The Classical Mechanics Project: Kinematicsphysics.tpjc.net/lessons/kinematics/2008/Kinematics_Tut...Worked Example 3: A racing car accelerates uniformly through three gear changes

The Classical Mechanics Project: Kinematics

Physics @ Tampines Junior College, 2008

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Name : ____________________________________ ( )

Civics Group : __________ Date : ______________

The Classical Mechanics Project Overview of Key Concepts

1 Kinematics Mr. Elvin Yeo 1 Basics:

Vectors (a) Vector addition (Parallelogram

Law, Triangle Law) (b) Vector subtraction (c) Perpendicular components of a

vector (d) + and - signs

2 Basic Kinematics Concepts

(a) Displacement vs. Distance (b) Velocity vs. Speed (c) Acceleration

3 Graphical Skills � Displ-time graph � Vel.-time graph � Acc – time graph

(a) Instantaneous displacement, instantaneous velocity and average velocity

(b) Instantaneous velocity, instantaneous acceleration and change in displacement

(c) Instantaneous acceleration, change in velocity

(d) Link between acceleration and displacement – velocity.

4 Signs for acceleration (a) Conventions taken.

Be proficient in both methods of finding acceleration. (b) ∆v vector diagram (c) Gradient of v-t graph

5 Equations for constant

acceleration (a) Condition (b) Derivation (c) Applications to 1D problems,

including Problem-Solving Heuristics

Vertical: g is value only. a = g or a = – g depending on sign conventions

6 Projectile Motion (without air resistance)

(a) Independence of horizontal and vertical

(b) Horizontal: No acceleration Vertical: a = g or a = – g

Connect via time variable (c) Horizontal launch vs. launch at an

angle (d) Other examples: Free fall on

Moon, Jupiter where g is less than 9.81

(e) Other examples: Charged particles in electric field



2

2.5

4.0 1350

4.7km

450 θ2 2 2 0

0

0

2.5 4.0 2(2.5)(4.0)cos135

6.03 6.0

sin135 sin

4.0

28.0

x

x

x

θ

θ

= + −= =

=

=

4.0

450

2.5

1 Basics: Vectors

(a) Vector addition (Parallelogram Law, Triangle Law)

Q1: Given that the sum of vector X and vector Y is the vector Z, for each of the Fig 1.1 below, which figure gives the wrong vector representation of Z in terms of X and Y. Which of the following does not show the correct vector diagram?

Worked Example 1 Find the sum of the following pairs of vectors: s1 = 2.5 km due east, s2 = 4.0 km 450 South of East

Guide: 1. Draw the individual vectors. 2. Draw the vector sum of the individual vectors to obtain the

resultant vector. 3. Use mathematical operations such sine rule, cosine rule etc

to find magnitude and direction.

Q2: Find the sum of the following pairs of vectors: (a) v1 = 30 m s-1 at bearing 45°, v2 = 40 m s-1 at bearing 90° (b) F1 = 40 N vertically upwards, F2 = 30 N horizontal

X Y

X

Y

Y

Z Z

X

Y

Z

X

X

Y

Z

A B

C D

Fig 1.1

Fig 1.2



3

2 2 2 0

1

0

( ) 25 30 2(25)(30)cos90

39.1

30tan 50.2

25

v

v ms

θ θ

−

∆ = + −∆ =

= => =

Vf= 25

Vi =30

∆Vθ

Basics: Vectors

(b) Vector subtraction - kinematics: change in velocity - kinematics: relative velocity

Worked Example 2 A car changes its velocity from 30 m s-1 due East to 25 m s-1 due South.

(i) Draw a vector diagram to show the initial and final velocities and the change in velocity.

(ii) Calculate the change in speed. (iii) Calculate the change in velocity.

[Ans: (ii) -5 m s-1; 39.1ms-1 at 50.20 southwest] Ans: Vf = Vi + ∆V

Change in speed = 25 – 30 = - 5 Change in velocity = 39.1ms-1 at 50.20 southwest or west of south [Basically, there is no difference in method of vector subtraction and vector addition, except the fact that the vector diagram is different!

Q3: Fig 1.4 shows a particle with an initial velocity of 15 m s-1 in the Ox direction. Its velocity, at a later time, is 15 m s-1 at an angle of 60° to Ox. (Directions are given as angles measured anticlockwise from the direction Ox.) What change of velocity has taken place in this interval ? A zero. B 26 m s-1 at an angle of 30° to Ox. C 15 m s-1 at an angle of 120° to Ox. D 15 ms -1 at an angle of 360° to Ox.

Q4: Fig 1.5 shows an object with an initial velocity of 20 m s-1 at an angle of 180o to OX. At a later time, its velocity is 20 m s-1 at an angle of 120o to OX. (Directions are indicated by measuring angles anticlockwise from the direction OX.)

(a) Sketch a vector diagram to show the initial and final velocities; and

the change in velocity of the object.

O X

60o

15 m s-1

15 m s-1

Fig 1.4

O X

120o

15 m s-1

20 m s-1 180o

Fig 1.5

1. Use the vector equation to draw the vector diagram!

2. Draw the vector diagram 3. Use mathematical tools to solve for unknowns.



4

θ

Fx

Fy

F

θ

x y

F

Fig 1.7

Fig 1.8

Q5: Suppose Adam and Ben were running towards each other along the same line. If Adam’s speed is 2.5 m s-1 and Ben’s speed is 5.5 m s-1. What is the magnitude of the relative velocity of Adam with respect to Ben? Assuming that they maintain the same speeds and how long do they take to be 100m apart?

Optional: relative velocity pointed in non-parallel direction Q: Suppose Adam and Ben were 100m apart and there were moving at a speed of 15 ms-1 in the direction as shown by vA and vB respectively in the Fig 1.6. Find the relative velocity of Adam with respect to Ben and determine if they will meet? If they meet, how long do they take to meet?

Ans: vAB = vA – vB

The triangle is an isosceles triangle. Therefore by symmetry, the magnitude of vAB is 15 ms-1 in the direction of their distance apart. Therefore, they will meet. Time they take to meet = 100 / 15 = 6.7 s

1 Basics: Vectors

(c) Perpendicular components of a vector

By Referring to page 5 of lecture notes section 1(c), attempt the questions below. Q6: The vector F shown in Fig 1.7 is made up vector Fx and Fy. Which of the following correctly describes the magnitude Fx and Fy in terms of F and θ? Fx Fy

A F sin θ F sin θ B F sin θ F cos θ C F cos θ F sin θ D F cos θ F cos θ Q7: Resolve the vector F shown in Fig 1.8 in the direction of x and y. You should indicate the direction and magnitude of the forces in the direction of x and y.

60o

15 m s-1 15 m s-1

vA -vB

120o vAB

60o

15 m s-1

120o

15 m s-1

100 m

vA vB

Fig 1.6

2.5 m s-1 5.5 m s-1 vA vB



5

300

x y

10N

Fig 1.10

1 Basics: Vectors

(d) + and - signs

By referring to page 6 of lecture notes section 1(d), attempt the question below. Q8: Taking the right and downwards as positive directions for x and y. Resolve the 10N force shown in Fig 1.10 into its x and y components. Q9: Comparing two objects A and B with velocity, denoted by 10 ms-1 and -15ms-1 respectively. Which object is moving faster, why? 2 Basic Kinematics

Concepts (a) Displacement vs. Distance

By referring to page 6 of lecture notes section 2(a), attempt the questions below. Q10: A car travels 6km due east from O and makes a U-turn back to travel a further distance of 2km. Find its displacement. Q11: A man travels from a point O in this manner: 2km North, 2km East and 4km West, find the its displacement.

2 Basic Kinematics Concepts

(b) Velocity vs. Speed

Worked Example 3: A racing car accelerates uniformly through three gear changes with the following average speeds: 20 m s-1 for 2.0 s 40 m s-1 for 2.0 s 60 m s-1 for 6.0 s What is the overall average speed of the car in the 10.0 seconds? Ans:

1tan 20 2 40 2 60 648

2.0 2.0 6.0

total dis ce x x xAverage speed ms

total time−+ += = =

+ +

Q12: The following table below shows the timing that a student took to

complete each of the six laps of his 2.4 km run.

Lap 1 2 3 4 5 6

Timing/s 97 104 115 129 132 122

What was the student’s average speed for the run?

Q13: A rabbit, nervously trying to cross a road, first moves 80 cm to the right, then 60 cm to the left, then 90 cm to the right, and then 310 cm to the left. (a) What is the rabbit's total displacement? (b) If the elapsed time was 18 s, what was the rabbit's average speed? (c) What was its average velocity?



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Fig 2.1

By referring to QED 8 of lecture note, attempt the question below.

Q14: A man walks 5.0 km due east and a further 5.0 km due south in 1.0 h. Determine

(a) his average speed and (b) his average velocity. 2 Basic Kinematics

Concepts (c) Acceleration

Worked Example 4: A person kicks a ball, which rolls up a sloping street, comes to a halt, and rolls back down again. The ball has constant acceleration. The ball is initially moving at a velocity of 4.0 m s-1, and after 10.0 s it has returned to where it started. At the end, it has sped back up to the same speed it had initially, but in the opposite direction. What was its acceleration?

Ans: By giving a positive number for the initial velocity, the statement of the question implies a coordinate axis that points up the slope of the hill. The ball is moving in the opposite direction after 10.0 s, so its velocity then is - 4.0 m s-1. (note the negative sign) Since acceleration is constant, we have

a = tuv

∆

-

= -1((- 4.0) - (4.0)) m s

10 s

= - 0.80 m s -2 up the slope (i.e. 0.80 m s -2 down the slope) The acceleration was no different during the upward part of the roll than on the downward part of the roll. It is always negative. Notice how the observations for this worked example are consistent with QED 9 on page 8 of lecture notes. � As the ball is rolling up the slope, we observe that the ball is slowing

down. Indeed, velocity and acceleration are in opposite directions. Velocity is positive (i.e. up the slope) while acceleration is negative (i.e. down the slope).

� As the ball is rolling down the slope, we observe that the ball is speeding up.Indeed, velocity and acceleration are in the same direction. Both are negative (i.e. down the slope)

Incorrect solution : Acceleration is ( v-u)/∆�t, and at the end it’s not moving any faster or slower than when it started, so v – u =0 and a = 0. [The velocity does change, from a positive value to a negative value!]

Q15: A ball that hits a wall at a speed of 14 m s-1 and bounces back with a speed of 12 m s-1. The duration of impact is 0.2 s. Find the magnitude of the average acceleration of the ball and determine the direction of acceleration of the ball.

Q16: The following table records the increase in speed of an accelerating car over a period of 7.0 s.

Time/s 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0

v /(m s-1) 0.0 3.4 6.8 12.2 17.6 22.0 24.9 27.5

What is the average acceleration of the car over the period of 7.0 s?



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Fig 2.2

Fig 2.4

Fig 2.5

3 Graphical Skills � Displ-time graph

(a) Instantaneous displacement, instantaneous velocity and average velocity.

Q17: An object is moving along a straight line. The graph in Fig 2.2 above shows its displacement from the starting point as a function of time. (a) In which section(s) of the graph does the object have the highest speed? (b) At which time(s) does the object reverse its direction of motion? (c) How far does the object move from t = 0 to t=3 s? 3 Graphical Skills

� Vel.-time graph (b) Instantaneous velocity,

instantaneous acceleration and change in displacement.

Q18. If Fig 2.4 above shows four graphs of velocity versus time, a) which graph shows a constant velocity? b) which graph shows constant and positive acceleration? c) which graph shows constant and negative acceleration? d) which graph shows a changing acceleration that is always positive?

Q19: Fig 2.5 shows a velocity-time graph for a jogger running in the north direction and has a journey lasting 65 s. It has been divided up into six sections for ease of reference. a) What distance does the jogger travel during the first 10.0 min (t = 0 to 10.0 min)? A) 8.5 m B) 510 m C) 900 m D) 1020 m b) What is the displacement of the jogger from t = 18.0 min to t = 24.0 min? A) 720 m, south B) 720 m, north C) 2160 m, south D) 3600 m, north c) What is the displacement of the jogger for the entire 30.0 min? A) 3120 m, south B) 2400 m, north C) 2400 m, south D) 3840 m, north d) What is the total distance traveled by the jogger in 30.0 min? A) 3840 m B) 2340 m C) 2400 m D) 3600 m e) What is the average velocity of the jogger during the 30.0 min? A) 1.3 m s-1 , north B) 1.7 m s-1, north



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C) 2.1 m s-1, north D) 2.9 m s-1, north f) What is the average speed of the jogger for the 30 min? A) 1.4 m s-1 B) 1.7 m s-1 C) 2.1 m s-1 D) 2.9 m s-1 g). In which region of the graph is a positive? A) A to B B)C to D C) E to F D) G to H h). In which region is a negative? A) A to B B) C to D C) E to F D) G to H. i) In which region is the velocity directed to the south? A) A to B B) C to D C) E to F D) G to H

Worked Example 6: The graph below shows the variation with time of the velocity of a trolley, initially projected up an inclined runway.

Use the above graph to determine (a) the initial projection velocity of the trolley, (b) the time taken for the trolley to reach its highest point on the runway, (c) the deceleration of the trolley as it moves up the runway, (d) the maximum distance up the slope which the trolley reaches. (e) What is the displacement of the trolley at t = 5.0 s? (d) Describe the motion of the trolley throughout the time interval shown. Ans: (a) 0.80 m s-1

(b) 2.5 s (c) a = gradient of v-t graph = - 0.32 m s-2∴∴∴∴ deceleration = 0.32 m s-2 (d) maximum distance = area under v-t graph between t = 0 & t = 2.5 s = ½ × 0.80 × 2.5 = 1.0 m (e) displacement = total area under graph = 1.0 + (-1.0) = 0 m (f) Trolley travels 1.0 m up the runway with uniform deceleration, stops momentarily at t = 2.5 s and then accelerates uniformly down the runway.

1 2 3 4 5 –0.80 –0.60

–0.40

–0.20

0

0.20

0.40

0.60 0.80

time / s

velocity / m s–1



9

Fig 2.6

Fig 2.7

Use worked example 6 as a guide if you need help. Q20: Figure 2.6 shows the vertical velocity component of an elevator versus time. (a) How high is the elevator above the starting point (t = 0 s) after 20 s have elapsed? (b) If the elevator starts to move at t = 0 s, when is the elevator at its highest location above the starting point? (c) Find the acceleration of the lift during the period 14 s to 16 s? Is the lift undergoing acceleration or deceleration?

Refer to QED 13 and 14 if you need guidance to Q24 and/or Q25. Q21: The graph of velocity against time for an object moving in a straight line is shown.

Which of the following is the corresponding graph of displacement against time?

displacement

time 0

time 0

time 0

time 0

A B C D

displacement displacement displacement

velocity

time 0



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Fig 2.8

Q22: The figure below shows the velocity of a particle plotted as a function of time. The total distance travelled by the particle over the interval t = 0 s to t = 3 s is 30 m. 0 (a) What is the average velocity over the interval t = 0 s to t = 1.5 s? (b) What is the particle’s maximum displacement from its position at t = 0 s? (c) Sketch the particle’s displacement and acceleration as a function of time. Ans: a) Average velocity = total displacement / time = 0 m s-1 since the area on top and below are equal. [Total displacement is the sum of the area under the v-t graph.] b) Particle max displacement = 30/2 = 15 m since the distance traveled is 30m c) See graph on the right! To Teachers: If you felt that students need more practice, make use of Q23 to plot the displacement time graph.

1 3 2

20

- 20

v / m s-1

t / s



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Worked Example 7 The velocity-time curves of two cars A and B are shown below: Both cars started at the same point.

Just as car A starts up, car B passes it, moving at a steady velocity. (a) How long does it take car A to be going as fast as car B? (b) At that time, by how far is car B ahead of car A? (c) Which car is ahead, and by how much, at the end of 40 s? (d) At what time does car A catch up with car B?

Ans

(a)By using similar triangle, 60 40

40 t= , Solving t = 26.67 s

b) Till time 26.67 s, For car A, the displacement travelled s = ½ (40) (26.67) = 533.3 m For car B, the displacement travelled s = 40 (26.67) = 1066.6 m Therefore car B is ahead of car A by 533.3 m

c) After 40 seconds

Distance travelled by A = 12

at2 = 12

(60)(40) = 1200m

Distance travelled by B = 40(40) = 1600m Car B is ahead of car A by 1600 – 1200 = 400m d) For Car A to catch up with car B, Distance travelled by A = Distance traveled by B Let the time where they meet be T 40T = 1200 + 60 (T-40) T = 60 s Q23: The velocity-time curves of a car and a lorry are shown in Fig 2.9: Both vehicles started at the same point.

20

40

60

20 40 60 0

v / m s-1

t / s

Lorry

Car

Fig 2.9

30

50

80 70

20

40

60

20 40 60 0

v / m s-1

t / s

Car A

Car B

Since they start at the same point, to meet, they must travel the same distance.



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Just as the lorry starts up, the car passes it, and later slows down again. (a) at what times does the car and the lorry have the same speed? (b) At that earlier time, by how far is car of the lorry? (c) when the car slows down, determine whether the two vehicles meet again before 70 s?

Q24:

The graph in Fig.2.10 is the vertical velocity component v, of a bouncing ball as a function of time. The y-axis points up. Answer these questions based on the data in the graph. (a) At what time does the ball reach its maximum height? (b) For how long is the ball in contact with the floor? (c) What is the maximum height of the ball? (d) What is the acceleration of the ball while in the air? (e) What is the average acceleration of the ball while in contact with the floor?

Fig 2.10



13

R

S

Fig 2.11

3 Graphical Skills � Acc – time graph

(c) Instantaneous acceleration, change in velocity.

Q25: For Q24, using the ans derived for (d) and (e), draw the acceleration time graph from the velocity-time graph shown Q26: Fig 2.11 shows the acceleration time graph of a car. Given that the area under the acceleration time graph in the 2 loops shown in Fig 2.11 are R and S respectively and the car starts from rest. What is the velocity of the car at t = x? A) R + S B) R C) R - S D) S

3 Graphical Skills

(d) Link between displacement and acceleration - velocity

Q27: An object starts from rest and accelerates uniformly at 8ms-2 for 5s. Draw the displacement time graph of the object?

a

t

s

v

t

t

8

5

a

t



14

A B C D

Fig 2.12

4 Signs for displacement, velocity, acceleration

(a) Convention taken

Worked example 8: A body is under the influence of the earth’s gravitational field and subject to the acceleration g = 9.81 m s-2. Considering only the motion in the vertical direction, complete the following table:

Velocity Acceleration Example of situation

Is the object getting faster or slower or neither

+ + Downward taken positive Body thrown downwards

Faster

- + Downward taken positive Body thrown upwards

Slower

0 + Downward taken positive Body released

Faster

+ - Upward taken positive Body thrown upwards

Slower

- - Upward taken positive Body thrown downwards

Faster

0 - Upward taken positive Body released

Faster

When the 2 signs are opposite in direction, the ball slows down. When there is only one sign or 2 –ve or 2 +ve signs, balls accelerates

Q28 Fig 2.12 shows 4 velocity time graphs. By considering the motion of a ball, give the situation of the ball which gives the graphs of A, B, C and D and the positive direction The first one is done for you. Situation A: Ball thrown downwards. Downwards as positive. Situation B: Situation C: Situation D: 4 Signs for acceleration (b) ∆v vector diagram

[To find change in velocity: refer to worked example 2, Q3 and Q4] Q29: What is the change in velocity when an object moves with an initial velocity of 4 m s-1 due South changes to 12 m s-1 due North? Hence determine the direction of the acceleration of the object?



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4 Signs for acceleration (c) Gradient of v-t graph Q30: Refer to Q28 and determine the graphs where the acceleration is negative. 5 Equations for constant

acceleration (a) Condition

Worked example 9: What are the conditions that must be satisfied in order for the kinematics equations to be valid? Ans: There must be uniform acceleration. Motion must be in a straight line. 5 Equations for constant

acceleration (b) Derivation

Q31: The kinematics equations can be derived from 2 information of the velocity time graph. What are they?

5 Equations for constant acceleration

(c) Applications to 1D problems, including Problem-Solving Heuristics Vertical: g is value only. a = g or a = – g depending on sign conventions

Refer to page 17 and 18 of lecture notes to help you Q32: An aeroplane at rest accelerates at 3.1 m s-2 to reach its take-off speed of 100 m s-1.

(a) How long must the runway be? (b) How long does it take before the plane takes off? Q33: A car is travelling along the motorway at 35ms-1. The driver does not notice a stationary police car, which then gives chase. The police car accelerates at 2.5 m s -2 until it overtakes the speeding car. (a) If the speed limit of the road is 80 km h-1, is the car speeding? If it is, how long does it take for the police car to draw level with the speeding car? (b) How far does the police car travel before this happens? (c)* Sketch a graph of distance against time, showing the positions of both cars.



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Acceleration due to free fall Q34: Complete question (a) to (f) (a) You hold a ball in your hand at a fixed height and release it. Its initial velocity is _______

A. up B. zero C. down

(b) If you drop an object in the absence of air resistance, it accelerates downward at 9.8 m s-2. If instead you throw it downward, its downward acceleration after release is

A. less than 9.8 m s-2. B. 9.8 m s-2. C. more than 9.8 m s-2.

(c) A ball is thrown straight up. At the top of its trajectory the ball is A instantaneously at rest. B instantaneously in equilibrium. C Both A and B are true. D Neither A nor B is true. (d) You are throwing a ball straight up in the air. At the highest point, the ball's

A. velocity and acceleration are zero. B. velocity is nonzero but its acceleration is zero. C. acceleration is nonzero, but its velocity is zero.

(e) A ball is thrown straight up into the air. Neglect air resistance. While the ball is in the air its acceleration A increases B remains constant D decreases on the way up and increases on the way down E changes direction Ans: B. It is a constant of about 9.8 m s-2 downward, independent of the motion of the ball.

(f) A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed. Neglecting air resistance, the ball to hit the ground below the cliff with the greater speed is the one initially thrown A. upward. B. downward. C. neither-they both hit at the same speed.

Refer to page 19 of the lecture notes to help you in Q35 Q35: A man throws a ping pong ball vertically upwards with a velocity of 20.0 m s-1. Neglecting air resistance, determine (a) the maximum height the ball reaches, and (b) the time taken for it to return to the man’s hand.



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Q36: A 55-kg lead ball is dropped from the leaning tower of Pisa. The tower is 55 m high. (a) How far does the ball fall in the first 3.0 s of flight? (b) What is the displacement of the ball in the 3rd second? (c) What is the speed of the ball after it has traveled 2.5 m downward? (d) What is the speed of the ball 3.0 s after it is released? Q37: A stone is released from rest at the top of a 40 m high tower. Air resistant ay be taken as negligible. (Take g as 10 ms-2). What is the time taken for the stone to fall the last 10 m to the ground? A 0.38s B 1.4s C 2.5s D 2.8s

Worked example 10: A balloon is ascending at the rate of 12 m s-1 at a height of 80 m above the ground when a package is dropped. How long does it take for the package to reach the ground? In this case, we adopt the upward direction as positive. u g + (a) u = 12.0 m s-1 v = unknown s = - 80 m a = - 9.81 m s-2 t = unknown Plan the solution

1. v = u + at 2. s = 21

(u + v) t

3. s = ut + 21

at 2 4. v 2 = u 2 + 2 as

Execute the solution

s = ut + 21

at 2 , solving t = 5.4 s (Try this yourself!)

Check your answer: SUM S: Answers can leave in 3 s.f if in doubt. U: Units are correct for a and s M: Makes sense?

1. Sketch out the path. Indicate 2 initial and final points. Jot down available, implicit and unavailable info.

2. Choose the most appropriate formula.

The initial velocity of the balloon is 12 m s-1 upwards.

Do you know why this is negative?



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Q38: A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 10.0 m s-1, releases a sandbag when the balloon is 40.8 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground? Assume g = 9.81 m s-2. 6 Projectile Motion

(without air resistance) (a) Independence of horizontal and

vertical Worked example 11: The projectile motion may be regarded as a combination of a horizontal motion and a vertical motion. What is the difference between these two components of motion (apart from the fact that they are in different directions)? Ans: The magnitude of the horizontal velocity does not change with time. The magnitude of the vertical velocity changes with time. [Vertical velocity component varies linearly with time in the absence of air resistance.]

Q39: Answer (a) to (b) (a) Two balls, identical except for color, are projected horizontally from the roof of a tall building at the same instant. The initial speed of the red ball is twice the initial speed of the blue ball. Ignoring air resistance, A the red ball reaches the ground first. B the blue ball reaches the ground first. C both balls land at the same instant with different speeds. D both balls land at the same instant with the same speed. (b) A person stands on the roof garden of a tall building with one ball in each hand. If the red ball is thrown horizontally off the roof and the blue ball is simultaneously dropped over the edge, which statement is true? A Both balls hit the ground at the same time, but the red ball has a higher speed just before it strikes the ground. B The blue ball strikes the ground first, but with a lower speed than the red ball. C The red ball strikes the ground first with a higher speed than the blue ball. D Both balls hit the ground at the same time with the same speed. 6 Projectile Motion

(without air resistance) (b) Horizontal: No acceleration Vertical: a = g or a = – g Connect via time variable

Make use of the example in page 21 to help you in Q40. Q40: Jane stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m s-1. The cliff is 50.0 m above a flat horizontal beach. (a) How long after being released does the stone strike the beach below the cliff? (b) With what speed and angle of impact does it land? (3.19 s, 36.1 m s-1 and 60.1° below horizontal)



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(c) Horizontal launch vs. launch at an angle vs. vertical lauch

Q41: Answer (a) and (b) (a) A ball is thrown into the air and follows a parabolic trajectory. At the highest point in the trajectory, A the velocity is zero, but the acceleration is not zero. B both the velocity and the acceleration are zero. C the acceleration is zero, but the velocity is not zero. D neither the acceleration nor the velocity are zero. (b) A ball is thrown into the air and follows a parabolic trajectory. Point A is the highest point in the trajectory and point B is a point as the ball is falling back to the ground. Choose the correct relationship between the speeds and the magnitudes of the accelerations at the two points. A vA > vB and aA = aB B vA < vB and aA > aB C VA = VB and aA ~ aB D vA < vB and aA = aB Q42: A student throws a ball from point S to a friend at point F. The path of the ball is shown in the diagram.

The points S and F are on the same horizontal level. Air resistance is negligible. The ball is thrown from point S with velocity v, represented by the vector arrow shown in the diagram On the diagram,

(a) draw arrows from point S to represent the initial horizontal and vertical components of the velocity v (label these components vH and vV respectively),

(b) draw arrows at A and at B to represent the horizontal and vertical components of the velocity of the ball at these two points.

Q43: A ball is thrown from a point 1.0 m above the ground. The initial velocity is 19.6 m s-1 at an angle of 30.0° above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory. (c) the velocity of the stone just before it hits the ground (d) the horizontal distance covered by the stone when it reaches the ground.

S

A

B

F

v

S

A

B

F

v



20

Q44: A battleship simultanously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first?

A. A B. both at the same time C. B D. need more information

6 Projectile Motion

(without air resistance) (d) Other examples: Free fall on

Moon, Jupiter where g is less than 9.81.

Q45: a) Describe the motion represented by the velocity-time graph in Figure 8.7. Comment on each of the lettered sections or points. b) What distance is travelled in the first 2 s? c) Sketch a graph showing the first 6 s of a similar motion on the Moon, where g=1.6 m s-2. d) What distance will be travelled before the velocity becomes zero on the Moon?


(e) Other examples: Charged particles in electric field

Q46: The diagram shows part of the deflection system of a cathode-ray tube. An electron moving with a speed of 1.5 × 107 m s–1 approaches the region between two parallel metal plates, which are 20 mm apart and 60 mm long. The upper plate is at a steady positive potential of + V and the lower plate is at a potential of – V. The magnitude of the acceleration of the electron is 7.0 × 1014 m s–2

(a) On the diagram, sketch the path of the electron as it passes between and beyond the plates. (b) Hence, find the vertical and horizontal components of the velocity of the electron when it emerges from the plates. [2.8 × 106 m s–1; 1.5 × 107 m s–1] (c) Use your answer to (c) above to find the angle through which the electron beam has been deflected as a result of passing between the plates. [10.6°]

60 mm

– V

+ V

20 mm



21

25o

15°

81 m s-1

D

Optional Section Q1: A rocket launcher is sitting on a surface that is inclined at 25o as shown in the figure. The launcher is inclined at an angle of 15o with respect to the surface. The initial speed of the projectile is 81 m s-1. Find the distance D up the incline where the rocket lands. (323 m) Q2: An introduction to air resistance [ optional ] Please visit this website for this question: http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html. Using velocity = 100 m s-1, angle = 60o and mass = 10 kg, sketch the path of the motion of the object. a) Change the mass of the object to 20 kg. Are there any changes in the path of the motion? Introduce air resistance to the motion now by checking the air resistance box. Are there any changes in the path of the motion of the object now? Q3: A bomber, flying at a horizontal speed of 360 km h-1 and at an altitude of 2 km above sea level, wishes to attack an enemy vessel. Calculate the angle of sight θ, at which the bomber should release its bomb so that it would most probably hit the vessel.

Q4: : An archer fires an arrow and hits a target which is a distance away and on the same horizontal level as the bow. The arrow is aimed so that on release it makes an angle θ with the horizontal. (a) Calculate the angle θ for an arrow with initial speed of 32 m s-1 and a

target at a distance of 94 m from the bow. (b) Suggest, with a reason, whether the angle θ would, in practice, be

larger or smaller than that calculated in [a] for the arrow to hit the target. (32.1o)

**Q5: A rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 20.0 m s-2. However, after 50.0 s of flight the engine fails. Neglect air resistance. (a) What is the rocket's altitude when the engine fails? (b) When does it reach its maximum height? (c) What is the maximum height reached? (d) What is the velocity of the rocket just before it hits the ground? (assuming no air resistance) Q6: A stone is thrown horizontally with an initial velocity of 15 m s-1. After 2.0 seconds, the horizontal and vertical components of its final velocity are 15 m s-1 and 20 m s-1 respectively, as shown in Fig 1.9.

(i) Find the resultant velocity of the stone after 2.0 seconds.

(ii) Find the change in velocity of the stone after 2.0 seconds.

Q7*: A marble is rolled so that it is projected horizontally off the top landing of a staircase. The initial speed of the marble is 3.0 m s-1. Each step is 0.18 m high and 0.30 m wide. Which step does the marble strike first?

2 km

360 km h-1

x

θ



22

2. [Ans: 64.8 m s-1 at bearing 071°, 50 N at 53.1° above horizontal] 4. [18.0 ms-1 at 46.10 to OX] 10. [displacement: 4m due east of O] 11. [displacement: 2.8km bearing of 1350] 12. 3.43 m s-1

14. Average speed = 10 km h-1 Average velocity = 5.0 2 km h-1 16. 3.93 m s-2 19. D, A, B, A, A, C, A, D, C 22. 0 m s-1, 15 m 23. (a) 20 s and 55 s (b) 600 m (c) Yes, they meet again, 70 s 24. (a) 0.30 s (b) 0.05 s (c) 0.45 m (d)– 10.0 m s-2 (e)120 m s-2 29. 16 m s-1 Due North. 32. s = 1613 m, t = 32 s 35. s = 20.4 m, t = 4.07 s 38. v = -29.7 m s-1

40. (3.19 s, 36.1 m s-1 and 60.1° below horizontal) 45. b) 20 m; (d) 125 m

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The Classical Mechanics Project: Kinematicsphysics.tpjc.net/lessons/kinematics/2008/Kinematics_Tut...Worked Example 3: A racing car accelerates uniformly through three gear changes