, in terms of i and j, the acceleration of P. (b) Calculate the magnitude of F.

Connected particles moving in the same direction Problems concerning connected particles moving in the same direction should usually be solved by considering the particles separately Eg Two particles P and Q, of masses 5 kg and 3 kg respectively, are connected by a light, inextensible string. Particle P is pulled by a horizontal force of 40 N along a rough horizontal plane. The coefficient of friction is 0. 2 and the string is taut. P Q 3 kg 5 kg Find the acceleration of each particle and the tension in the string For Q: (1) + (2) For P: Sub in (1) Now try Ex 3 F, Q 1 -3

Ex 3 F Q 1 For P: For Q: P Q 2 kg 8 kg Sub T = 0. 8 in (1) Q 2 For P: P 5 kg For Q: Q 5 kg Sub R = 45 in (2)

Q 3 500 kg 1500 kg For the car: For the trailer: Sub a = 3 in (2) NB: if the tensions are acting in opposite directions then the whole system can be considered as one particle, sometimes making the algebra simpler For the whole system: 500 kg 2000 kg 1500 kg

WB 16 This figure shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is at an angle of 15° to the road. The lorry and the car experience constant resistances to motion of magnitude 600 N and 300 N respectively. The lorry’s engine produces a constant horizontal force on the lorry of magnitude 1500 N. Find (a) the acceleration of the lorry and the car, (b) the tension in the towbar. For the car,

When the speed of the vehicles is 6 m s– 1, the towbar breaks. Assuming that the resistance to the motion of the car remains of constant magnitude 300 N, (c) find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest. After towbar brakes, for the car: (d) State whether, when the towbar breaks, the normal reaction of the road on the car is increased, decreased or remains constant. Give a reason for your answer. Before After The vertical component of T is no longer present, so the normal reaction increases Now try Ex 3 F

Connected particles moving in different directions Problems concerning connected particles moving in different directions must be solved by considering the particles separately Eg two particles connected by a light, inextensible string over a smooth pulley Find the acceleration of the particles For P: For Q: Find the tension in the string For P, using (1): Find the force exerted by the string on the pulley 2 m P 3 m Q

WB 13 A particle A of mass 0. 8 kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1. 2 kg which hangs freely below the pulley, as shown in the diagram above. The system is released from rest with the string taut and with B at a height of 0. 6 m above the ground. In the subsequent motion A does not reach P before B reaches the ground. In an initial model of the situation, the table is assumed to be smooth. Using this model, find (a) the tension in the string before B reaches the ground, For A: For B: Sub in (1)

(b) the time taken by B to reach the ground. In a refinement of the model, it is assumed that the table is rough and that the coefficient of friction between A and the table is Using this refined model, (c) find the time taken by B to reach the ground. For A: For B:

WB 14 A particle A of mass 4 kg moves on the inclined face of a smooth wedge. This face is inclined at 30° to the horizontal. The wedge is fixed on horizontal ground. Particle A is connected to a particle B, of mass 3 kg, by a light inextensible string. The string passes over a small light smooth pulley which is fixed at the top of the plane. The section of the string from A to the pulley lies in a line of greatest slope of the wedge. The particle B hangs freely below the pulley, as shown in the diagram above. The system is released from rest with the string taut. For the motion before A reaches the pulley and before B hits the ground, find (a) the tension in the string, For A: For B: Sub in (1)

(b) the magnitude of the resultant force exerted by the string on the pulley. (c) The string in this question is described as being ‘light’. (i) Write down what you understand by this description. (ii) State how you have used the fact that the string is light in your answer to part (a). (i) The string has no weight/mass (ii) The tension in the string is constant (same at A as B)

WB 15 The diagram above shows two particles A and B, of mass m kg and 0. 4 kg respectively, connected by a light inextensible string. Initially A is held at rest on a fixed smooth plane inclined at 30° to the horizontal. The string passes over a small light smooth pulley P fixed at the top of the plane. The section of the string from A to P is parallel to a line of greatest slope of the plane. The particle B hangs freely below P. The system is released from rest with the string taut and B descends with acceleration g. (a) Write down an equation of motion for B. (b) Find the tension in the string. Using (a) (c) Prove that m = For A:

(d) State where in the calculations you have used the information that P is a light smooth pulley. The tension in the string is constant (same at A as B) On release, B is at a height of one metre above the ground and AP = 1. 4 m. The particle B strikes the ground and does not rebound. (e) Calculate the speed of B as it reaches the ground. P (f) Show that A comes to rest as it reaches P. When B hits the floor, A is moving at Between then and A stopping, only gravity is acting: at P

Ex 3 F Q 4 For A: For B: Sub in (1): 3 kg 4 kg A For A after falling for 2 metres: For B after A hits the ground: Total height gained by B = B

Q 5 For A: For B: A 5 kg Sub in (1): 3 kg B

Q 6 For P: For Q: P m For P: m Q

Q 7 Q m For P: For Q: Sub in (2): m P

Q 8 For the whole system: 0 90 For the trailer: 0 50 kg kg

Q 9 For Q: P For P: 2 kg 3 kg Q

Q 10 For whole system: 300 kg 900 kg For the trailer: When decelerating, the tow bar exerts a thrust on the car and trailer

Ex 3 F Q 4 3 kg 4 kg A B

Q 5 A 5 kg 3 kg B

Q 6 P m m Q

Q 7 Q m m P

Q 8 0 90 0 50 kg kg

Q 9 P 2 kg 3 kg Q

Q 10 300 kg 900 kg

Review Exercise 1, Q 41 m Sub (1) in (2)

Review Exercise 1, Q 45 50 kg

Review Exercise 1, Q 47 10 k g

Momentum The momentum of a body of mass m kg which is moving with velocity v ms-1 is mv N By Newton’s 3 rd law, when two bodies collide, each exerts an equal and opposite force on the other. By the impulse – momentum principle, these forces are equal & opposite, hence the total momentum of the system is unchanged. This is called the principle of conservation of momentum In words, Total momentum before impact = Total momentum after impact Eg A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s-1. The truck P collides with a truck Q of mass 3000 kg, which is at rest on the same track. Immediately after the collision Q moves with speed 5 m s-1. Calculate the speed of P immediately after the collision, Before I After Conservation of momentum P Q I

Impulse If a constant force F acts for time t then we define the impulse of the force to be Ft Substituting in gives This is called the impulse - momentum principle In words, Impulse = change in momentum Eg Calculate the magnitude of the impulse exerted by P on Q during the collision. Before I After P Q For Q: I NB: the impulse Q exerts on P must be the equal and opposite For P:

Eg A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s-1. The truck P collides with a truck Q of mass 3000 kg, which is at rest on the same track. Immediately after the collision Q moves with speed 5 m s-1. Calculate the speed of P immediately after the collision, Before I Conservation of momentum P Q I After Eg Calculate the magnitude of the impulse exerted by P on Q during the collision. NB: the impulse Q exerts on P must be the equal and opposite For Q: For P:

WB 17 Two particles A and B have mass 0. 12 kg and 0. 08 kg respectively. They are initially at rest on a smooth horizontal table. Particle A is then given an impulse in the direction AB so that it moves with speed 3 m s-1 directly towards B. (a) Find the magnitude of this impulse, stating clearly the units in which your answer is given. Before I A B After I Immediately after the particles collide, the speed of A is 1. 2 m s-1, its direction of motion being unchanged. (b) Find the speed of B immediately after the collision. Conservation of momentum (c) Find the magnitude of the impulse exerted on A in the collision.

WB 18 A particle P of mass 2 kg is moving with speed u m s– 1 in a straight line on a smooth horizontal plane. The particle P collides directly with a particle Q of mass 4 kg which is at rest on the same horizontal plane. Immediately after the collision, P and Q are moving in opposite directions and the speed of P is one-third the speed of Q. (a) Show that the speed of P immediately after the collision is Before Conservation of momentum P Q After the collision P continues to move in the same straight line and is brought to rest by a constant resistive force of magnitude 10 N. The distance between the point of collision and the point where P comes to rest is 1. 6 m. (b) Calculate the value of u. For P:

WB 19 Two trucks A and B, moving in opposite directions on the same horizontal railway track, collide. The mass of A is 600 kg. The mass of B is m kg. Immediately before the collision, the speed of A is 4 m s– 1 and the speed of B is 2 m s– 1. Immediately after the collision, the trucks are joined together and move with the same speed 0. 5 m s– 1. The direction of motion of A is unchanged by the collision. Find (a) the value of m, Before I After A Conservation of momentum B A&B I (b) the magnitude of the impulse exerted on A in the collision. Now do Ex 3 I, Q 1, 3, 7, 14, 15

Before Q 1 Gun Conservation of momentum Bullet After Before Q 3 Conservation of momentum P Q After For P:

Q 7 For the driver: Before Driver After Pile driver Conservation of momentum

Before Q 14 Conservation of momentum P Q After As v <0, direction is unchanged For P: Before Q 15 After A B A&B Conservation of momentum

Assignment 4, Q 2 For A: Before I After For B: A B I

Assignment 4, Q 3 For B: A For A: P m 2 m B For B when it hits the ground: For A after B hits the ground: A m Giving Information :

Assignment 4, Q 4 P 2 kg