Answers to Grade 12U Physics Key Questions Unit 1

Answers to Grade 12U Physics Key Questions Unit 1: Lesson 1 Answers: 1. A ladybug with a velocity of 10.0mm / s[W ] crawls on a chair that is being pulled [W 50  N ] at 40.0mm / s. What is the velocity of the ladybug relative to the ground? (9 marks) Solution: First we must identify what we are given.   vbc  10.0mm / s[W ], v cg  40.0mm / s[W 50  N ],   50 

where the subscript bc represents the velocity of the bug relative to the chair and cg  represents the velocity of the chair relative to the ground. Now we’ll create the value vbg which represents the velocity of the ladybug relative to the ground.  v bg  ?  vbg 

  (v xnet ) 2  (v ynet ) 2



Here, v xnet represents the sum of all the horizontal component vectors in the problem  and v ynet represents the sum of all the vertical component vectors.  v xnet  ?    v xnet  v xchair  v xbug

 v ynet  ?

   v ynet  v ychair  v ybug

 v xbug  10.0mm / s[W ]

 v ybug  0.00mm / s[ N ]

 v xchair  ?

 v ychair  ?

  v xchair  vcg cos 

  v ychair  v cg sin 

 v xchair  40.0mm / s[W 50  N ]  cos 50 

 v ychair  40.0mm / s[W 50  N ]  sin 50 

 v xchair  25.7 mm / s[W ]  v xnet  25.7 mm / s[W ]  10.0mm / s[W ]

 v xnet  35.7 mm / s[W ]

 v ychair  30.6mm / s[ N ]



 v ynet  30.6mm / s[ N ]

 vbg 

(35.7 mm / s[W ]) 2  (30.6mm / s[ N ]) 2

 v bg  47.0mm / s[ NW ]

 ?  v ynet   tan (  ) v xnet 1

  tan 1 (

30.6mm / s[ N ] ) 35.7 mm / s[W ]

  41  vbg  47 mm / s[W 41 N ]

Therefore, the velocity of the ladybug relative to the ground is

47 mm / s[W 41 N ].

2. An airplane is flying to a city due west from its current location. If there is a slight wind blowing to the southwest, in what direction must the plane head (that is, in what direction must it point)? Explain your answer using a diagram. (3 marks) Solution:

As demonstrated by the diagram, the initial vector of the plane represents the airspeed of the plane (its speed relative to the air). When the wind vector is added with a speed relative to the ground, the resultant vector represents the plane’s bearing relative to the ground. The airspeed and airplane’s resultant bearing are both pointing west, but one is a vector relative to the air, the other is relative to the ground. 3. Do research to find out how relative velocity is related to the direction in which rockets are launched to send them into space. Explain the benefits (to society and/or the environment) of using a specific direction for launching rockets. (6 marks: 3 marks for explanation of relative velocity and 3 marks for explanation of benefits) Solution: Rockets are launched at a velocity relative to the rotating Earth below them. The Earth rotates on its axis with a west to east direction at a maximum velocity of approximately 1000 mi/h (at the equator, with respect to the distant stars). When space rockets are launched, they are launched pointing in an eastward direction so that they can add the rotation of the earth to their ground velocity (i.e., they gain the extra 1000 mi/h of speed from the earth’s rotation). NASA launches its rockets from Cape Canaveral in Florida because its proximity to the equator allows rockets to exploit this as much as possible within the CONUS (Continental United States). This means the rockets are permitted to consume less fuel to achieve the necessary velocity, and so emit less greenhouse warming CO2 (as carbon dioxide is a product of the combustion of rocket fuel). It also means that less fuel needs to be produced for the rocket, reducing the mission cost, and allowing for the fuel to be diverted to other sectors of a society. It must be noted that rockets are only initially launched vertically relative to the local ground before rotating to find an Eastward direction, eventually flying parallel to the direction of Earth’s rotation (Adcock and Davison, 2011). The diagram below will help clarify this phenomenon (Cape Canaveral, Florida is where all NASA rockets and the space-shuttle are launched from):

Therefore, it can be concluded that the net or final velocity of the rocket is equal to the sum of its velocity relative to an unmoving object and the rotational velocity of the Earth.    In equation form, this can be stated as v net  v rocket  v Earth . References Adcock, G. and Davison, RC. (2011, October 31). Rocket Science-The Very Basics. Retrieved March 1, 2016, from Bright Hub Website: www.brighthub.com/science/space/articles/29607.aspx

Lesson 2 Answers: 4. Using the data you have just collected from the projectile experiment, complete the Analysis questions and Conclusion statement for the projectile motion investigation. Analysis: a) Construct a labelled drawing of your investigation, indicating the object projected, the height from which it was projected, and its horizontal range. (4 marks) 



b) Using the known vertical height ( d y ) and the average horizontal range ( d x ), perform the necessary calculations to determine the initial velocity of your object. (3 marks) c) State two possible sources of error (reviewed here, or others that you can think of) and how you worked to reduce the possibility or effects of these errors. (2 marks)

d) Describe any safety precautions taken when carrying out this investigation. (1 mark) Conclusion: Write a conclusion by filling in the blanks in the following statement. Note that the statement is concise and is directly related to the objective of the investigation. When projecting a ___________________ from a height of ____________________, the average horizontal range was measured to be _____________________ and the magnitude of the initial velocity was calculated to be ____________________. (2 marks for complete conclusion) Solution: Analysis: From the experiment it was observed that when the paint bottle hit the ground, it would bounce back up before hitting the ground for a final time. Consequently, the paint bottle was covered in red paint to act as a marker so that when the projectile hit the ground for the first time it would leave a paint mark to indicate how far from the table it landed.

a)

 d y  0.825m[ down ]

b)

 d xave  1.56m[ right ]

 g  9.8m / s 2 [down]

 v1  ?

Vertical

Horizontal  d xave  1.56m[ right ]

 d y  0.825m[ down ]

 a  0.00m / s 2

 g  9.8m / s 2 [down]  v1 y  0.00m / s

t  ?



 v1 x  ?

As v1 y  0.00m / s we can use the shortened formula below to solve for t.  1  d y  gt 2 2

t 

 2 d y  g

2(0.825m[down]) 9.8m / s 2 [down]

t 

t  0.410 s  v1 x  ?

 d xave  v1 x  t

 1.56m[right ] v1 x  0.410s  v1 x  3.81m / s[ right ]



 v1  3.81m / s[right ]

Therefore, the initial velocity of the paint bottle is 3.81m / s[ right ]. c) Below are the horizontal range measurements obtained from the experiments:  d 1 x  d 2 x  d 3 x  d 4 x  d 5 x

 1.275m[ right ]  1.2725m[ right ]

 1.975m[ right ]  1.65625m[ right ]

 1.63m[ right ]

The large deviation in the above values is the result of a number of possible sources of error. One possible source was that the torsion rope in the catapult did not give off the same amount of force for each throw. It is possible that each time the arm of the catapult was brought down and released, a strain was placed on the rope that altered its ability to apply a certain amount of force. As a result, the force applied to the bottle on each throw varied, resulting in varying degrees of acceleration, and therefore, velocity and range. Unfortunately, there was no real way to limit this source of error. A second source of error was the distance measurements may not have been 100% accurate. When the bottle hit the ground, the red paint would have splattered an area of about 1.00cm in diameter, and the bottle could have hit the ground anywhere in between that 1.00cm, potentially shifting the range value over by a small margin. Therefore, the horizontal range values may only have been accurate up to a certain degree of accuracy (90% for example). This would not have created a large deviation, but is still considered a source of error. This

could be made more accurate by having a slow motion camera record the missile’s flight and record exactly where it landed at the instant of impact with the ground. d) No real safety precautions needed to be taken with this experiment because there was nobody else in the room during the experiment or any delicate or dangerous devices in the area. Conclusion: When projecting a paint bottle from a height of 0.825m, the average horizontal range was measured to be 1.56m[right] and the magnitude of the initial velocity was calculated to be 3.81m/s. 5. A projectile is launched so that its point of launch is lower than its landing point. (5 marks: 1 mark each) a) b) c) d) e)

When is the vertical velocity at a maximum? When is the horizontal velocity at a maximum? When is the vertical velocity at a minimum? What is the acceleration of the object? Which will take longer: the upward motion or the downward motion?

Solutions: a) The vertical velocity is greatest at the point of launch (as its upward velocity will drop trying to counteract gravity). This is because the launch point is lower than the landing point. b) The horizontal velocity is greatest at the projectile’s launch point because the projectile hasn’t experienced any amount of drag at the point of launch to slow it down. c) The projectile reaches its minimum vertical velocity at its apogee (the point of maximum height where the projectile will begin to fall down). d) The acceleration of the projectile is the acceleration due to gravity ( 9.8m / s 2 [downward ]).

e) The upward motion will take longer because it has more distance to travel upward than downward (due to the greater height of the landing point than the launch point). 6. A child sitting in a tree throws his apple core from where he perched (4.0 m high) with a velocity of 5.0m / s[35  above the horizontal], and it hits the ground right next to his friend. a) How long does it take for the apple core to hit the ground? (3 marks) b) How far from the base of the tree will the apple core land? (3 marks) c) What is the velocity of the apple core on impact? (4 marks)

Solutions: 



a) v1  5.0m / s g  9.8m / s 2 [down]

 d  4.0 m[ down]

  35 

t  ?   1  d  v1 y t  gt 2 2  v1 y  ?   v1 y  v1 sin   v1 y  (5.0m / s )(sin 35  )

 v1 y  2.87 m / s[upward ]

1 (9.8m / s 2 [ down])t 2 2 ( 4.9m / s 2 [down ])t 2  ( 2.87m / s[ down]) t  4.0m[ down]  0 4.0m[down ]  (2.87 m / s[upward ])t 

Now we’ll use the quadratic formula. b  2.87 c  4.0

a  4 .9

 ( 2.87) 

x

x

x

2.87 

2.87 

x

x

( 2.87) 2  4(4.9)(4.0) 2(4.9)

8.3769  78.4 9 .8

86.7769 9.8

2.87  9.32 9.8

2.87  9.32 9.8

x

2.87  9.32 9.8

x  1.2 x  0.66 A negative value for time would not be sensible in this calculation. Therefore,  t  1 .2 s

It takes 1.2 seconds for the apple core to hit the ground.

 d x  ?

b)

 d x  v1x  t   d x  v1 x t

 v1 x  ?

  v1x  v1 cos   v1x  (5.0m / s )(cos 35  )

 v1 x  4.1m / s[ right ]  d x  ( 4.1m / s[ right ])(1.2 s )  d x  4.9m[ right ]

The apple core will land 4.9 m to the right from where it was launched.  v2  ?

c) Vertical  g  9.8m / s 2 [ down ]  d y  4.0m[ down ]

 v1 y  2.87 m / s[ down ]

Horizontal t  1.2 s

 a x  0.0m / s 2 [right ]  d x  2.7 m[ right ]

 v1x  4.9m / s[ right ]

 v2 y  ?

 v2 x  4.9m / s[ right ]

The horizontal velocity does not change because there is no force to accelerate the projectile to the right (except possibly, drag which is absent in this case).

    v 2 2 y  v 21 y  2 gd y

 v 2 2 y  (2.87m / s[down]) 2  2(9.8m / s 2 [ down])(4.0m[ down])  v 2 2 y  8.4369m 2 / s 2 [down]  78.4m 2 / s 2 [down]

 v 2 2 y  86.8369m 2 / s 2 [down]  v2 y 

86.8369m 2 / s 2 [ down]

 v2 y  9.3m / s[ down ]

 v2  ?

   v 22  v 22x  v 22y

 v 2 2  (4.9m / s[right ]) 2  (9.3m / s[down]) 2  v 2 2  24.01m 2 / s 2 [right ]  86.49m 2 / s 2 [ down]  v 2 2  110.5m 2 / s 2 [downright ]  v2  110 .5m 2 / s 2 [ downright ]

 v2  11m / s

 ?  v2 y   tan (  ) v2 x 1

  tan 1 (

9.3m / s[down] ) 4.3m / s[right ]

  65 

Therefore, the velocity of the apple core upon hitting the ground is 11m / s[65 below the horizontal]. 7. Describe three ways that understanding projectile motion and relative velocity could help you improve your success in a basketball game. (6 marks) Solution: Firstly, understanding projectile angles can help a basketball player determine where he or she will toss from. If I intend to pass a ball through the hoop from a substantial distance I should take into consideration the force of gravity will cause the ball to drop with time. I should therefore, increase the angle at which the ball is thrown to compensate for the drop of the ball; the angle would vary depending on the distance from the hoop.

Secondly, I would also need to consider the ball’s speed. A faster throw would need to be done at a greater distance from the hoop as the position the ball lands in is determined by the speed of the ball (distance is proportional to speed). If I shoot right below a net, I would have to throw at a lower vertical speed to avoid overshooting the hoop due to the greater distance a faster ball would travel in a given amount of time. Thirdly, I would need to consider the velocity of the ball relative to my velocity. If the ball is thrown while I am moving forward, the net, or final velocity of the ball will be increased due to the extra speed it gains from my forward velocity. The increased velocity would cause the ball to travel a greater distance and the ball could possibly overshoot the hoop. Consequently, I would want to throw the ball with reduced initial velocity for it so my velocity and the ball’s will add to have the ball travel at the right speed to pass through the hoop.

Lesson 3 Answers: 8. If an object is in motion, does that mean that the object has a net force in the direction of that motion? Explain. (2 marks) Solution: An object in motion does not need to have the net force acting in the same direction because a net force could be acting opposite to the direction of an object’s motion in slowing the object down until the object comes to a stop. 9. At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half full, while the other box is new. The boxes, including the nails, weigh 10kg and 20kg, respectively, and are the same size. a) As the plank tilts towards the heavier box, predict which box of nails will start to slide first. Explain your prediction. (1 mark) b) If the coefficient of static friction is 0.4, draw an FBD for each box of nails and use it to calculate the angle at which each box begins to slide. (6 marks0 c) If the coefficient of kinetic friction is 0.3, how fast will the boxes accelerate along the plank, once they start to slide? (5 marks) Solution:

a) Both boxes will slide at the same time because being the same size; they both have the same centre of gravity (the masses have no effect on which object will slide first). b)

Diagram for box 1: The y-axis is defied as the normal to the plane, while the x-axis is parallel to the plane, as shown in the diagrams below:

Diagram for box 2:

Because both boxes are the same size, they will start to slide at the same time, and thus we can use one equation to solve for the slide angle of both boxes. For the following calculations we can assume that the net force acting on the boxes is 0.00N. We’ll also only use the mass of box 1 for the calculation. In the y-direction: m1  10kg

 g  9.8m / s 2

 Fnet  0.00 N

 Fg  ?   Fg  m1 g

 Fg  10kg  9.8m / s 2 [down]  Fg  98 N [down]    Fnety  FN  Fgy

 0  FN  98 N [ down]

 FN  98 N [up ]

In the x-direction:    Fnetx  Fgx  F f    Fnetx  Fg sin    s FN   FN   Fg cos 

   Fnetx  Fg sin    s (  Fg cos  )

0  98 N [down] sin   (0.4)( 98 N [ down] cos  ) 0  98 N [ down] sin   39.2 N cos  98 N [ down] sin   39.2 N cos 

sin  

39.2 N [ down] cos  98 N [ down ]

sin   0.40 cos  sin   0.40 cos 

tan   0.40

  tan 1 0.40   22 

Therefore, the two boxes will start to slide at an angle of 22  to the horizontal. c)

 k  0.3

 a ?    Facceleration  Fg sin   F f

(Force of kinetic friction)

  Facceleration  ma    ma  Fg sin    k FN cos 

   Fg sin    k FN cos  a m

The acceleration of box 1 will be given by the equation:   Fg sin    k FN cos   a1  m1

and the acceleration of box 2 will be given by the equation:   Fg sin    k FN cos   a2  m2  98 N [ down] sin 22   (0.3)( 98 N [down]) cos 22  a1  10 kg

 a1  6m / s 2

First we must calculate the value of

 Fg

for box 2.

 Fg  ( 20kg )(9.8m / s 2 [down])  Fg  196 N [ down]



 FN  196 N [ down ]

 196 N [ down ] sin 22   (0.3)(196 N [ down]) cos 22  a2  20kg

 a2  6 m / s 2

Notice that both boxes accelerate at the same rate. Therefore, the boxes will accelerate at a rate of 6m / s 2 down the slope. 10. Design a simple experiment that you could carry out in your home to i) determine the coefficient of static friction between an object and a surface. ii) prove that the coefficient of static friction is dependent only on the surfaces in contact, and is not affected by any change in the mass of your object. (Hint: You might want to look back at the last couple of Support Questions in this lesson, to help you.) a) Describe your plan. It must include a list of materials, a diagram of the set-up, and an explanation of the steps you would take and the data you would collect. (5 marks) b) Explain how you would analyze the collected data to determine the coefficient of static friction and prove that it is unaffected by any change in the mass of your object. (2 marks) c) State one possible source of error that you might encounter in this experiment and state the steps you took to minimize or eliminate this source of error. (1 mark) Solution: a) The plan is to prove that a wooden block of greater mass will have the same static coefficient of friction as a wooden block of smaller mass based on similarities in the angle at which they begin to slide down a wooden plank. Below is a list of the materials used in the experiment: - a 32.2cm long plank of wood 4.8cm wide. - a small block of wood - a larger block of wood - a wooden plank 4.8cm long - a protractor Once the materials were obtained, the protractor was held against the 4.8cm long plank of wood and the longer plank was raised by the experimenter’s brother with a block on it as it pivoted around the pivot point. This was to measure the angle at which either block of wood would begin to slide.

Below are the angles at which the blocks began to slide for each test: For the Smaller Block Experiment 1, Experiment 2, Experiment 3, Experiment 4, Experiment 5,

    

 26.5   18   22   18.2   21

For the Larger Block Experiment 1,   21 Experiment 2,   20  Experiment 3,   23 Experiment 4,   21 Experiment 5,   22.5 

Now the average angle for the smaller block will be taken:  ave 

(26.5   18   22   18.2   21 ) 5

 ave  21.14 

for the smaller block

Now the average angle for the larger block will be taken:  ave 

(21  20   23  21  22.5  ) 5

 ave  21.5  for the larger block

b) To calculate the coefficient of static friction, the average or net value of the two average slide angles must be obtained as follows:



( ave forblock1   ave forblock 2) 2

 

(21.5   21.14  ) 2

  21.32 

The coefficient of static friction can be calculated as follows:

s 

sin   tan  cos 

 s  tan 21.32 

 s  0.4

Therefore, the coefficient of static friction for the blocks was 0.4. c) A possible source of error was different characteristic coefficients of static friction for the two different blocks of wood. Although, both blocks were wood, the fact that they were different types of wood and had slightly different lengths meant they would have likely had slightly different static friction coefficients. As a result, the two blocks tended to slide at slightly different angles. This source of error was not considered while the experiment was conducted, and so no measures were taken to avoid it. However, in the future, better selection could be taken to use two blocks of the exact same length of the same type of wood, with the same height. 11. Efficient and safe transportation depends on friction being either minimized or maximized as necessary. Using research, find a) one example in which friction is maximized to aid in transportation. (2 marks) b) one example in which friction is minimized to aid in transportation. (2 marks) For each situation, explain how the friction is maximized/minimized and why this is necessary or beneficial. c) Give at least one source that you used for your research. (1 mark)

Solutions: The best example in which friction is maximized is the brakes on a car which can change from applying low brake pressure (little friction) to high brake pressure (high friction) to slow a car at different rates. This is useful in that it allows cars to stop immediately if a sudden emergency came up, or gradually if a driver needs to begin braking far from an intersection (“Friction-Real life applications”, n.d.). To reduce friction, bearings that include cylindrical balls are placed in annular spaces between the axle and the hub of a wheel in modern vehicles and electric motors to allow the axle to turn more smoothly and rapidly without losing energy to friction. This is necessary for wheels to turn more rapidly without wearing away, and thus allowing vehicles to move faster (Radhakrishnan, 2016). References Radhakrishnan, V. (1998, February 10). Locomotion: Dealing with friction. Retrieved March 8, 2016 from PNAS website: http://www.pnas.org/content/95/10/5448.full#sec-4 Science Clarified. (n.d.). Friction-Real-life applications. Retrieved March 8, 2016 from Science Clarified website: http://www.scienceclarified.com/everyday/Real-LifeChemistry-Vol-3-Physics-Vol-1/Friction-Real-life-applications.html Lesson 4 Answers: 12. Does the label “centripetal force” ever appear in an FBD? Explain? (2 marks) Solution: The label “centripetal force” is avoided in FBDs because specific kinds of force need to be indicated, where centripetal force is a vague description of force because it could be any kind of force that keeps an object in a rotational motion. For example, the centripetal force acting on a horizontally swinging yo-yo would be the tension force in the string. Centripetal force could be a frictional, gravitational, normal, or applied force, as long as it’s directed between an object and the centre it is rotating around. 13. Sometimes, road surfaces have banked curves. Use an FBD to explain how this helps cars to make turns more safely. (3 marks) Solution:



If a curve is banked, the horizontal component of the normal force FN sin  is added to the frictional force already present to increase the centripetal force acting on the car, this extra centripetal force counteracts factors like ice or water to provide more traction when a vehicle makes a turn (see for instance Holzner, 2011, p. 123.) Holzner, S., PhD. (2011). Physics I for Dummies (2nd ed. ). Hoboken, NJ: Wiley Publishing, Inc. 14. A bus passenger has her laptop sitting on the flat seat beside her as the bus, travelling at 10.0m/s, goes around a turn with a radius of 25.0m. What minimum coefficient of static friction is necessary to keep the laptop from sliding? (5 marks) Solution:   v  10.0m / s, r  25.0m, g  9.8m / s 2 [down ]

s  ?  v2 s   gr

s 

(10.0m / s ) 2 (9.8m / s 2  25.0m)

 s  0.41

Therefore, the minimum coefficient of sliding friction required is 0.41. 15. Keys with a combined mass of 0.100kg are attached to a 0.25m long string and swung in a circle in the vertical plane. (9 marks) a) What is the slowest speed that the keys can swing and still maintain a circular path? b) What is the tension in the string at the bottom of the circle? Solutions: a)

 m  0.100kg , r  0.25m, g  9.8m / s 2

The slowest speed the swinging keys would have would be at the peak of their climb when they swing as they lose their momentum. At this point, only the fore of gravity is acting on the keys and therefore,   Fc  Fg

  mv 2  mg r  v 

 gr

 v 

(9.8m / s 2 )(0.25m)

 v  1 .6 m / s

Therefore, the slowest speed the keys could swing yet still maintain a circular path is 1.6m / s.

b)

 Ft  ?    Ft  Fc  Fg  Fg  ?   Fg  mg

 Fg  0.100kg  9.8m / s 2 [ down]  Fg  0.98 N [ down]

 Fc  ?

  mv 2 Fc  r

 (0.100kg (1.6m / s ) 2 ) Fc  0.25m  Fc  1.024 N [ down]

 Ft  1.024 N [ down ]  0.98 N [ down]  Ft  2.0 N [ down ]

Therefore, the tension in the string at the bottom of the circle is 2.0 N [ down]. 16. Do research to find out what artificial gravity is and how it is related to centripetal motion. Explain how artificial gravity could be created in a weightless environment and give a reason why we want to do this. Give at least one source that you used for your research. (5 marks) Solution: Artificial gravity is the inertial reaction to centripetal acceleration that acts on the human body in circular motion (like when one sees astronauts swinging around in a space craft) (Hall, 2012). Artificial gravity relates to centripetal motion in that the most realistic method of producing it is by creating a centripetal force to produce a pulling sensation toward the floor of a spaceship to mimic the effects of gravity (Feltman, 2013). The centripetal force would be created by the rotation of a space station in orbit, as the reference frame for astronauts is the spaceship (if it starts to spin, the people inside move accordingly, like a driver who feels no motion driving a car). One major reason artificial gravity is necessary on a space station is that many astronauts suffer from Space Adaptation Syndrome (SAS), a condition that includes nausea and disorientation because the brain is made to function under gravity and gets confused in its absence (Weir, 2014). References Feltman, R. (2013, May 3). Why Don’t we Have Artificial Gravity? Retrieved March 10, 2016 from Popular Mechanics’ Website: http://www.popularmechanics.com/space/rockets/a8965/why-dont-we-haveartificial-gravity-15425569/ Hall, T. W. (2012, July 6). SpinCalc. Retrieved March 10, 2016 from website: http://www.artificial-gravity.com/sw/SpinCalc/ Weir, A. (2014, March 4). If We’re Serious About Going to Mars, We Need Artificial

Gravity (Op-Ed). Retrieved from Space.com Website: http://www.space.com/24904-gravity-for-mars-missions.html