the classical mechanics project:...

The Classical Mechanics Project: Kinematics

Physics @ Tampines Junior College, 2008

1

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

A

B velocity

v~

Force F

AB

Fig 1.1

1 Basics: Vectors

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

Vectors and scalars A vector is a physical quantity which has both magnitude and direction. A scalar is a physical quantity which has magnitude only. Below are some examples of vector and scalar quantities:

Vectors Scalars

Displacement, velocity, acceleration, force, momentum, magnetic flux density

Distance, time, length, temperature, volume, mass, density, energy, power, speed

Vector diagram To distinguish vectors from scalars, vectors are written with special symbols. The following are some common examples of symbols used to represent vectors:

(i) with an arrow above (ii) In bold (iii) with a small sine curve below

Specifying a vector To specify a scalar, only its magnitude and its unit need be given; e.g. speed of car = 10 m s-1. To specify a vector, its magnitude, unit and direction must all be given; e.g. velocity of car = 10 m s-1 due west.

Vector notation : Fr

or F or F%

Magnitude : Fr

or F or F%

Vector Addition There are two ways of adding vectors, namely (i) and (ii). (i) Parallelogram law .

To add vectors P and Q, draw them such that they form the adjacent sides of a parallelogram. The sum of P and Q is the vector R which is given by the diagonal of the parallelogram. The vector R is called the resultant vector

(ii) Vector triangle

Draw vectors P and Q to form the two sides on a triangle. P and Q are joined nose-to-tail. The resultant vector R is represented by the third side of the triangle. Note that R joins P and Q nose-to nose and tail-to -tail.

The magnitude and direction of the resultant can be found either by scale drawing or calculation using the cosine equation. R2 = P2 + Q2 - 2PQ cos θ

Direction is important in a vector quantity!

P P

Q

Q

R

Fig 1.2

P

Q

Q

P R

Q

P R

θ Fig 1.3



3

.(a) .(b)

P P

P Q

-Q

-Q

R R

R = P – Q

Fig 1.5 Fig 1.6

Fig 1.7

X Y

X X X X

Y Y

Y Y

Z Z Z Z

X Y

X X X X

Y Y

Y Y

Z Z Z Z

When more than 2 vectors are to be added, the resultant vector R can be found by drawing a polygon. The result is the same whatever the order in which the vectors are added. QED 1 In Fig 1.5, given that the sum of vector X and vector Y is the vector Z, for each of the figures below, which figure gives the correct vector representation of Z in terms of X and Y. A B C D

1 Basics: Vectors

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

Vector Subtraction Let R = P – Q P – Q can be considered as the vector sum of P and – Q. So R = P + (- Q) - Q, is a vector of the same magnitude but opposite in direction to Q. R can be found by either the parallelogram law or the vector triangle as shown in the Fig 1.6 (a) and Fig. (b) QED 2 In Fig 1.7, given that Z = X – Y. For each of the figures below, which figure gives the correct vector representation of R in terms of X and Y. A B C D

B

A

D C

R

E

A

B D

C

E Fig 1.4



4

Fig 1.8

Fig 1.9

Fig 1.10

Fig 1.11

Kinematics Applied Change in speed and change in velocity To find the change, we normally use the final – initial. So change in speed is usually final speed – initial speed. v∆ = f iv v−

Change in velocity is final velocity – initial velocity.

v∆uuv

= f iv v−uuv uv

Similarly, f iv v v= + ∆uuv uv uuv

is another expression used!

In the below 2 situations, we will see how we use the vector equation to solve problems. Situation 1: 2 vectors which lie along the same line. In Fig 1.8, a car travels at 5 m s-1 due north. After a while it travels 10 m s-1 due south. Find the change in speed and change in velocity? Initial Final Change in speed = 10 – 5 = 5 m s-1 5 m s-1 10 m s-1 Change in velocity = 10 - (-5) = 15 m s-1 due south Situation 2: 2 vectors which do not lie along the same line. In Fig 1.9, a car goes round a right-angled corner so that its velocity changes from 15 m s-1 due east to 20 m s-1 due south. Find the change in speed and change in velocity? Initial Final Change in speed = 20 – 15 = 5 m s-1 15 m s-1 20 m s-1

From Fig 1.10, using a vector diagram,

(i) Change of velocity = f iv v−uuv uv

20 m s-1 From the diagram drawn, Magnitude = √(152 + 202) = 25 m s-1 15 m s-1

Direction θ = tan-1 (15/20)

= 36.9° The change in velocity is ________ with a bearing of _______. QED 3 The velocity of a car, rounding a bend, changes from 40 m s-1 due North to 30 m s-1 due East, as shown in Fig 1.11. Which of the following gives the change in speed and change in velocity of the car? Initial Final 40 m s-1 30 m s-1 Change in speed Change in velocity A 10 m s-1 50 m s-1, 53.10 South of East B 10 m s-1 50 m s-1, 36.10 East of North C -10 m s-1 50 m s-1, 53.10 South of East D -10 m s-1 50 m s-1, 36.10 East of North Ans:

Speed is a scalar!

Velocity is a vector!

A case of subtraction of scalars. (magnitude)

A case of subtraction of vectors. (magnitude and direction)

How do we express the direction of a vector? Bearing!



5

Fig 1.12

θ

Fx

Fy F

Relative Velocity If object A travels with a constant velocity of vA and object B travels with constant velocity of vB , then the relative velocity of B with respect to A is defined as

vBA = vB - vA Obviously, velocity of A relative to B is given by

vAB = vA – vB = -vBA

Situation 1 In a 100 m race, Adam is running aside Ben at the 10 m mark. If Adam’s speed is 10.3 m s-1 and Ben’s speed is 10.0 m s-1 at that instant, what is the relative velocity of Adam with respect to Ben? Assuming that they maintain the same speeds, how long will Adam take to be ahead of Ben by 3.0 m? Ans: QED 4 Suppose Adam and Ben were running towards each other along the same line. If Adam’s speed is 10.3 m s-1 and Ben’s speed is 10.0 m s-1. What is the relative velocity of Adam with respect to Ben? Assuming that they maintain the same speeds A 0.3 m s-1 B 10.3 m s-1 C 10.0 m s-1 D 20.3 m s-1 Ans: [Note: If velocity vectors are pointed in non parallel directions, then a vector diagram is needed to find the relative velocity vector, as done in the Additional Mathematics in Secondary School.]

1 Basics: Vectors

(c) Perpendicular components of a vector

Components of a Vector A vector can be split up into two parts which are called the components of the vector. The components are usually chosen to be along two mutually perpendicular directions. The process of splitting vectors into their components is known as resolution of vectors.

Fx = F cos θ Fy = F sin θ F2 = Fx

2 + Fy2

We always resolve a vector into perpendicular components because

• these components will be independent of each other i.e. they do not interfere with each other.

• these components can be neatly represented using trigonometric

functions .

• the magnitude of the resultant (original vector) and those of its components are related by the familiar Pythagoras Theorem .

F is the vector that is resolved!

Direction is important!



6

300

10 m s-1

vx

vy vy vy vy

vx vx vx

43°

y

F

x

Fig 1.14

Fig 1.16

Fig 1.15

F F

F

Fig 1.13

Most of the time, resolution is done in x and y direction. At times, it is better to resolve in 2 other perpendicular directions as shown in Fig 1.13. Helpful tips:

1. Construct rectangle. 2. Resolve in 2 mutually perpendicular vectors 3. 2 Vectors point outwards

QED 5 Consider a ball that has been projected at a velocity of 10.0 m s-1 at an angle of 300 below the horizontal, shown in Fig 1.14. Which of the following shows the mutually perpendicular components of the velocity?

A B C D

1 Basics: Vectors

(d) + and - signs

In vector, when 2 parallel vectors of the same magnitude are in opposite direction, we will distinguish these two vectors by a positive and negative sign. So which vector do we allocate the positive sign and which the negative sign. That depends on where is taken as the positive direction! Many a times, you have to state your positive direction. +ve 5.0 m s-1 5.0 m s-1

For example, a boy is running to the right at 5.0 m s-1 and a girl is running to the left at 5.0 m s-1, shown in Fig 1.15. If you take the right as positive, then the boy has velocity of 5.0 m s-1 while the girl has velocity of - 5.0 m s-1. Obviously, if you take the left as positive, the answer would be exactly opposite. QED 6 Fig 1.16 shows a vector with its magnitude F and direction shown. Which of the following shows resolved components in the x and y direction? X - direction Y - direction A + F sin 43 o + F cos 430 B - F sin 43 o - F cos 430 C + F cos 430 + F sin 43 o D - F cos 430 - F sin 43 o

The direction of axis gives you an indication of the positive x and y direction!



7


(a) Displacement vs. Distance

Mechanics is that part of physics which deals with the motion of objects. It consists of two branches: Dynamics, which deals with the motion of bodies with reference to the forces that act on the system, and Kinematics, which deals with the motion of bodies without reference to the forces that act on the system. In kinematics, we are concerned with distance, displacement, speed, velocity and acceleration. Displacement vs. Distance Distance Distance is the length of a straight line or curve. Displacement, s Displacement is the distance travelled from a fixed point along a specified direction.It is vector quantity. SI unit : metre , m QED 7 A man from a fixed point O moves 40 m due east and then 30 m due north. Which of the following gives the distance and displacement of the man (from O)? distance displacement A 70m 70m somewhere in the NE direction B 70m 50m somewhere in the SE direction C 70m 50m somewhere in the NE direction D 50m 50m somewhere in the NE direction


(b) Velocity vs. Speed

Speed, v Speed is the rate of change of distance travelled. It is a scalar quantity. SI unit : m s -1

Average speed, < v > = timetotal

distancetotal .

Note: A body that travels equal distances in equal time intervals is said to be moving with constant or uniform speed. Velocity, v Velocity is the rate of change of displacement. It is a vector quantity. SI unit : m s -1

v = ts

dd [Mathematical representation]

Average velocity, < v > = total displacement

total time .

Note: A body that travels equal distances in the same direction in equal time intervals is said to be moving with constant or uniform velocity. Situation: A man walks to the right 100 m from a point O and walks to the left 50 m and walks to the right 30 m in 4 minutes. Find the average speed and average velocity of the man. Average speed, < v > = total distance

total time=

Average velocity, < v > = total displacement

total time=

Note: <speed> is not always equal to <velocity>



8

QED 8 A man walks 5.0 m due east and a further 5.0 m due south in 10 s. Which of the following gives his average speed and magnitude of his average velocity?

Average speed Average velocity A 1.0 1.0 B 1.0 0.7 2 Basic Kinematics

Concepts (c) Acceleration

Acceleration, a

Acceleration is the rate of change of velocity.

It is a vector quantity. SI unit : m s -2

a = tv

dd

[Mathematical representation]

A change in velocity results in acceleration and it can be caused by:

[i] a change in its magnitude only, [what you learn in secondary] [ii] a change in its direction only, or [iii] a change in both its magnitude and direction. Note: A body that travels with equal increase/decrease in speed in the same direction in equal time intervals is said to be moving with constant or uniform acceleration/deceleration. In other words, its velocity changes at a constant rate.

QED 9 By taking the right direction as positive. Consider the following, a car is moving to the left at a speed of 10 m s-1. It undergoes a steady change in speed and 4 seconds later, it moves at a speed of 30 m s-1 to the left. Which of the following gives the correct initial velocity, final velocity and acceleration of the car. Is the car accelerating or decelerating?

Initial velocity / m s-1

Final velocity / m s-1

Acceleration / m s-2

Accelerating / deceleraing

A 10 30 5 Accelerating

B -10 -30 5 Decelerating

C -10 -30 - 5 Accelerating

D -10 -30 - 5 Decelerating

Initial velocity =

Final velocity =

Change in velocity =

Acceleration

Clearly, the car is ______________ by seeing the change in magnitude.

The sign of the acceleration or velocity _________ tell you whether the object is accelerating or decelerating. The velocity _____ the acceleration do! When they are in the same direction, the car is accelerating.

Note: Deceleration or retardation describes a decrease in the _________________________ with time. Thus a decelerating body is slowing down. It occurs when the acceleration and velocity (initial) act in __________ directions.

When change in velocity is negative, this implies acceleration is also negative!



9

time

displacement

time

displacement

t B

Fig 3.1

Fig 3.2

Fig 3.3

3 Graphical Skills � Displ-time graph

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

Displacement-time graph

The displacement-time graph gives the following information: [i] the displacement of a body at any instant of time, and [ii] the gradient of the graph at any time gives the instantaneous

velocity of the body. v = ts

dd [Mathematical representation]

From Fig 3.1, 1. average velocity between O and A

to ta l d isp la cem en t

to ta l tim e= =

s1

t1

2. slope or gradient at A = instantaneous velocity at A = ts

dd

QED 10 A train car moves along a long straight track. The graph in Fig 3.2 shows the displacement as a function of time for this train. The graph shows that the train: A. speeds up all the time. B. slows down all the time. C. speeds up part of the time and slows down part of the time. D. moves at a constant velocity. QED 11 The graph in Fig 3.3 shows displacement as a function of time for two trains running on parallel tracks. Which of the following is true? A. At time tB, both trains have the same velocity. B. Both trains speed up all the time. C. Both trains have the same velocity at some time before tB. D. Somewhere on the graph, both trains have the same acceleration.

s1 A

O

displacement, s

time, t t1

ds

dt



10

A B

C O

x

t Fig 3.4

QED 12 Consider a body moving along a straight line. Its displacement from the origin O is shown in Fig 3.4. Which of the following describes the velocity during the period OA, AB and BC? OA AB BC

A Uniform Zero Decreasing

B Uniform Zero Increasing

C Increasing Zero Decreasing

D increasing Zero Increasing

3 Graphical Skills � Vel.-time graph

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

Velocity-time graph The velocity-time graph gives the following information: [i] instantaneous velocity is velocity of a body at any instant of time. [ii] the gradient of the graph is the instantaneous acceleration of the body.

a = tv

dd

[Mathematical representation]

[iii] area under the graph is the change in displacement of the body.

s = vdt∫ [Mathematical representation]

A B

C

O

v

t

For body moving in straight line, negative value only means its moving in opposite direction!

The gradient of the graph gives instantaneous acceleration!

The area under graph gives change in displacement!



11

Fig 3.5

Consider a body moving along a straight line. Its velocity from the origin O is v.

Analysis of the graph O to A v increases from zero at a constant rate

⇒ body is moving from rest with uniform acceleration A to B v increases at a decreasing rate

⇒ body continues to move faster but with decreasing acceleration

B to C v remains constant [no change in v]

⇒ body experiences zero acceleration C to D v decreases at a constant rate but still positive

⇒ body experiences constant but negative acceleration ie. it is decelerating

D to E v is zero [value of v is zero]

⇒ body is stationary E to F v is negative but magnitude increases at a constant rate

⇒ body is moving from rest in opposite direction and speeds up ⇒ uniform negative acceleration

F to G v remains constant [no change in v]

⇒ body continues to move in opposite direction but with constant velocity ⇒ zero acceleration

G to H v is negative but magnitude decreases at a constant rate ⇒ body slows down (still in opposite direction) and comes to rest at H ⇒ body experiences uniform deceleration (positive acceleration) [ velocity is negative but acc is positive]

Note: Total distance travelled = A1 + A2 Net displacement = A1 − A2

O

D

C B

Area = A2

H

F

A

v

G

E

Area = A1

t O

Gradient of graph is changing, the acceleration is changing.

Gradient of graph is constant, the acceleration is constant.

Area covered below the axis means the object is moving towards O.

Is the object undergoing acceleration or deceleration?



12

displacement

time 0

time 0

time 0

time 0

A B C D

displacement displacement displacement

Fig 3.6

Fig 3.7

Fig 3.8

A

B

v

D

C

t O

s

0 t

QED 13 The graph of velocity against time for an object moving in a straight line is shown in Fig 3.6. Which of the below shows that the object is decelerating? A OA and BC B AB and BC C AB and BC D AB and CD From QED 13, we can also use velocity time graph to draw the displacement time graph, shown in Fig 3.7.

Guide to getting displacement time graph from veloc ity time graph. 1. As long as v is positive, displacement is increasing. 2. When v is zero, displacement is maximum or minimum. 3. Concave or Convex or straight line: 4. For constant gradient: When gradient is positive, concave, when gradient is negative, convex. When gradient is zero, straight line. 5. When non constant gradient: Complications arise, consult your tutors. QED14 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? .

velocity

time 0



13

A

B

C

O

a

t D

Fig 3.9

Fig 3.10

3 Graphical Skills � Acc – time graph

(c) Instantaneous acceleration, change in velocity.

From this graph, you can find useful information about the object’s (i) instantaneous acceleration (the acceleration of a body at any instant of time) (ii) change in velocity (area under the graph)

O-A The area under the graph is positive, the object experiences increase in speed. Since the gradient is positive, the speed of the object increases at an increasing rate.

A-B The area under the graph is positive, the object experiences increase in speed. Since the gradient is negative, the speed of the object increases at a decreasing rate.

B-C The area under the graph is negative, the object experiences decrease in speed. Since the gradient is negative, the speed of the object decreases at an increasing rate.

C-D The area under the graph is negative, the object experiences decrease in speed. Since the gradient is positive, the speed of the object decreases at a decreasing rate.

QED 15 A car is traveling along a straight road from rest. The graph shows the variation with time of its acceleration during part of the journey. At which point on the graph does the car have its greatest velocity?

acc

t

A

B

C D

The area under the graph tells you whether the object increase in speed or decrease in speed..

The gradient and value of a tells you the rate at which the speed is changing!



14

A B C D E

disp

lace

men

t

time

Fig 3.11

Fig 4.1

Fig 4.1

3 Graphical Skills

(d) Link between displacement and acceleration : velocity

At times, we can indirectly obtain information of the acceleration based on the displacement time graph and vice versa. The link is velocity. The graph represents how displacement varies with time for a vehicle moving along a straight line. During which time interval does the acceleration of the vehicle have its greatest numerical value? Acceleration is rate of change of velocity. For acceleration to be present, there must be a change in velocity. Also, the gradient of the given graph gives the instantaneous velocity.

A The gradient is zero => the velocity remains zero and hence no acceleration.

B The gradient increases => the velocity increases. Hence there is acceleration during this period.

C The gradient stays constant => velocity stays constant. Hence no acceleration.

D The gradient is zero which means the velocity remains zero and hence no acceleration.

E The gradient stays constant which means velocity stays constant. Hence acceleration is zero.

So answer is B.

4 Signs for displacement, velocity, acceleration

(a) Convention taken

We have touch on signs for vectors in 1(d). Now, we will apply this u, v, a and s, which are vector quantities. Their directions must be taken into account. Appropriate signs must be assigned to them when solving problems. If, say, upward direction is taken as positive, then • displacement is negative for points below the starting position, • velocity of body moving up is positive, and • downward directed acceleration is negative. Acceleration of object/s under gravity [gravitational force]. For an object on earth surface, it will experience a constant gravitational pull of the Earth. (See Fig 4.1). The downward acceleration of such a body is known as acceleration due to gravity, g, and is assumed to be a constant equal to 9.81 m s-2 near the Earth’s surface.

Fig 4.1 shows only the direction of acceleration. If upward direction is taken as positive, then acceleration of the ball is negative for the projected ball at ALL points.

Q: When upwards is taken as positive, indicate position of ball with negative displacement, negative velocity and negative acceleration?



15

2

A

4

5

1 3

v

t

v t

A B

A B

C O

v

t Fig 4.2

Fig 4.3

QED 16 A ball is projected vertically upwards from point A, shown in Fig 4.2. Fill in the blanks the sign of the vertical displacement, velocity and acceleration during its flight, taking the upward direction as positive.

s v a 1

2

3

4

5

4 Signs for acceleration (b) Using ∆v vector diagram

Acceleration is a vector. v

at

∆=∆

. The direction of ∆v vector diagram tells

you the direction of the acceleration. See page 4 and QED 9 also on page 8.

a) acceleration b) deceleration c) no acceleration

4 Signs for acceleration (c) Using gradient of v-t graph and direction of motion

A change in the magnitude of the velocity indicates an acceleration or deceleration. The gradient of the v-t graph tells you the signs for the acceleration. When the gradient of the graph is positive, the sign for acceleration is positive. [Reminder: the sign of the acceleration alone does not tell you whether the object is accelerating or decelerating, see QED 9 on page 8 if you do not understand this statement.] QED 17 Fig 4.3 shows the

velocity time graph of an object. Which region shown that the acceleration is negative? A OA B AB C BC

Direction of motion How can you tell from a velocity-time graph which direction is taken as the positive direction? It depends on the scenario considered. Suppose an object being released from rest on the surface of the earth. What is the velocity time graph of the object? Will its motion be represented by A or B? Ans: .



16

u

v

0 t

velocity

time

The velocity-time graph of a uniformly accelerated body is a straight line. u and v are the initial and final velocities respectively.

Fig 5.1

5 Equations for constant acceleration

(a) Condition for using kinematics equations

In order for kinematics equation to be used, there are two conditions required

1. Constant acceleration

2. Motion must be in a straight line.

QED 18 Which of the following described below does not undergo uniform acceleration? A Object released at the top of a building, ignoring air resistance. B Object moving in a circle in constant motion. C Object is thrown upwards on the surface of the earth, ignoring air resistance. D Object undergoing steady increase in velocity. Free fall. A body is said to be in free fall if the only force acting on it is the gravitational pull of the Earth. The downward acceleration of such a body is known as acceleration due to gravity, g, and is assumed to be a constant equal to 9.81 m s-2 near the Earth’s surface. True free fall only exists in a vacuum where there is no air resistance. All bodies falling freely undergo this constant acceleration irrespective of their masses.


(b) Derivation

Equations representing uniformly accelerated motion in a straight line Suppose that a body is moving with constant acceleration a and that in a time interval t, its velocity increases from u to v and its displacement increases from 0 to s.

Since a = tv

dd

= tu-v

Hence v = u + at --- [1] Since the displacement is the area under the velocity time graph in Fig 5.1

s = 21

(u + v ) t ------------ [2]

Substituting [1] into [2],

s = ut + 21

at 2 ----------- [3]

From [1], t = a

uv - and substituting it into [2], s =(

2uv +

) x (a

u- v)

Therefore v 2 = u 2 + 2 as --------- [4] Equations 1, 2, 3 and 4 are called kinematic equations applicable for uniformly accelerated motion in a straight line .



17

Fig 5.2


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

In section 4, we have gone through the importance of direction and sign notation. Now we will apply this to the kinematics equations which are used for uniformly accelerated motion in a straight line. Variables and their symbols Displacement: s Velocity: u for initial velocity, v for final velocity Acceleration: a 4 Equations of Motion:

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

(u + v ) t

3. s = ut + 21

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

General Strategy for Solving Kinematics Problems Step 1: Understand the Problem � Sketch out the path . � Indicate 2 initial and final points � sign convention � Jot down , on the sketch, all the

available info , implicit info , and info that is not available .

� Determine the objective of the problem.

Step 2: Plan the solution � Choose the most

appropriate formula out of the 4 given to apply.

� Decide on the steps to arrive at the final solution

Step 3: Execute the solution � Substitute values into formula to

solve for required quantity. � Be careful about signs .

Step 4: Check your answer Check SUM � S: Significant figures � U: Units � M: Makes sense

Scenario 1 A car is moving at a speed of 5.0 m s-1 and accelerates uniformly to 15.0 m s-1 in 8.0 seconds, shown in Fig 5.2.

(i) Find the acceleration of the car. (ii) Find the distance moved by the car.

In section 5a, we learned about conventions. It is important to state the positive direction so as to avoid confusion later. Understand the problem Take right as positive u = 5 m s-1 v = 15 m s-1 t = 8s a = ? s = ? 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 (i) v = u + at 15.0 = 5.0 + a (8.0) => a = 1.25 = 1.3 m s-2

(ii) s = 21

(u + v) t

s = 21

( 5.0 + 15.0 )(8.0) = 80 m

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?

u v

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

2. Choose the most appropriate formula.

3. Substitute values into formula and solve.

For (ii) equation 4 can also be used since a is known!

+



18

Fig 5.3

Scenario 2 Suppose that the car in Scenario 1 decelerates uniformly to rest through a distance of 100 m, shown in Fig 5.3.

(i) Find the acceleration of the car. (ii) Find the time taken by the car to come to a stop.

Take right as positive

u = 15.0 m s-1 v = 0.0 m s-1 s = 100m a = ? t = ? 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 (i) v 2 = u 2 + 2 as (0.0)2 = (15.0)2 + 2 a (100.0) => a = -1.125 = -1.1 m s-2

(ii) s = 21

(u + v) t

100 = 21

(15.0 + 0.0) t => t = 13.3 = 13 s

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 to decelerate at 1.1 m s-2

An approach to do checking: graphical approach. Note the kinematics equations are derived using the velocity time graph.

QED 19 A racing car accelerates uniformly along a straight road. Marker posts are placed at 100 m apart along the road. The car has a speed of 10 m s-1

when it passes one post and 20 m s-1 when it passes the next. What is the acceleration of the car? A 0.67 m s-2 B 1.5 m s-2 C 2.5 m s-2 D 6.0 m s-2

15.0

0 t

velocity

time

Area = 100m

Fig 5.4

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



For (ii) equation 4 can also be used since a is known!

+



19

Fig 5.5

Falling objects or objects thrown upwards? (Under influence of gravity) We all know that the g value is 9.81 m s-2 near the surface of the earth. If we consider objects falling or thrown upwards, they all undergo a constant ‘acceleration’ of 9.81 m s-2. u g Consider an object being thrown upwards at 30.0 m s-1, shown in Fig 5.5. Take the constant g as 10 m s-2. Find (i) the time to reach the maximum height (ii) the velocity of the object after 5 seconds. (ii) the displacement of the object after 7 seconds. Ans: Remember about conventions. It is important to state the positive direction so as to avoid confusion later. In this case, we adopt the upward direction as positive. u g + (i) u = 30.0 m s-1 v = 0.0 m s-1 s = unknown a = - 10 m s-2 t = ?

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 v = u + at 0.0 = 30.0 + (-10) t => t = 3.0 s It takes 3.0 s to reach to maximum height, it will take another 3.0 s to go from maximum height to zero displacement. 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? (ii) u = 30.0 ms-1 v = ? s = unknown a = - 10 ms-2 t = 5 s After choosing the formula, substitute values into the formula v = u + at v = 30.0 + (-10) 5 => v = -20.0 ms-1 Ans: It tells you that the ball is traveling in the opposite direction to the positive direction adopted! (iii) u = 30.0 m s-1 v = unknown s = ? a = - 10 m s-2 t = 5 s After choosing the formula, substitute values into the formula.

s = ut + 21

at 2

s = 30.0 (7.0) + 21

(-10) (7)2 = - 35 m

Ans: It tells you that the ball’s is below the point where the displacement is taken to be zero.




At the maximum height, the velocity is zero.

Since the upward direction is positive and acceleration due to gravity is downwards, a negative sign is required!

What does the negative sign of the velocity tell you?

What does the negative sign of the displacement tell you?



20


(a) Independence of horizontal and vertical

What about objects projected horizontally? We have seen how acceleration due to gravity affects objects in the vertical component, how about objects projected horizontally. Let consider this scenario

As you can see from Fig 6.1, only the vertical component is affected by gravity (constant acceleration). The horizontal velocity remains the same (No acceleration) by looking at the horizontal distance travelled at equal time intervals. In other words, GRAVITY AFFECTS ONLY VERTICAL MOTION AND DOES NOT AFFECT HORIZONTAL MOTION. HORIZONTAL MOTION DOES NOT AFFECT VERTICAL MOTION. So what does it mean when I project a ball horizontally at 10 m s-1. It means that it has a initial horizontal velocity of 10 m s-1 and a initial vertical velocity of 0 m s-1. Note: the x and y direction’s initial velocity are different! To help us to differentiate the 2 directions, we give a subscript x for horizontal component and subscript y for vertical component. In other words, we give ux for initial horizontal velocity and uy for initial vertical velocity and sx for horizontal displacement etc.

QED 20 Two identical balls roll off a tabletop. Just before rolling off, one is rolling at 0.5 m s-1 and one is rolling at 1.0 m s-1. Which ball hits the ground first? (Assuming no air resistance)

A. The ball with initial speed = 0.5 m s-1. B. The ball with initial speed = 1.0 m s-1. C. Both land at the same time.

QED 21 Two identical balls roll off a tabletop. Just before rolling off, one is rolling at 0.5 m s-1 and one is rolling at 1.0 m s-1. After some time, which ball travels a further horizontal distance? (Assuming no air resistance and the ball does not bounce after landing.)

A. The ball with initial speed = 0.5 m s-1. B. The ball with initial speed = 1.0 m s-1. C. Both travel the same horizontal distance.

Answer: QED 22 Suppose a rescue airplane, shown in Fig 6.2 drops a relief package while it is moving with a constant horizontal velocity at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane? A. below the plane and behind it. B. directly below the plane C. below the plane and ahead of it

gravity free path

Fig 6.1

Fig 6.2



21

θ

2 2 1

0

39.6

39.24tan 82.7

5.0

R x y

y

x

V u v m s

vbelow horizontal

uθ

−= + =

= = =

6 Projectile Motion

(without air resistance) (b) Horizontal: No acceleration

Vertical: a = g or a = -g and connect via time variable

Consider the 2 components separately. Vertical component

Horizontal component

Adopt +ve direction and follow the strategies listed in 5c for the vertical direction.

Adopt +ve direction and follow the strategies listed in 5c for the horizontal direction.

Constant acceleration 4 equations of motion 1. vy = uy + ay t

2. sy = 21

(uy + vy ) t

3. sy = uy t + 21

ay t 2

4. vy 2 = uy

2 + 2 ay sy

No acceleration a. ux = vx 1. sx = ux t

Let’s try to make use these equations on an example: A stone is thrown horizontally at a speed of 5.0 m s-1 from the top of a cliff 78.4 m high, shown in Fig 5.8. Taking the value of g as 9.81 m s-2 (a) How long does it take the stone to reach the bottom of the cliff? (b) How far from the base of the cliff does the stone hit the ground? (c) What are the horizontal and vertical components of the stone’s

velocity just before it hits the ground? What is the resultant velocity?

For an object projected horizontally, initial vertical velocity uy = 0 ms-1. Taking downwards as positive. (a) sy = 78.4 m ay = 9.81m s-2 t = ? uy = 0 m s-1 vy = unknown

sy = uy t + 21

ay t 2

78.4 = 21

(9.81) t 2

t = 4.00 s Remember: 4. Check your answer: SUM (b) As you know, the stone takes a certain amount of time to reach the ground. In that amount of time the stone will also travel horizontally, and covers a horizontal displacement. That is why in the equations on the left, we do not attempt to differentiate the time t for the vertical and horizontal component, it’s the same t! Taking right as positive. sx = ? ux = 5.0 m s-1 t= 4.00 s sx = ux t sx = 5.0 (4.00) = 20.0 m (c) Horizontal velocity ux = 5.0 m s-1 Vertical velocity vy = uy + a t = 0 + 9.81 (4.00) = 39.24 m s-1 Resultant velocity

Why isn’t the acceleration negative?

Remember: You have taken downward direction as positive.




Using subscripts can help you to differentiae the 2 directions.

This is the resultant velocity ! Find its magnitude and direction!



ux

vy

VR

Refer to Section 1c if required!

Fig 6.3



22

6 Projectile Motion

(without air resistance) (c) Horizontal launch vs. vertical

launch vs. launch at an angle Projectile Motion at an angle In the earlier sections, we have done vertical and horizontal projection, shown in Fig 6.4 (a) and (b). Now, how do we attempt to tackle objects projected at an angle θ with respect to the horizontal, shown in Fig 6.1 (c)? Simple, we still separate them into their vertical components and horizontal components. The horizontal component still remains unchanged during its flight. (in the absence of air resistance) The vertical components will undergo constant acceleration.

(a) Vertical (b) Horizontal (c) Launch at an angle A summary of the types of projection is shown in the table below. In order to differentiate the direction, we need to adopt a positive direction. In this case, we adopt upwards and right as the positive direction for vertical and horizontal direction respectively. Case (a) Case (b) Case (c) left Case (c) right ux = 0 uy = v

ux = v uy = 0

ux = v cosθ uy = v sinθ

ux = v cosθ uy = - v sinθ

Let’s consider an example: A stone is projected with an initial velocity of 12 m s-1 at angle of 30° above the horizontal from a cliff top which is 75 m above the sea level. Find

(a) the time taken for the stone to reach the sea, and

(b) the range of the stone from the cliff when it reaches the sea.

(a) Taking upwards and rightwards as positive. ay = -9.81m s-2 t = ? uy = 12sin300 m s-1 vy = unknown sy = -75.0 m ux = 12cos300 m s-1 sx = ? Based on the above information,

sy = uy t + 21

ay t 2

- 75.0 = 12sin300 t + 21

(-9.81) t 2

t = 4.50 s

Note: solving for time in one of the dimensions, vertical or horizontal, automatically gives you the time for the other dimension!

(b) The range of the stone also implies the maximum horizontal

displacement of the stone sx = ux t sx = 12 cos300 (4.5) = 46.8 m Remember: 4. Check your answer: SUM

θθ

V

V

V

V Fig 6.4

75.0 m

sx

12 m s-1

30°


Do you know why sy is negative?

Consider horizontal motion to find the range.

2. Choose the most appropriate formula. 3. Substitute

values into formula and solve.

Fig 6.5



23

030

20 m s-1

Fig 6.6

Fig 6.7

X

QED 23 There are two TPJC boys who threw a ball each from the top of the building, shown in Fig 6.6. One of them threw at 20 m s-1 at an angle of 300 below the horizontal. The other threw vertically downwards. Given that they threw their ball at the same time and the balls touched the floor at the same time. Determine the initial velocity of the vertical ball thrown?

A 20 B 20 sin 300 C 20 cos 300

D 20 tan 300 QED 24 Consider Fig 6.7: A stone is projected with an initial velocity of 20 m s-1 at angle of 30° above the horizontal from a cliff top which is 10 m above the sea level. Determine the velocity of the ball at the highest point, marked X? A 20 ms-1

B 0 ms-1 C 20 cos 300 ms-1 D 20 sin 300 ms-1

General Problem Solving tips for Projectile Motion questions (in the absence of air resistance and thrust force) 1 “Divide and Conquer”: Consider Vertical and Horizon tal components separately!

� Motion in two dimensions can be solved by breaking the problem into two interconnected one-dimensional problems. (Recall, forces have no perpendicular effects).

� Hence, projectile motion can be divided into a vertical motion problem and a horizontal motion problem.

2 Vertical : The usual free fall with acceleration g

� The vertical motion of a projectile is exactly that of an object dropped or thrown straight up or down.

� The vertical acceleration is therefore the famous g as a result of the gravitational force acting on it.

3 Horizontal : No acceleration, so simple!

� Since a projectile has no thrust force and air resistance is neglected, there are no forces acting in the horizontal direction and thus no horizontal acceleration.

� Therefore, analysing the horizontal motion of a projectile is the same as solving a constant velocity problem. Not very difficult!

4 Let time do the connection!

� Vertical motion and horizontal motion are connected through the variable time t (clock reading).

� The time interval from the launch of the projectile to the time it hits the target is the same for vertical and horizontal motion.

� Therefore, solving for time in one of the dimensions, vertical or horizontal, automatically gives you the time for the other dimension!

10.0 m

sx

20 m s-1

30°



24

+q -q

acceleration

acceleration

+ + + + +

- - - - -


(d) Other examples: Free fall on Moon, Jupiter where g is less than 9.81.

In the cases of projectile considered previously, we take g as 9.81 m s-2 since projection is done near the surface of the earth. What happens when projection is done near the moon or some other planets? Obviously, the object will experience an acceleration that is not equal to 9.81 m s-2. Consider an object projected from the surface of the moon, where the acceleration on the surface of the moon is one-sixth that on the surface of the earth. Perhaps the only difference is that g becomes about 1.52 m s-2. QED 25 An object is projected horizontally at 20 m s-1 at the top of the cliff 50m high on the surface of the earth. The time taken to reach the ground is T and the range is X. If the same experiment is done on the surface of the moon. What can be said about the time taken to reach the ground and the range? A shorter than T and shorter than X B shorter than T and further than X C longer than T and shorter than X D longer than T and longer than X


(e) Other examples: Charged particles in electric field

For an object in a uniform gravitational field, we take the acceleration of an object to be a constant Likewise, in other uniform fields such as uniform electric field, a charged object will also experience constant acceleration. However, its value may not be 9.81 m s-2 but very large, say about 1012 - 1014 m s-2. At this moment, we will not touch on the direction of the electric field. We just need deal with the kinematics of it. We just need to know the basics of electrical attraction or repulsion so that to help us decide whether the charged particle undergoes acceleration upwards or downwards! QED 26 An electron is projected into a uniform electric field. Which of the following indicates the direction of the electron in the field?

+ + + +

– – – –

e

charged plate

A

C

D

B

the classical mechanics project:...

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