Final Science Revision Sheet-Gr8 1. Answer the following ...

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Final Science Revision Sheet-Gr8

Unit1 lesson 3 Newton's laws of motion

Exploration 1 + exploration 2 + exploration 3 + exploration 4

Unit 1 lesson 4 collision between objects

Explorations 1+2

1. Answer the following questions :

1- A spaceship is moving in a straight line on its way to a planet. The pilot sees that the

planet is ahead and to the left. She claims that she does not need to use the engines to

push the ship toward the planet.

Explain why the pilot is correct or incorrect.

NGSS Constructed Response Answer – 1 Point

The answer Sample answer:

The pilot is correct because the planet will exert a gravitational force on the

spaceship, pulling it toward the planet on its own.

2- A 1 kg object accelerates at a rate of 5 m/s2.

Calculate the net force on the object and explain how its motion is changing.

NGSS Constructed Response Answer – 1 Point

The answer Sample answer:

Newton's second law says F = ma.

Therefore the net force is F = (1 kg)(5 m/s2) = 5 N.

The object is accelerating, which means its speed is increasing.

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3- State the Newton's first law of motion :

4- What are the two situations where we can find the total amount of forces =

0N

5- State the Newton's second law of motion

6- What are the factors that Newton's second law of motion depends on and

how they related to each other .

7- State the Newton's third law of motion

If the object at rest it will remain at rest , and if the object in motion it will remain

in motion (in a constant velocity) unless, unbalance force act on it.

1- When the object at rest

2- When the object moves in a constant velocity

The acceleration of an object is related to the amount of force and the mass of the

object

F= m.a

that’s mean the acceleration directly proportional to the amount of force applied

and inversely proportional to the amount of the mass

For each action there is an equal and opposite reaction

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8- Identify two examples of force pairs (action and reaction)

Solve the following problems :

1- Find the mass of the object in gram if you know that this object

moves for 50 meters in 15 seconds and there was an applied force

act on it = 55N.

2- A car forward velocity increases from 4.0m/s to 36m/s over a 4.0s time

interval. What is its average acceleration?

3- An object starts from rest with an acceleration of 5 m/s2 along a straight- line.

Find the speed and distance traveled after 6 s.

4- An object has a mass of 2000 grams accelerate by 20 m/s2 , find the amount of

force applied.

1- When the swimmer push the wall in order to swim

he applied a force in the direction of the wall and the wall applied a force in

the direction of the swimmer

2- When the boy hit the ball by the bat , the ball applied force in the bat and

the batt applied force in the ball in opposite direction.

First we have to find the acceleration , in order to find the acceleration we

need to find the velocity

V= X/t = 50/15= 3.33 m/s

Now we have to use the velocity in order to find out the acceleration

A= v/t = 3.33 / 15 = 0.22 m/s2

Now we can use the new data in order to calculate the mass

M = f/a = 55/0.22 = 250 kg

No we have to convert the answer to grams

250 X 1000 = 250000 grams

The acceleration is the change in velocity over time , so a= v2-v1/t

A= 36-4/4 = 32/4 = 8 m/s2

The acceleration is the change in velocity over time ,

Then we have to calculate the velocity

v= a*t = 5*6 = 30 m/s as same as speed , now we can calculate the distance

Since the s= D/t then the D= S*t = 30*6 = 180 m

Its an immediate application for newton’s second law , but you have to

convert the mass in to kg

M= 2000g = 2000/1000 = 2 kg

Then f= m*a = 2*20 = 40 N

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Also you can find a lot of practices by clicking over the following links :

https://www.liveworksheets.com/pf74239hy ,

https://www.liveworksheets.com/as10699rt

Complete the following sentences:

a- __Motion_______ is a change in objects position over time.

b- ____reference point___ any point could use to describe object position or

mevement.

c- _________position_____ the exact location of an object according to reference

point .

d- ____reference frame __________ a place that the motion happen inside it.

e- ____speed__________ change of distance over time.

f- ______velocity_ is a speed with direction.

g- ______distance________ is how far objects move.

Answer the following questions : ( collision between objects)

1- Two bumper cars move toward each other and collide. In what situation would they

have the same acceleration during the collision?

NGSS Constructed Response Answer

Evidence

of

Mastery

Sample answer:

According to Newton’s third law, objects exert equal, opposing forces on each

other when they collide. If the two bumper cars have the same mass, they would

have the same acceleration because, according to Newton’s second law, “F = ma

and a = F/m.”

2- Ben has two carts with masses of 100 g and 200 g. He pushes the smaller cart toward the

larger cart and they collide. The smaller cart rebounds back in the direction it came from.

He measures the force on the smaller cart from the larger cart to be 5 N.

What was the magnitude of the acceleration of the smaller cart?

Circle the letter of the correct answer.

A. 20 m/s2

B. 50 m/s2

C. 200 m/s2

D. 500 m/s2

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What force does the smaller cart exert on the larger cart?

Circle the letter of the correct answer.

A. 2.50 N

B. 5 N

C. 10 N

D. 15 N

A golfer has a choice of several golf clubs with different masses that he can use to

hit a golf ball. He can swing all the clubs at the same speed. If he wants to hit the

golf ball as far as possible, which club should he choose?

A. the club with the greatest mass because it will have a greater acceleration

B. the club with the smallest mass because it will have a greater acceleration

C. the club with the greatest mass because it will provide the largest force to the golf ball

D. the club with the smallest mass because it will provide the largest force to the golf ball

Remember in collision we have 2 types of collision

1- Same mass collision :

the objects will accelerate in the same magnitude but in opposite direction until their velocity

decreased.

2- Different mass collision :

the object which has less mass will accelerate more according to newton’s second law , the

larger mass object may accelerate but not like the smaller mass object.

3- collision of one mass hit two other masses stick together:

The third ball will move away because of energy transfer

When you analyse the collision you have to consider the following :

1- Before the collision

all the object experience a total forces = 0N according to newton’s first law

of motion even if the object at rest or the object moves in a constant

velocity.

2- During the collision

you have to recall the newton second law of motion in order to describe the

amount of force and the acceleration

3- After the collision

you have to recall newton’s third law of motion in order to describe the

action and reaction forces pairs and you have to recall Newton's first law

(the inertia law) in order to describe why the objects moves forward or

backward.

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Look at the graph below and answer the following question :

This graph represent a motion of 3 objects A, B and C

Q1 :

Which object moves in a constant velocity ?

Object B

Q2:

Which object increase its acceleration?

Object c

Q 3:

Which object is the fastest ?

Object A

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Complete the following table:

Phenomenon Stability and

change

Cause and effect Real life example

Acceleration Acceleration

represents a

change in velocity.

Unbalanced force

acting on an

object.

Cars increasing in

speed are

accelerating.

Collision When an object of

a very small mass

collides with an

object of a great

mass, the smaller

object will

experience a

larger change in

motion than the

larger object.

Collisions result in

equal forces acting

in opposite

directions

A marble hitting

another marble, a

hammer striking a

nail, two

pendulums

striking one

another

Solve the following problem:

Christian is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s2 for 5 s

to reach a speed of 2.2 m/s. The total mass of Christian and his bicycle is 54 kg. Later, he

straps some cargo onto the back of his bicycle. The mass of the cargo is 12 kg.

Calculate the force that Christian can exert on his bicycle before picking up the

cargo and exerts the same force on his bicycle.

to calculate the force that is applied on the system without the cargo. Force is equal to mass times

acceleration, so F = (54 kg)(0.44 m/s2) = 23.8 N. The following response, or an equivalent, is acceptable.

23.8 N

Calculate the acceleration of the bicycle when Christian adds the cargo and exerts the same force

on his bicycle.

To calculate the acceleration on the system with the cargo. Acceleration is equal to force divided by mass,

so a = (23.8 N)/(54 kg + 12 kg) = 0.36 m/s2. The following response, or an equivalent, is acceptable.

0.36 m/s2

Identify how Christian could reach the same speed on his bicycle when he is carrying the

cargo and explain your answer.

After he adds the cargo, the total mass is larger, and therefore Christian cannot reach the same

acceleration with the same force. He will need to exert a larger force to reach the same speed. OR

After he adds the cargo, the total mass is larger, and therefore Christian cannot reach the same acceleration

with the same force. Since the acceleration is smaller, he needs to exert the force for a longer amount of

time to reach the same speed.

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Newton's first law of motion.

NGSS : Plan an investigation to provide evidence that the change in an object’s motion depends on the sum of the Forces on the object and the mass of the object.

Newton’s First Law

An object in motion stay in motion , and object at rest stay at rest unless unbalance force act on the object. This simply means that balanced forces, acting on the same object, will have no effect on the motion of an object.

In practice this means:

it could be stationary, or;

it could be moving at constant speed in a straight line (constant velocity).

Example

The forces acting on this car are balanced.

Car moving forward at 20 m/s

The thrust (force applied) from the engine is equal and opposite to the drag caused by air

resistance and friction between the road and car tyres.

There is no (net) force as the forces add up to zero.

The car will continue to travel forward with a speed of 20 m/s in a straight line.

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Example

These forces balance. The car is still.

The upward force equals the downward force and they both act on the car.

The car remains at rest. It does not move upwards or downwards.

Balanced forces have no effect on an object.

If it is at rest, it remains at rest.

If it is moving at constant speed in a straight line, it continues to move at the same speed

in the same straight line.

Questions :

1- The net force acting on an object at rest is = __________

2- The net force acting on an object moves in a constant velocity = _________

Newton’s second Law

Force, mass and acceleration

Newton's second law of motion can be described by this equation:

resultant force = mass × acceleration

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This is when:

force (F) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

acceleration (α) is measured in metres per second squared (m/s2)

The equation shows that the acceleration of an object is:

proportional to the resultant force on the object

inversely proportional to the mass of the object

In other words, the acceleration of an object increases if the resultant force on it increases, and decreases if the mass of the object increases.

Example

Calculate the force needed to accelerate a 22 kg cheetah at 15 m/s2.

Questions:

1- Calculate the force needed to accelerate a 15 kg gazelle at 10 m/s2. ____________________________________________________________________________________________________________________________________________________________________________________________________________________

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Estimations

It is important to be able to estimate speeds, accelerations and forces involved in road

vehicles. The symbol ~ is used to indicate that a value or answer is an approximate one.

The table gives some examples.

Vehicle Maximum legal speed on a single carriageway

in m/s

Mass in

kg

Acceleration in

m/s

family

car ~27 ~1,600 ~3

lorry ~22 ~36,000 ~0.4

Example

Estimate the force needed to accelerate a family car to its top speed on a single

carriageway.

Using values of ~1,600 kg and ~3 m/s2, and F = m a:

1,600 × 3 = ~4,800 N

Question:

Estimate the force needed to accelerate a lorry to its top speed on a single carriageway. ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Newton's third law According to Newton's third law of motion, whenever two objects interact, they exert

equal and opposite forces on each other.

This is often worded as 'every action has an equal and opposite reaction'. However, it is

important to remember that the two forces:

act on two different objects

are of the same type (eg both contact forces)

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Examples of force pairs

Newton's third law can be applied to examples of equilibrium situations.

A cat sits on the ground

There are contact gravitational forces between Earth and the cat:

the cat pulls the Earth up

the Earth pulls the cat down

These forces are equal in size and opposite in direction.

Pushing a cart

There are contact forces between the person and the cart:

the person pushes the cart forwards

the cart pushes the person backwards

These forces are equal in size and opposite in direction.

Car tire on a road

There are contact forces between the tyre and the road:

the tyre pushes the road backwards

the road pushes the tyre forwards

These forces are equal in size and opposite in direction.

A satellite in Earth orbit

There are non-contact gravitational forces between Earth and the satellite:

the Earth pulls the satellite

the satellite pulls Earth