Exercise P2 Topic 3 and P2 Topic 4_questions

7/30/2019 Exercise P2 Topic 3 and P2 Topic 4_questions

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Surname Name

QUESTIONS & EXERCISES

P2 Topic 3 Motions & Forces

P2 Topic 4 Momentum, Energy, Work & Power

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FORMULAE

You may find theses formulae useful

charge = current time Q = I t

potential difference = current resistance V= I R

electrical power = current potential difference P = I V

energy transferred = current potential difference time E= I V t

speed =distance

time v=

s

t

acceleration =change in velocity

time taken a =

v - u

t

force = mass acceleration F= m a

weight = mass gravitational field strength W= m g

momentum = mass velocity p = m v

force =change in momentum

time F=

(mv- mu)

t

work done = force distance moved in the direction of the force E= F d

power =work done

time taken P =

E

t

gravitational potential energy = mass gravitational field strength vertical height GPE = m g h

kinetic energy =1

2 mass (velocity)

2 KE =

1

2 m v2

Do not forget to include units in all your answers.


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1. (a) Physical quantities can be divided into vectors and scalars.

(i) Describe the difference between the two types of quantities.

[2 marks]

(ii) Put a cross ( ) in the box next to the correct answer.

[1 mark]

Which row gives the correct type of the physical quantities?

vector scalar

A force displacement

B distance speed

C acceleration work

D energy momentum

(b) John cycles to the nearby lake and then back home.

Below you can see the distance-time graph of Johns motion.

(i) Use the graph to answer the following questions very briefly.

[3 marks]

1. How far away is the lake from Johns house?

2. How long did John stay at the lake?

3. What do you need to calculate to find the speed in a distance-time

graph?

0

1

2

3

4

5

6

7

0 10 20 30 40 50

Distance

from home

(km)

Time (min)


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(ii) Use the graph to calculate Johns speed (in m/s) while cycling to the lake.

Be careful with units.

[2 marks]

Speed = m/s

(iii) Johns speed is the same while cycling to and away from the lake.

Explain whether Johns velocity is the same as well.

[2 marks]

(iv) Calculate the average speed for the whole of Johns 50-min journey to the lake and

back.


[2 marks]

Speed = m/s

(c) In another journey John travels for 15 min at a constant speed of 6 m/s.

He rests for 5 min and then travels back at a speed of 5 m/s.

In the space below draw the graph of Johns journey

[5 marks]

[Total for Question 1 = 17 marks]

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0

1

2

3

4

5

6

7

0 10 20 30 40 50

Distance

from home

(km)

Time (min)


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2. (a) Join each quantity to its unit.

[3 marks]

acceleration kg m s

1

force m s1

momentum m s2

velocity N

(b) The following forces are acting on a car.

Put a cross ( ) in the box next to the correct answer.

[1 mark]

The resultant force on the car is

A 400 N to the left

B 400 N to the right

C 2000 N to the left

D 2000 N to the right

(c) Motion cameras record the velocity of a racing car at different times.

(i) Calculate the acceleration of the racing car.

[2 marks]

Acceleration =

1200 N 800 N

time = 2 sec

speed = 6 m/s

time = 6 secspeed = 30 m/s


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(ii) The racing car has a mass of 650 kg.

Calculate the resultant force on the car between t= 2 sec and t= 6 sec.

State the direction of the resultant force.

[2 marks]

Force =

Direction =

(iii) The race car then continues to move at 30 m/s in a straight line.

State the resultant force on the racing car.

[1 mark]

(iv) Calculate the momentum of the racing car when moving at 30 m/s.

[2 marks]

Momentum =

(v) The racing car takes a turn along the track while maintaining the speed of 30 m/s.

Explain why the car is still accelerating.

[2 marks]


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3. The graph below shows how the velocity of a car changes with time.

(a) (i) Put a cross ( ) in the box next to the correct answer.

[1 mark]

Which row gives the correct way of calculating acceleration and distance from the graph?

acceleration distance

A area under the graph total length of the line

B gradient of the line y-intercept

C gradient of the line area under the graph

D y-intercept total length of the line

(ii) Calculate the acceleration of the car in part C.

[2 marks]

Acceleration =

(iii) The car has a mass of 900 kg.

Calculate the resultant force on the car in part C.

[2 marks]

Resultant force =

(iv) The resistive forces on the car in part C are equal to 600 N and assumed to stay

constant.

Calculate the forward force on the car from the engine on the car in part C.

[2 marks]

Forward force =

0

2

4

6

810

12

14

0 5 10 15 20 25

Velocity

(m/s)

Time (sec)

A

B

C

D

E


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(v) Describe two differences that you can deduce from the gradient between

the acceleration of the car in part C and the acceleration of the car in part E.

[4 marks]

1.

2.

(b) Calculate the total distance covered by the car in 23 sec.

[4 marks]

Total distance =

(c) (i) Calculate the kinetic energy of the car in part D.

[2 marks]

Kinetic energy =

(ii) State the work done by the braking force in part E.

[1 mark]

(iii) Using the distance travelled by the car in E calculate the braking force.

[2 marks]

Braking force =

(iv) Explain why the braking force from the car is actually smaller.

[1 mark]


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4. The picture below shows a plane in flight.

The dot in the centre represents the Centre of Mass of the plane.

(a) The plane maintains constant speed and height.The mass of the plane is 10 000 kg.

The Thrust of the plane is equal to 75 kN.

(i) Calculate the weight of the plane.

Complete the free-body force diagram below with all the missing forces.

Write down the values of all the forces.

Careful with the arrows.

[4 marks]

(ii) Explain what must happen in terms of forces in order for the plane to

start moving downwards.

[2 marks]

(b) The Thrust increases to 100 kN.

(i) Calculate the acceleration of the plane.


[2 marks]

Acceleration =


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(ii) The plane can be loaded with weapons.

Explain in terms of forces what must happen in order to for the plane to have the same

acceleration as in (i).

[2 marks]

(iii) A short time after the thrust increases and is maintained at 100 kN the plane reaches

a higher steady speed.

Explain this fact in terms of forces involved.

[3 marks]

(c) (i) The plane is flying at a height of 2 km.

Calculate the GPE of the plane.

[2 marks]

GPE =

(ii) Give two reasons why the total work done by the planes engine is greater

than your answer in (i).

[2 marks]

1.

2.

(iii) State the original form of energy that the plane uses.

[1 mark]


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5. (a) The picture below shows a rocket taking off.

The total mass of the rocket at take of is 400 000 kg.

The upward thrust is equal to 5 000 kN.

(i) Assuming drag forces are negligible at lift off, complete the free-body force diagram.

[3 marks]

(ii) Calculate the resultant force on the rocket.

[1 mark]

Resultant force =

(iii) It takes 10 minutes for the rocket to reach the end of the atmosphere.

Calculate the change in momentum of the rocket during that time.Calculate the velocity of the rocket, assuming it starts from rest.

[4 marks]

Change in momentum =

Velocity =

(iv) In reality the rocket burns fuel as it ascends.

Explain the effect this has on the speed of the rocket.

[3 marks]


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(b) To get its thrust the rocket pushes the gases downwards.

(i) State Newtons 3rd

Law.

[2 marks]

(ii) Put a cross ( ) in the box next to the correct answer.

[1 mark]

The rocket moves upwards because

A the gases push against the atmosphere

B the gases push against the Earth

C the gases push the rocket with an equal force upwardsD the gases push the rocket upwards with a larger force

(c) After 10 min the rocket reaches a height of 100 km and a speed of 10 km/s.

Assume that g = 10 m/s2.

Take the average mass of the rocket to be 200 000 kg.

(i) Calculate the total energy of the rocket.

Careful with units.

[5 marks]

Total energy =

(ii) Calculate the average power developed by the rocket.

[2 marks]

Power =


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6. (a) In an experiment on the Moons surface an American astronaut released a hammer

and a feather at the same time.

Explain why they reached the Moons surface at the same time.

Why is such a demonstration impossible on Earth? [3 marks]

(b) The graph below shows how velocity changes with time for a parachutist.

(i) The parachutist has a weight of 1000 N.

Draw the free body diagram of the parachutist at t= 25 sec.

[2 marks]

(ii) State the acceleration of the parachutist at

[2 marks]

t= 0 sec

t= 45

sec

0

10

20

30

40

50

60

0 5 10 15 20 25 30 35 40 45 50 55

Velocity

(m/s)

Time (sec)


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(iii) At a certain instant during the fall the acceleration of the parachutist is 4 m/s2.

Show that the mass of the parachutist is 100 kg.

Calculate the drag force on the parachutist at that instant.

[4 marks]

Drag force =

* (iv) State the two terminal velocities of the parachutist.

Explain why the parachutist reaches a terminal velocity.

Explain why the second terminal velocity is lower than the first.

[6 marks]

(v) Calculate the change in momentum of the parachutist when he lands.

[2 marks]

Change in momentum =

(vi) It takes 0.8 sec for the parachutist to come to a complete stop when he lands.

Calculate the resultant force on the parachutist.

[2 marks]

Resultant force =


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7. (a) The chart below shows the thinking and stopping distances for various speeds.

(i) Tick the correct column for each of the changes.

[5 marks]

change

thinking distance braking distance stopping distance

decreases increases decreases increases decreases increases

The car is loaded

The speed is smaller

The driver is tired

The road is wet

The tyres are new

(ii) Describe the relationship between speed and thinking distance.

[1 mark]

(iii) State what happens to the braking distance when speed doubles?

Calculate the stopping distance for 80 km/h according to the chart above.

[4 marks]

Stopping distance =

6 m30 km/h

8 m/s

40 km/h11 m/s

50 km/h14 m/s

60 km/h16 m/s

6 m

12 m

8 m 11 m

10 m 17 m

24 m

Thinking

Distance

BrakingDistance


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(b) A car of mass 900 kg has kinetic energy equal to 45 000 J.

(i) Calculate the speed of the car.

[2 marks]

Speed =

(ii) State the work that the braking force needs to do in order to stop the car completely.

[1 mark]

(iii) The car needs 10 m to stop while braking.

Calculate the force developed by the brakes.

[2 marks]

Force =

(iv) What will be the kinetic energy of the car at twice the speed?

Try figuring it out without detailed calculations.

[2 marks]

Kinetic energy =

(v) If the force and mass are the same what is the new braking distance?

You can do the calculations or figure it out the easy way -.

[2 marks]

Braking distance =

(vi) Describe the relationship between braking distance and speed.

[2 marks]


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8. (a) Below you can see a car of mass 1200 kg being raised to the top of a roller coaster.

(i) Calculate the gravitational potential energy of the car at point X.[2 marks]

GPE =

(ii) Assuming no energy losses, calculate the force F.

[2 marks]

Force, F=

(iii) Using energy conservation, calculate the speed of the car at Y.

[5 marks]

Speed =

(iv) Explain whether you expect the car to pass over point Z.

[2 marks]

40 m

30 m

50 m

10 m

F

25 mY

Z

X


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(v) The actual speed at Y will be less than the one you calculated in (iii).

Explain why.

[2 marks]

(b) The two cars shown below collide.

(i) Calculate the total momentum of the two cars.

[3 marks]

Total momentum =

(ii) State the total momentum of the two cars after collision.

[1 mark]

(iii) Right after collision the two cars are locked and move together.

Calculate their common speed.

[2 marks]

Speed =

(iv) State the principle you used to find the answer.

[1 mark]

(v) Kinetic energy is not conserved in this collision.

State briefly where energy is transferred to.

[1 mark]

car A

mass = 800 kg

speed = 20 m/s

car Bmass = 2500 kgspeed = 15 m/s


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* (vi) State the formula that connects momentum and force.

Explain why crumple zones protect the driver during an accident.

Name one more safety feature that is incorporated in modern cars

[4 marks]

(c) The two balls shown below collide.

(i) Calculate the velocity (speed and direction) of the 600 g ball after collision.

[4 marks]

Speed = , Direction =

(ii) The time of collision of the two balls is 0.1 sec.

Calculate the force (magnitude and direction) acting on the 400 g ball.

[4 marks]

Force = , Direction =

(iii) State the magnitude and direction of the force on the 600 g ball.

[2 marks]

Force = , Direction =


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before collision after collision

mass = 400 g

speed = 6 m/s

mass = 600 gmotionless speed = 4 m/s v?

Documents

Exercise P2 Topic 3 and P2 Topic 4_questions