STAGE 2 PHYSICSSkills Assessment Task:

PROJECTILE MOTION and UNIFORM CIRCULAR MOTION

Purpose

This assessment provides you with the opportunity to demonstrate your ability to represent, analyse, and interpret investigations in physics through the use of technology and numeracy skills, communicate knowledge and understanding of the concepts and information of physics using the appropriate literacy skills of physics and demonstrate and apply knowledge and understanding of physics to a range of applications and problems relating to electricity and magnetism.

Description of assessment

In this assessment you need to: communicate your knowledge and understanding clearly and concisely use physics terms correctly present information in an organised and logical sequence include only information that is relevant to the question use clearly labelled diagrams that are related to your answer show all steps and reasoning in your answer give answers with appropriate units and direction.You may use the formula sheet provided to select appropriate formulae.

Assessment conditions

This is a supervised 70 minute closed-book assessment completed under test conditions in one double lesson. A calculator may be used.

YEAR 12 PHYSICS: SUMMATIVE TEST ONE: Name:

PROJECTILE MOTION AND UNIFORM CIRCULAR MOTION

QUESTION 1

The hammer throw is an olympic event.

The athlete throws a heavy metal ball attached by a wire to a handle.

(a) The hammer thrower swings the hammer round in a circle with a constant radius (0.75 m) before letting go.

He swings the hammer slowly at first and then faster.

(i) Identify the force causing the centripetal acceleration in the above example (1 mark) A2

(ii) State the change to the magnitude of this force as the speed increases.

(1 mark) A2

(b) Determine the period of rotation if the hammer is swung 10 complete times over a period of 6.00 seconds

(1 mark) AE1

Tension∈thewire

Increases (F=mv2

rso F ∝v2) . e.g. if the speed is doubled, the force will be quadrupled.

T= total time¿revolutions

T=6.0010

T=0.60 s (2 sig.fig)

(c) Show that the magnitude of the centripetal acceleration is equal to 82 ms -2

(3 marks) A2

(d) The magnitude of the force causing the centripetal acceleration in this example is 600 N.

(i) Determine the mass of the hammer given the magnitude of the centripetal acceleration in (c) above.

(2 marks) A2

(ii) Identify the letter corresponding to the direction of the centripetal force on the diagram below:

(1 mark) A2(iii) State the magnitude of the centripetal force if the radius is halved.

(1 mark) A2

v=2πrT

v=2π 0.750.60

v=7.9m.s−1 (2 sig.fig)

a= v2

r

a=¿¿

a=82m.s−2 (2 sig.fig)

F=ma

m= Fa

m=60082

m=7.3 kg (2 sig.fig)

B

F∝ 1r

r→ 12r ∴F→2F=2×600=1200 N

QUESTION 2

The diagram shows the apparatus used by two students to determine the changes in the magnitude of the speed as the radius of the circular path was increased.

(a) Complete the table below by determining the values of v2

Radius (m) speed (m.s-1) speed squared (m2.s-2)

0.300 2.05 4.20

0.350 2.11 4.45

0.400 2.37 5.62

0.450 2.62 6.86

0.500 2.65 7.02

0.550 2.77 7.67

0.600 3.05 9.30

(2 marks) I4

(b) State the independent variable in this experiment

(1 marks) I4

Rubber stopper (0.130 kg)

Radius

Weight (0.200 kg)

(c) Plot the data on the space provided below ensuring that you label the axis, state the units and use an appropriate scale.

(5 marks) I4

(d) State the feature of your graph that shows that the data obtained is of low precision

(1 mark) AE1

(e) Identify the type of error which lowers the precision of the experimental data

(1 mark) AE1

(f) Use the graph to determine the speed of the rubber stopper at a radius of 0.420 m

You must show a ruled interpolation on the graph to obtain the value

(3 marks) AE1

scatter of data pointsaway ¿ the lineof best fit

Random

¿ the graph :0.420mcorresponds¿a v2 value of 6.00 .

v=√6.00

v=2.45m. s−1

(g) Calculate the slope of the line of best fit. (Remember to include units).

(3 marks) AE1

(h) Determine the equation of this line, i.e. find the relationship connecting v2 and r.

(1 mark) AE1

(i) Explain why Mg=mv2

r given the diagram opposite:

(2 marks) AE1

(j) Use the slope to determine the value of g in the experiment

(2 marks) AE1

(k) Comment on the accuracy of the experimentally determined value of g given the true value of 9.81 m.s-2

(1 mark) AE1

(l) Identify the type of error which effects the accuracy of the experimental data

(1 mark) AE1

slope=y2− y1x2−x1

slope= 9.30−4.200.600−0.300

m2 s−2

m−1

slope=17m .s−2

y=(slope) x

v2=17 r

The weight of the slotted masses provides the centripetal acceleration allowing the rubber stopper to move with a

constant speed in a circular path.

Therefore, the weight (Mg) is the force providing the centripetal

acceleration (F=mv2

r)

So: Mg=mv2

r

slope = Mgm

∴g= slope x mM

g=17 x 0.1300.200

g=11m.s−2

(0.130 kg)

(0.200 kg)

Lowaccuracy

Systematic error

QUESTION 3

Construction began on building the famous Daytona Speedway in 1957. The radius of the track is 150 m.

To build high banking curves (310), crews had to use millions of tons of soil from the tracks infield.

(a) On the diagram below, draw and label the forces acting on the car, of mass m, as it travels around a banked curve of radius r without relying on friction.

You must use a ruler for your vector diagram

(4 marks) KU3

(b) Derive the expression tanθ= v2

rg given the forces acting on the car.

(3 marks) KU1

(c) Determine the maximum speed at which a car can travel around the banked curve (310) without relying on friction.

(3 marks) A2

FN y=mg

Fg=mg

FN H=mv2

r

FN

tanθ=oppadj

=FN H

FN y

tanθ=mv2

rmg

tanθ= v2

rg

tanθ= v2

rg

v=√ tanθrg

v=√ tan 31x 150 x9.81

v=30ms−1(2 s . f )

QUESTION 4

A tennis player serves a ball from a height of 2.51 m at 18.0 ms-1 in a horizontal direction. The ball just clears the net which is 1.00 m high.

In this question, assume air resistance is negligible and that g = 9.81 ms -2

(a) Show that the ball takes approximately 0.555 seconds to reach the net after being served

(3 marks) A2

(b) Determine the vertical component of velocity as it passes over the net

(2 marks) A2

(c) Calculate the magnitude and direction of the resultant velocity of the ball as it passes over the net.

Draw a fully labelled vector diagram (with a ruler) in the space provided.

s=v0 t+12a t 2

t=√ 2 sat=√ 2 x (2.51−1.00)9.81

=0.555 s

Using vertical values

v⃗=v⃗0+at

v⃗=0+9.81 x 0.555

v⃗=5.45m. s−1 down

v=√vH2+vV

2

v=√5.452+18.02

v=18.8m.s−1

tanθ=v⃗V

v⃗H=5.4518.0

θ=16.90 below thehorizontal

v⃗V

v⃗H

v⃗

θ

QUESTION 5

The shot put is an Olympic event which involves the throwing of a large heavy sphere (the shot) as far as possible.

(a) In one event, a shot put thrower launched the shot with an initial velocity of 10.5 ms -1 at an angle of 440 above the horizontal. The shot had a time of flight of 3.02 seconds

(i) Calculate the horizontal component of velocity of the shot

(2 marks) A2

(ii) Determine the maximum horizontal range of the shot

(2 marks) A2

(b) Explain why a greater horizontal range is achieved by a shot put thrower who launches the shot from a greater height above the ground.

(3 marks) KU1

v⃗H=s⃗H

t

s⃗H=v⃗H x t

s⃗H=7.55×3.02

s⃗H=22.8m

If launched from a greater initial height above the surface, the shot will spend a greater magnitude of time in the air.

This results in an increase in the maximum horizontal range since sH∝ t (given the equation sH=vH t ¿

v⃗H= v⃗ cosθ

v⃗H=10.5× cos44

v⃗H=7.55m .s−1

QUESTION 6

EXTENDED RESPONSE

Two balls are launched with the same initial velocity, at an angle of θ above the horizontal. Air resistance is a force which opposed the motion of the balls as they move through the air. If different forces of air resistance act on the balls, the motion of the balls will be different.

Write an extended response discussing the following:

Identify and explain three factors that affect the magnitude of the force of air resistance opposing the motion of each projectile.

Explain the effect that air resistance has on the maximum height and maximum horizontal range achievable by the balls.

Credit will be given for answers which are clear and expressed in a concise and coherent manner

Air resistance occurs because as a projectile moves through the air, it pushes against air molecules. As a result, the air molecules push back onto the projectile (Newton’s Third Law) and slow it down. This is why air resistance always opposes a projectile’s motion.

The magnitude of the force of air resistance is affected by the density of the air in the given volume of space where the object is moving, the velocity of the object itself and the cross sectional area of the object.

Air density is defined as the ratio of the mass of air molecules within a given volume of air. The density of air will vary based on a number of contributing factors such as air temperature and altitude. The greater the density of the air, the greater the number of air molecules which can apply an opposing force on the projectiles as they move through the air. Therefore, in regions of greater air density, a greater opposing force is applied on each projectile.

The faster the object is travelling, the more air molecules it will come into contact with during a given time interval. As a result, a projectile moving with greater velocity will experience more air resistance opposing its motion.

The cross sectional area of the projectile is the area over which air molecules can act and apply a force on that surface. A projectile with a greater cross-sectional area will be opposed by a greater magnitude of air resistance as it moves through a body of air.

Both the horizontal and vertical components of the velocity of a projectile will be affected due to air resistance throughout its flight. Since the horizontal component of velocity is drastically reduced throughout the flight, the maximum horizontal range will also be greatly reduced as sH=vH t . Note that air resistance also slightly reduces the projectile’s overall time of flight as it shortens the time for the projectile to rise but slightly lengthens its descent. This is because air resistance acts with the gravitational force and then against it as the projectile moves upwards and then downwards (see diagram).

Gravitational force

Air resistance

Gravitational force

Air resistance

The maximum height achievable by a projectile is also reduced. This is because air resistance acts against the vertical component of velocity of the projectile as it moves upwards. Hence, the projectile rapidly slows to a stop (in the vertical direction) in a much shorter time. The projectile is therefore not able to travel as high.

Air resistance occurs because as a projectile moves through the air, it pushes against air molecules. As a result, the air molecules push back onto the projectile (Newton’s Third Law) and slow it down. This is why air resistance always opposes a projectile’s motion.

The magnitude of the force of air resistance is affected by the density of the air in the given volume of space where the object is moving, the velocity of the object itself and the cross sectional area of the object.

Air density is defined as the ratio of the mass of air molecules within a given volume of air. The density of air will vary based on a number of contributing factors such as air temperature and altitude. The greater the density of the air, the greater the number of air molecules which can apply an opposing force on the projectiles as they move through the air. Therefore, in regions of greater air density, a greater opposing force is applied on each projectile.

The faster the object is travelling, the more air molecules it will come into contact with during a given time interval. As a result, a projectile moving with greater velocity will experience more air resistance opposing its motion.

The cross sectional area of the projectile is the area over which air molecules can act and apply a force on that surface. A projectile with a greater cross-sectional area will be opposed by a greater magnitude of air resistance as it moves through a body of air.

Both the horizontal and vertical components of the velocity of a projectile will be affected due to air resistance throughout its flight. Since the horizontal component of velocity is drastically reduced throughout the flight, the maximum horizontal range will also be greatly reduced as sH=vH t . Note that air resistance also slightly reduces the projectile’s overall time of flight as it shortens the time for the projectile to rise but slightly lengthens its descent. This is because air resistance acts with the gravitational force and then against it as the projectile moves upwards and then downwards (see diagram).