Momentum and Newton s Laws

Transcription

1 Momentum and Newton s Laws Newton s Laws Newton s Laws of Motion were covered in detail in Grade 11, so this will be a summary of the content needed for the Grade 12 examination. Forces Lessone s s o n 1 A force is defined as a push or a pull which may change the state of rest or uniform motion of a body in a straight line. Force is measured in newtons (N). Forces can be considered as either contact or non-contact forces. Contact forces are forces where the objects involved come into direct contact with each other. For example, when a tennis racket hits a tennis ball to change the ball s motion. Non-contact forces are forces which can be experienced over a distance. For example, magnetic forcs, electrostatic forces and gravitational attraction. Forces always act in pairs as explained by Newton s Third Law of Motion. Newton s Third Law of Motion If body A exerts a force on body B, then body B exerts a force on body A that is equal in magnitude and opposite in direction. Remember in Newton III pairs, the forces: have the same magnitude but opposite directions are in line with each other act on different bodies can be caused by the same event Examples of Newton III 1. A book on a table: the book exerts a force on the table due to gravity and the table exerts a force of the same size against the book in the opposite direction. F of table on book EXAMPLE e = mc² F of book on table 2. A donkey pulling a cart: the donkey pulls the cart forward and the cart pulls backward against the donkey with the same force. F of cart on donkey F of donkey on cart Page 1 Science Catalyst Workbook.indb /12/13 01:32:02 PM

2 3. Walking: the foot pushes against the earth and the earth pushes against the foot with the same force but in the opposite direction. F of ground on foot F of foot on ground Inertia The property of an object that causes it to resist a change in its state of motion. Consider trying to push a large, heavy box across the floor. It is very difficult and requires a large force for this to happen; this is due to the box s inertia. If the box however was small and light, it would be much easier to move it across the floor. This means that it is easier to overcome the box s inertia. The mass of an object is a measure of the object s inertia. The inertia of an object is also explained by Newton s First Law of Motion. Newton s First Law A body will continue its state of rest or uniform velocity unless acted on by an external net force. For example, when a car brakes suddenly, the driver will continue to move forward due to their inertia. The force and acceleration of the car is in the opposite direction to the movement of the driver. The seat belt stops the movement/inertia of the driver. F of seat belt on man F of man on seat belt Momentum An object s momentum is defined as the product of the mass and velocity of a body. Momentum is a vector and therefore has both magnitude and direction. We use the following equation to calculate momentum: p = mv p = momentum. Measured in kg m s 1 m = mass. Measured in kg v = velocity. Measured in m s 1 Page 2 Science Catalyst Workbook.indb /12/13 01:32:02 PM

3 Change in Momentum If the velocity of an object changes, then the momentum of the object will change too. We can calculate the change in momentum as follows: Δp = p final p initial = mv f mv i = m(v f v i ) Δp = change in momentum. Measured in kg m s 1 m = mass. Measured in kg v f = final velocity. Measured in m s 1 v i = initial velocity. Measured in m s 1 How quickly the momentum of an object changes is equal to the resultant force acting on the object. This is explained by Newton s Second Law of Motion. Newton s Second Law of Motion In terms of momentum, Newton s Second Law says that the rate of change of momentum of a body is directly proportional to the net force applied and is in the direction of the net force. This can be expressed by the following equation: = Δp _ Δt Remember that: Δp = m(v f v i ) = mδv So the equation becomes: = mδv Δt but a = Δv F Δt net = net force. Measured in newtons (N) Therefore: m = mass of object. Measured in kilograms (kg) = ma = net force. Measured in newtons (N) Δp = change in momentum. Measured in kg m s 1 Δt = change in time. Measured in seconds (s) a = acceleration of the object. Measured in m s 2 The relationship between the net force, mass and acceleration of an object can be expressed by an alternative form of Newton s Second Law which states: The acceleration of a body is directly proportional to the net force and inversely proportional to the mass of the body and is in the direction of the net force Conservation of Momentum The Law of Conservation of Linear Momentum states that the total linear momentum of an isolated system remains constant in both magnitude and direction for interacting bodies. The law can be understood another way: In an isolated system, the total momentum before a collision is equal to the total momentum after the collision providing there is no external resultant force. = = mass of object 1 m 2 = mass of object 2 Page 3 Science Catalyst Workbook.indb /12/13 01:32:03 PM

4 = initial velocity of object 1 = initial velocity of object 2 = final velocity of object 1 after collision = final velocity of object 2 This equation can be manipulated to suit the specific question. Consider the following situations: Situation 1: Two objects move towards each other and after the collision they move off separately. = Situation 2: Two objects move towards each other and after the collision they move off together as one object. = ( )v f The final velocity of the objects after the collision is the same and the masses are added together. Situation 3: Two objects which are stuck together to begin with become two objects moving in opposite directions after an explosion. ( )v i = The initial velocity of the objects before the collision is the same and the masses are added together. e = mc² EXAMPLE Example A car of mass 1920 kg is stopped at a traffic light when a truck of mass 3300 kg and travelling at 20 m s 1 collides with the car. The car is pushed into the intersection and the truck continues with a velocity of 12 m s 1. Calculate the velocity of the car after the collision. v 3300kg i1 = 20m s -1 Before After v 1920kg i2 = 0m s -1 Page 4 v i1 = 12m s -1 v =? i2 Science Catalyst Workbook.indb /12/13 01:32:03 PM

5 Solution Original direction of the truck is positive Car: = 1920 kg Truck: m 2 = 3300 kg = 0 m s 1 = 20 m s 1 = unknown = 12 m s 1 = 1 920(0) (20) = (12) = = = = 13,75 m s 1 in the truck s original direction Note: Always specify direction Make sure you substitute in all values, even when v = 0 SOLUTION Example Jo pushes his skateboard of mass 5 kg ahead of him. The skateboard moves with a velocity of 2 m s 1. Jo runs and jumps on the skateboard. His mass is 60 kg and he runs with a velocity of 3 m s 1. Calculate the velocity of both Jo and his skateboard after he landed on it. = 60 kg = 3 m s 1 Before = 5 kg = 2 m s 1 After EXAMPLE e = mc² Solution Original direction of Jo s motion is positive Jo: = 60 kg Skateboard: m 2 = 5 kg = 3 m s 1 = 2 m s 1 = unknown = unknown = = v SOLUTION = ( )v 60(3) + 5(2) = (60 + 5)v = 65v v = = 2,92 m s 1 Activity 1 A car travelling to the right with a mass of 900 kg and a velocity of 16 m s 1 crashes head on into a car with a mass of 1300 kg and travelling to the left with a velocity of 22 m s 1. Both cars remain locked together after the collision. Calculate the final velocity and the direction in which the two cars move after the collision. Before After m = 900kg 1 m = 1300kg 2-1 = 16m s v =? f -1 = 22m s Page 5 Science Catalyst Workbook.indb /12/13 01:32:03 PM

6 Activity 2 Two girls want to rollerblade. They push off from each other. Janice has a mass of 60 kg and she moves with a velocity of 1,2 m s 1 right. Portia has a mass of 45 kg and moves in the opposite direction. Calculate Portia s velocity. Activity 3 A boy with a mass of 65 kg jumps onto a stationary rowing boat with a horizontal velocity of 12 m s 1. If the rowing boat has a mass of 180 kg, calculate the speed of the boat once the boy has landed. Assume that there is no friction between the boat and the water. Activity 4 A 4 kg gun fires a bullet of mass 0,35 kg at 200 m s 1. Calculate the velocity of the gun recoil (recoil is the term given to the motion of the gun as it moves after the bullet has left the gun). Page 6 Science Catalyst Workbook.indb /12/13 01:32:04 PM

7 Activity 5 A train of mass kg is moving west at 25 m s 1. Another train with a mass of 6, kg is moving east on the same track with a velocity of 28 m s 1. The trains collide and lock together. Calculate the velocity and direction of the trains after their collision. Activity 6 A block of mass 8 kg falls freely from rest through a distance of 40 m. The block lands on a second block of mass 15 kg resting on a ledge below. Immediately after the collision, both objects move at the same speed. Calculate: 1. The speed of the 8 kg block just before it collides with the 15 kg block 2. The speed of the two objects immediately after the collision. Impulse Newton II can be expressed in terms of momentum. From this we can measure how difficult it is to bring an object to rest, or to move it. This is known as impulse. The unit is N.s. Δt = impulse. Measured in N s Δt = Δp Δp = change in momentum. Measured in kg m s 1 = mv f mv i m = mass. Measured in kg = m(v f v i ) v f = final velocity. Measured in m s 1 v i = initial velocity. Measured in m s 1 Page 7 Science Catalyst Workbook.indb /12/13 01:32:05 PM

8 e = mc² EXAMPLE Example A car company is testing its new model in a crash test. The vehicle v i = 15 m s 1 mass is kg and it collides with a wall. The car moves towards the wall at 15 m s 1 initially and after the collision at 2,5 m s 1 away from the wall. 1. If the collision lasts for 0,10 s, calculate the impulse due to the collision. 2. If the car is in contact with the wall for 0,1 s, calculate the magnitude of the force that the wall exerts on the car. SOLUTION Solution 1. Take towards the wall as positive m = 1300 kg Δt = Δp v i = 15 m s 1 = mv f mv i v f = 2,5 m s 1 = (1300)(-2,5) (1300)(15) = = = N.s The impulse is 22750N.s away from the wall. 2. Δt = N.s Δt = 0,1 s Δt = Δp (0,1) = = N Activity 7 You are playing tennis. The ball with a mass of 400 g approaches you with a velocity of 30 m s 1 and you swing your racquet with a velocity of 40 m s 1 in the opposite direction. The ball moves back over the net and you follow through with a velocity of 32 m s 1. The mass of the racquet is 3, 6 kg. 1. Calculate the velocity of the tennis ball after you have hit it. 2. Calculate the impulse that the racquet exerts on the ball if they are in contact for 0,8 s. Page 8 Science Catalyst Workbook.indb /12/13 01:32:05 PM

9 Activity 8 A boy is playing soccer with his friends. He runs towards the ball which is stationary on the ground. After he kicks the ball it moves at 20m s 1 to the right. The boy s foot is in contact with the ball for 0,4 s. Calculate the force that the boy s foot exerts on the ball. The soccer ball has a mass of 500g. Elastic and Inelastic Collisions Although momentum is always conserved during a collision, kinetic energy of the system may not be conserved. An elastic collision is a collision where the total kinetic energy of the system remains constant. That is: Total K before = Total K after A collision can only be elastic if there are no external forces involved in the collision. External forces are forces such as friction and air resistance. In reality elastic collision are virtually impossible as friction and air resistance are almost always involved. An inelastic collision is a collision in which the total kinetic energy of the system is not constant. That is: Total K before Total K after According to the law of conservation of energy, energy cannot be created or destroyed but can be transferred to another form. With an inelastic collision the total kinetic energy after the collision is often less than the total kinetic energy before the collision. The kinetic energy that is lost is transferred to sound and heat and therefore not lost as such. Important: to prove if a collision is elastic or inelastic the total kinetic energy before and after the collision must be worked out separately and then the answers compared. Example Consider Jo jumping onto his skateboard again. Before Jo jumps onto the skateboard he is running at 3m s 1 and the skateboard is moving at 2 m s 1. After he has jumped onto the skateboard they move off together at 2,92 m s 1. Is this an elastic or inelastic collision. Justify your answer with the appropriate calculations. = 60 kg = 3 m s 1 Before = 5 kg = 2 m s 1 After e = mc² EXAMPLE Page 9 Science Catalyst Workbook.indb /12/13 01:32:06 PM

10 SOLUTION Solution Before: = 60 kg m 2 = 5 kg After: m = 65 kg = 3 m s 1 v f = 2,92 m s 1 = 2 m s 1 Total K before = 1 2 m v i1 2 m v Total K 2 i22 after = 1 2 mv 2 f = 1 2 (60)(3) (5)(2)2 = 1 2 (65)(2,92)2 = = 277,11 J = 280 J The collision is inelastic as Total K before Total K after Activity 9 You are watching your friends play pool. A stationary colour ball of 0,25 kg is struck by the white ball of mass 0,2 kg with a velocity of 0,4 m s 1. The white ball is left stationary. Before After m = 0,2 kg m = 0,25 kg v = 0,4 m s 1 v = 0 m s 1 v = 0 m s 1 v =? 1. Calculate the velocity and direction of the colour ball. 2. Is the collision elastic or inelastic? Justify your answer with the relevant equations Activity 10 A toy car, with mass 100 g is travelling at 3 m s 1 to the right. It collides with another toy car with mass 150 g travelling at 4 m s 1 to the left. After the collision the two toy cars move off together. Page 10 Science Catalyst Workbook.indb /12/13 01:32:06 PM

11 1. Calculate the velocity of the two cars after the collision. 2. Is the collision elastic or inelastic? Justify your answer with the relevant calculations. Solutions to Activities Activity 1 To the right is positive. Car 1: = 900 kg Car 2: m 2 = 1300 kg = 16 m s 1 = 22 m s 1 = ( )v 900(16) ( 22) = ( )v ( 28600) = 2200v = 2200v v = = 6,45 m s 1 v =? v =? The vehicles move at 6,45 m s 1 to the left. Activity 2 Assume to the right is positive Janice: = 60 kg Portia: m 2 = 45 kg v i = 0 m s 1 v i = 0 m s 1 = 1,2 m s 1 =? ( )v i = ( )(0) = 60(1,2) + (45) 0 = = = 1,6 Portia moves with a velocity of 1,6 m s 1 left. Page 11 Science Catalyst Workbook.indb /12/13 01:32:07 PM

12 Activity 3 Assume movement away from land is positive. Boy: = 65 kg Boat: m 2 = 180 kg = 12 m s 1 = 0 m s 1 =? =? = = v = ( )v 65(12) + 180(0) = ( )v 780 = 245v v = = 3,18 The boy and the boat are moving at 3,18 m s 1 away from land. Activity 4 Assume movement away from you is positive. Gun: = 4 kg Bullet: m 2 = 0,35 kg = 0 m s 1 = 0 m s 1 =? = 200 m s 1 ( )v i = (4 + 0,35)(0) = 4 + 0,35(200) 0 = Hint: Remember that the gun and the bullet are one object before the explosion which causes the bullet to leave the gun. 70 = 4 = 70 4 = 17,5 The velocity of the recoil is 17,5 m s 1 toward you Activity 5 Hint: Draw a diagram of the situation described before answering the question east 3 6,5 x 10 kg west 3 2 x 10 kg Before 28m s -1 25m s -1 After Assume west is positive. Page 12 Science Catalyst Workbook.indb /12/13 01:32:08 PM

13 Train A: = 2 x 10 3 kg Train B: m 2 = 6,5 x 10 3 kg = 25 m s 1 = 28 m s 1 = v = v = ( )v (2 x 10 3 )(25) + (6,5 x 10 3 )( 28) = (2 x ,5 x 10 3 )v = 8,5 x 10 3 v = 8,5 x 10 3 v v = 8,5 x 10 3 = 15,53 The velocity of the trains is 15,53 m s 1 east. Activity 6 Assume down is positive. 1. g = 9,8 m s 1 v i = 0 m s 1 v f =? v 2 2 f = v i + 2aΔy Δy = 40 m v 2 f = 0 + 2(9,8)(40) v 2 f = 784 v f = 784 = 28 The velocity of Block A just before the collision is 28 m s 1 down. 2. Block A: = 8 kg Block B: m 2 = 15 kg = 28 m s 1 = 0 m s 1 = v v 1 = v 2 = v = v = ( )v 8(28) + 15(0) = (8 + 15)v 224 = 23v v = = 9,74 The blocks fall with a velocity of 9,74 m s 1 down Activity 7 30m s 400g -1?m s g 40m s -1 32m s -1 3,6kg racquet racquet Page 13 Science Catalyst Workbook.indb /12/13 01:32:09 PM

14 Assume movement toward you is positive. Ball: = 400 g = 0,4 kg Racquet: m 2 = 3, 6 kg = 30 m s 1 = 40 m s 1 =? = 32m s 1 1. = 0,4(30) + 3,6( 40) = 0,4( ) + 3,6( 32) = 0,4 115,2 132 = 0,4 115, ,2 = 0,4 16,8 = 0,4 = 16,8 0,4 = 42 The ball is travelling at 42m s 1 away from the racquet 2. Δt = Δp = ( ) = (0,4)( 42 30) = 28,8 The impulse is 28,8N.s away from the racquet. Note: You could also use the racquet as your point of reference because the force exert on it must be equal to the force exerted on the ball (Newton s third law) Activity 8 Take right as positive F =? Δt = 0,4 s m = 500 g = 0,5 kg v i = 0 m s 1 v f = 20 m s 1 Δt = Δp Δt = m(v f v i ) (0,4) = (0,5)(20 0) (0,4) = 10 = 10 0,4 = 25 N to the right Note: try draw a sketch to help you understand the question -1 v i = 20 m s Activity 9 Assume original direction of the white ball is positive. Colour ball: = 0,25 kg White ball: m 2 = 0,2 kg = 0 m s 1 = 0,4 m s 1 =? = 0 m s 1 Page 14 Science Catalyst Workbook.indb /12/13 01:32:09 PM

15 1. = 0,25(0) + 0,2(0,4) = 0,25 + 0,2(0) 0,08 = 0,25 = 0,08 0,25 = 0,32 The velocity of the colour ball is 0,32 m s 1 in the original direction of the white ball. 2. Total K before = 1 2 m v i1 2 m v 2 Total K 2 i2 after = 1 2 m v f1 2 m v 2 f22 = 1 2 (0,25)(0) (0,2)(0,4)2 = 1 2 (0,25)(0,32) (0,2)(0)2 = 0 + 0,016 = 0,013 J = 0,016 J The collision is inelastic since Total K before Total K after Activity Take right as positive Before: = 100 g = 0,1 kg After: v f =? m 2 = 150 g = 0,15 kg = 3 m s 1 = 4 m s 1 = ( )v f (0,1)(3) + (0,15)( 4) = (0,1 + 0,15)v f 0,3 0,6 = 0,25v f 0,3 = 0,25v f v f = 0,3 0,25 = 1,2 = 1,2 m s 1 left 2. Total K before = 1 2 m v i1 2 m v 2 Total K 2 i2 after = 1 2 m = 1 2 (0,1)(3) (0,15)( 4)2 = 1 2 (0,25)( 1,2)2 = 0,45 + 1,2 = 0,18 J = 1,65 J the collision is inelastic. Total K before Total K after Page 15 Science Catalyst Workbook.indb /12/13 01:32:10 PM