Grade 11 - Newton's Laws - Sample Workbookbubblegum.co.za/images/sc_wb_sample.pdf · Newton’s Laws of Motion ... Newton’s Second Law says that the rate of change of momentum of

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L e s s o nL e s s o nL e s s o n

11

Momentum and Newton’s LawsNewton’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

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.

2. A donkey pulling a cart: the donkey pulls the cart forward and the cart pulls backward against the donkey with the same force.

Fof donkey on cart

Fof cart on donkey

EXAMPLEe = mc²Fof table on book

Fof book on table

Science Catalyst Workbook.indb 1Science Catalyst Workbook.indb 1 2010/12/13 01:32:02 PM2010/12/13 01:32:02 PM

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3. Walking: the foot pushes against the earth and the earth pushes against the foot with the same force but in the opposite direction.

Fof foot on ground

Fof ground on foot

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.

Fof man on seat belt

Fof seat belt on man

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


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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 = pfinal − pinitial

= mvf − mv

i

= m(vf − v

i)

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:

Fnet = Δp

_ Δt

Remember that: Δp = m(vf − v

i)

= mΔv

So the equation becomes:

Fnet = mΔv ____ Δt but a = Δv __ Δt

Therefore:

Fnet = ma

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 mo-mentum 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.

Total pbefore = Total pafter

m1vi1 + m2vi2 = m1vf1 + m2vf2

m1 = mass of object 1 m2 = mass of object 2

Δp = change in momentum. Measured in kg·m·s–1


vf = final velocity. Measured in m·s–1

vi = initial velocity. Measured in m·s–1

Fnet = net force. Measured in newtons (N)


Δt = change in time. Measured in seconds (s)

Fnet = net force. Measured in newtons (N)

m = mass of object. Measured in kilograms (kg)

a = acceleration of the object. Measured in m·s–2


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vi1 = initial velocity of object 1 vi2 = initial velocity of object 2

vf1 = final velocity of object 1 after collision v

f2 = 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.


m1vi1 + m2vi2 = (m1 + m2)vf

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.


(m1 + m2)vi = m1vf1 + m2vf2

The initial velocity of the objects before the collision is the same and the masses are added together.

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.

EXAMPLEe =

mc²

Before

After

1920kg3300kg

v = 20m•si1-1

v = 12m•si1-1

v = 0m•si2-1

v = ?i2


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Solution

Original direction of the truck is positive

Car: m1 = 1920 kg Truck: m2 = 3300 kg

vi1 = 0 m·s–1 v

i2 = 20 m·s–1

vf1 = unknown v

f2 = 12 m·s–1



1 920(0) + 3 300(20) = 1 920vf1 + 3 300(12)

66 000 = 1 920vf1 + 39 600

66 000 − 39 600 = 1 920vf1

∴ vf1 = 26 400 _____ 1 920

= 13,75 m·s–1 in the truck’s original direction

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.

Solution

Original direction of Jo’s motion is positive

Jo: m1 = 60 kg Skateboard: m2 = 5 kg

vi1 = 3 m·s–1 v

i2 = 2 m·s–1

vf1 = unknown v

f2 = unknown

vf1 = v

f2 = v


m1vi1 + m2vi2 = (m1 + m2)v

60(3) + 5(2) = (60 + 5)v

180 + 10 = 65v

∴ v = 190 ___ 65

= 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.

SOLUTIONNote: Always specify direction

Make sure you substitute in all values, even when v = 0

EXAMPLEe = mc²

Before After

m1 = 60 kgvi1 = 3 m·s–1

m1 = 5 kgvi1 = 2 m·s–1

SOLUTION

ACTIVITY

v = 16m•si1-1 v = 22m•si2

-1

m = 900kg1 m = 1300kg2

v = ?f

Before

After


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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).

ACTIVITY

ACTIVITY

ACTIVITY


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Activity 5

A train of mass 2 × 103 kg is moving west at 25 m·s–1. Another train with a mass of 6,5 × 103 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.

FnetΔt = Δp

= mvf − mv

i

= m(vf − v

i)

ACTIVITY

ACTIVITY

FnetΔt = impulse. Measured in N·s



v

f = final velocity. Measured in m·s–1

vi = initial velocity. Measured in m·s–1


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Example

A car company is testing its new model in a crash test. The vehicle mass is 1 300 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

1. Take towards the wall as positive

m = 1300 kg FnetΔt = Δp

vi = 15 m·s–1 = mv

f − mv

i

vf = −2,5 m·s–1 = (1300)(-2,5) – (1300)(15)

= −3250 − 19500

= −22750

= 22750 N.s

The impulse is 22750N.s away from the wall.

2. FnetΔt = 22750 N.s

Δt = 0,1 s

FnetΔt = Δp

Fnet(0,1) = 22750

∴Fnet = 227500 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.

EXAMPLEe =

mc² vi = 15 m·s–1

SOLUTION

ACTIVITY


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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 Kbefore = Total Kafter

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 Kbefore ≠ Total Kafter

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.

ACTIVITY

EXAMPLEe =

mc²

Before After

m1 = 60 kgvi1 = 3 m·s–1

m1 = 5 kgvi1 = 2 m·s–1


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Solution

Before: m1 = 60 kg m2 = 5 kg After: m = 65 kg

vi1 = 3 m·s–1 v

f = 2,92 m·s–1

vi2 = 2 m·s–1

Total Kbefore = 1 __ 2 m1vi12 + 1 __ 2 m2vi2

2 Total Kafter = 1 __ 2 mvf 2

= 1 __ 2 (60)(3)2 + 1 __ 2 (5)(2)2 = 1 __ 2 (65)(2,92)2

= 270 + 10 = 277,11 J

= 280 J

∴ The collision is inelastic as Total Kbefore ≠ Total Kafter

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.

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 = ?

Before After

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.

SOLUTION

ACTIVITY

ACTIVITY


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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: m1 = 900 kg Car 2: m2 = 1300 kg

vi1 = 16 m·s–1 v

i2 = −22 m·s–1

v = ? v = ?


m1vi1 + m2vi2 = (m1 + m2)v

900(16) + 1300(−22) = (900 + 1300)v

14400 + (−28600) = 2200v

−14200 = 2200v

∴v = − 14200 _____ 2200

= −6,45 m·s–1

The vehicles move at 6,45 m·s–1 to the left.

Activity 2

Assume to the right is positive

Janice: m1 = 60 kg Portia: m2 = 45 kg

vi = 0 m·s–1 v

i = 0 m·s–1

vf1 = 1,2 m·s–1 v

f2 = ?


(m1 + m2)vi = m1vf1 + m2vf2

(60 + 45)(0) = 60(1,2) + (45)vf2

0 = 72 + 45vf2

∴vf2 = − 72 __ 45

= −1,6

Portia moves with a velocity of 1,6 m·s–1 left.


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Activity 3

Assume movement away from land is positive.

Boy: m1 = 65 kg Boat: m2 = 180 kg

vi1 = 12 m·s–1 v

i2 = 0 m·s–1

vf1 = ? v

f2 = ?

vf1 = v

f2 = v


m1vi1 + m2vi2 = (m1 + m2)v

65(12) + 180(0) = (65 + 180)v

780 = 245v

∴v = 780 ___ 245

= 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: m1 = 4 kg Bullet: m2 = 0,35 kg

vi1 = 0 m·s–1 v

i2 = 0 m·s–1

vf1 = ? v

f2 = 200 m·s–1


(m1 + m2)vi = m1vf1 + m2vf2

(4 + 0,35)(0) = 4vf1 + 0,35(200)

0 = 4vf1 + 70

−70 = 4vf1

∴vf1 = − 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 west

Before

After

6,5 x 10 kg3 2 x 10 kg3

28m•s-1 25m•s-1

Assume west is positive.

Hint: Remember that the gun and the bullet are one object before the explosion which causes the bullet to leave the gun.


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Train A: m1 = 2 x 103 kg Train B: m2 = 6,5 x 103 kg

vi1 = 25 m·s–1 v

i2 = −28 m·s–1

vf1 = v v

f2 = v


m1vi1 + m2vi2 = (m1 + m2)v

(2 x 103)(25) + (6,5 x 103)(−28) = (2 x 103 + 6,5 x 103)v

50 000 − 182 000 = 8,5 x 103 v

−132 000 = 8,5 x 103 v

∴v = − 132 000 _______ 8,5 x 103

= −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 vi = 0 m·s–1

vf = ? Δy = 40 m

vf 2 = v

i2 + 2aΔy

vf 2 = 0 + 2(9,8)(40)

vf 2 = 784

vf = √

____ 784

= 28

The velocity of Block A just before the collision is 28 m·s–1 down.

2. Block A: m1 = 8 kg Block B: m2 = 15 kg

vi1 = 28 m·s–1 v

i2 = 0 m·s–1

vf1 = v v

f2 = v

v1 = v2 = v


m1vi1 + m2vi2 = (m1 + m2)v

8(28) + 15(0) = (8 + 15)v

224 = 23v

∴v = 224 ___ 23

= 9,74

The blocks fall with a velocity of 9,74 m·s–1 down

Activity 7

racquet racquet

30m•s400g

-1

–40m•s-1–32m•s-1

3,6kgra

30m•s400g

-1

–40m•s-1

3,6kg–32m•s-

400g?m•s-1


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Assume movement toward you is positive.

Ball: m1 = 400 g = 0,4 kg Racquet: m2 = 3, 6 kg

vi1 = 30 m·s–1 v

i2= −40 m·s–1

vf1 = ? v

f2= −32m·s–1

1. Total pbefore = Total pafter


0,4(30) + 3,6(−40) = 0,4(vf1) + 3,6(−32)

12 − 144 = 0,4vf1 − 115,2

−132 = 0,4vf1 − 115,2

−132 + 115,2 = 0,4vf1

−16,8 = 0,4vf1

∴vf1 = − 16,8

____ 0,4

= −42

The ball is travelling at 42m·s–1 away from the racquet

2. FnetΔt = Δp

= m1(vf1 − v

i1)

= (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 vi = 0 m·s–1

vf = 20 m·s–1

FnetΔt = Δp

FnetΔt = m(vf − v

i)

Fnet(0,4) = (0,5)(20 − 0)

Fnet(0,4) = 10

Fnet = 10 ___ 0,4

= 25 N to the right

Activity 9

Assume original direction of the white ball is positive.

Colour ball: m1 = 0,25 kg White ball: m2 = 0,2 kg

vi1 = 0 m·s–1 v

i2 = 0,4 m·s–1

vf1 = ? v

f2 = 0 m·s–1

v = 20 m•si-1

Note: try draw a sketch to help you understand the question


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1. Total pbefore = Total pafter


0,25(0) + 0,2(0,4) = 0,25vf1 + 0,2(0)

0,08 = 0,25vf1

∴vf1 = 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 Kbefore = 1 __ 2 m1vi12 + 1 __ 2 m2vi2

2 Total Kafter = 1 __ 2 m1vf12 + 1 __ 2 m2vf2

2

= 1 __ 2 (0,25)(0)2 + 1 __ 2 (0,2)(0,4)2 = 1 __ 2 (0,25)(0,32)2 + 1 __ 2 (0,2)(0)2

= 0 + 0,016 = 0,013 J

= 0,016 J

The collision is inelastic since Total Kbefore ≠ Total Kafter

Activity 10

1. Take right as positive

Before: m1 = 100 g = 0,1 kg After: vf = ?

m2 = 150 g = 0,15 kg

vi1 = 3 m·s–1

vi2 = − 4 m·s–1


m1vi1 + m2vi2 = (m1 + m2)vf

(0,1)(3) + (0,15)(−4) = (0,1 + 0,15)vf

0,3 − 0,6 = 0,25vf

−0,3 = 0,25vf

∴vf = −0,3

____ 0,25

= −1,2

= 1,2 m·s–1 left

2. Total Kbefore = 1 __ 2 m1vi12 + 1 __ 2 m2vi2

2 Total Kafter = 1 __ 2 mvf2

= 1 __ 2 (0,1)(3)2 + 1 __ 2 (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 Kbefore ≠ Total Kafter


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Grade 11 - Newton's Laws - Sample Workbookbubblegum.co.za/images/sc_wb_sample.pdf · Newton’s Laws of Motion ... Newton’s Second Law says that the rate of change of momentum of