# Kinematics in One Dimension

August 8, 2017 | Author: Apex Institute | Category: Acceleration, Velocity, Speed, Kinematics, Geometric Measurement

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Kinematics in One Dimension...

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PHYSICS Kinematics in One Dimension

Kinematics in One Dimension

KINEMATICS IN ONE DIMENSION IIT JEE Syllabus: Kinematics in one dimension (Cartesian coordinaes only), Relative velocity. 1. INTRODUCTION Mechanics is the branch of physics that focuses on the motion of objects and the forces that cause the motion to change. There are two branches of Mechanics: Kinematics and Dynamics. The word kinematics comes from the Greek word "Kinema" meaning motion. Kinematics deals with the concepts that are needed to describe motion, without any reference to forces or the cause of the motion. The word dynamics comes from the Greek word "dynamis" meaning power. Dynamics deals with the relation between the forces and the motion. Scalar quantity : Any physical quantity, which can be measured on some scale, is said to be a scalar quantity. The values of physical quantities (such as mass, distance etc.) that have only magnitude add according to the rules of arithmetic. Mass, length, time, volume, density etc., are examples of scalar quantities. Vector quantity : Any physical quantity, which has a magnitude as well as the direction, is called a vector quantity. Displacement is a vector quantity. Vectors can be represented as a straight-line segment. Vectors are added by geometric addition. NOTE: To understand a physical quantity we should know:  Definition  Measuring Technique or Formula  Scalar or vector nature  Units

2. REVIEW OF BASIC CONCEPTS Particle An object of negligible dimensions, i.e., a point mass is known as a particle. This is only mathematical idealization. The earth can be treated as a particle if we consider the motion of the earth around the sun. The radius of the earth is very small, when we compare the distance between the earth and sun. Rest and Motion : The concept of motion is a relative one and a body that may be in motion relative to one reference system, may be at rest relative to another. REST :If the position of an object does not change as time passes, it is said to be at rest with reference to observer. MOTION : If the position of the object changes as time passes, it is said to be at motion with respect to observer. Motion is Relative : While sitting in a moving train, your distances from the walls, roof and the floor of the compartment do not change. That is, with respect to the compartment, your position does not change. You are at rest with respect to the compartment. But your distance from the platform, from which you boarded the train, changes as time passes. So you are moving with respect to the platform. This means that an object can be at rest with respect to one thing and in motion with respect to some other thing at the same time. So motion is not absolute; it is relative. Translational Motion : The motion in which all the particles of a body move through the same distance in the same time is called translational motion. Examples : An automobile moving along a road, freely falling body, firing of a bullet from a gun, etc., are some of the examples of translational motion.

The motion of translation is of two types : 1. Rectilinear Motion : When a body moves in a straight line, like a ball dropped from a height, the motion described by the body is rectilinear motion. Examples of rectilinear motion : A train moving along a straight track; a car moving on a straight road. 2. Curvilinear motion : When a body moves along a curved path, the motion described by the body is called curvilinear motion. Examples of curvilinear motion: The motion of a baseball hit in the air; a car negotiating a bend; a ball thrown upward at an angle.

3 MEASURING POSITION : Referense Systems and Co-ordinate System. The motion of a particle is always described with respect to a reference system. A reference system is 'made' by taking an arbitrary point as origin and imagining a co-ordinate system to be attached to it. This co-ordinate system chosen for a given problem constitutes the reference system for it. We generally choose a coordinate system attached to the earth as the reference system for most of the problems.

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Kinematics in One Dimension

C

A

C

B

B C

A

A

A B C

B Fig 1(a) Rectillinear Motion

Fig 1(b) Curvillinear Motion Fig. 1

The location of a point on a line can be described by one coordinate; a point on a plane can be described by two coordinates; a point in a three dimensional volume can be described by three coordinates. In general, the number of coordinates equals the number of dimensions. A coordinate system consists of : 1. a fixed reference point (origin) 2. a set of axes with specified directions and scales 3. instructions that specify how to label a point in space relative to the origin and axes. For Example : Cartesian coordinate system (rectangular coordinate system) : (x, y) y y0

P(x0,y0)

x

x0

O

Fig.2 Cartesian Co-ordinate System

Position : The position of a particle moving along a straight line is the distance of a particle specified in magnitude and direction from the origin of a reference system. The line shown in figure resembles with the x - axis, the position of a particle moving along a straight line is described by x. The details of the origin, positive and negative directions are shown in Fig.3. Origin 3

0 1 2 1 Negative Positive

2

3

Fig. 3

To describe the position of the particle at a given time, we have to specify (i) Its distance from the origin, i.e., numberical value with unit (ii) Whether the particle is moving towards right seen from the origin or moving towards left seen from the origin, i.e., the positive direction or the negative direction with the sign. Illustration 1. Consider the situation shown in Figure . (a) What is the position of a particle when it is at A and when it is at B? (b) Are the two position the same? (c) Are the two distances of the particle from the origin the same ? (d) What is the position of particle at 'O'? Solution. (a) The position of the particle is x = 2 m when it is at A, and x = -2 m when it is at B. (b) The two positions are not the same. (c) The distances of the particle from the origin in the two positions are the same, and equal to 2 m. (d) The position of the particle is x = 0 m when it is at O. 4 DISTANCE TRAVELLED AND DISPLACEMENT Trajectory : When a body is displaced from one point to another, it describes a certain curve, which is called a trajectory of a moving body. We call the motion of a particle as rectilinear or straight line if its trajectory is a straight line.

3

Kinematics in One Dimension The actual length covered by a moving body in between two points, irrespective of the direction in which the body is moving, is called distance. In the figure shown alongside, if the particle performs one complete rotation then the distance covered by it is 2  r..

r

O

The shortest distance covered by a moving body in between two points, in a particular direction, is called the displacement. The 'displacement' of a particle and the 'distance travelled' by it is two different quantities. Distance travelled in a given time has just a magnitude (numerical value) and no direction, whereas the displacement has magnitude as well as direction. It is important to note that the magnitude of the displacement is not always the same as the distance travelled. If a particle moves in a plane along a zig-zag path, the distance travelled in a given time interval may be much larger than the magnitude of its displacement in the same time interval. Displacement in straight-line motion If a particle moves along a straight line, its displacement may be obtained by subtracting its initial position, x1, from its final position, x2, i.e., the displacement is given by d = x2 – x1. It indicates both the quantities magnitude and direction - the straight - line distance between the initial and final position as well as the direction of the final position seen from the initial position. Consider the following example shown in figure. Origin 3

2

1

0

A 1

B 2

3

Fig. 4

Case I Suppose that at time t1, the particle is at A, and at a later time t2, it is at B. The position x1= 2 m and x2 = 4 m. The displacement of the particle is d = x2 – x 1 = 4 m – 2 m = 2 m Case II Suppose that at time t1, the particle is at B, and at a later time t2, it is at A. The position is at x1 = 4 m and x2 = 2 m. The displacement of the particle is d = x2 – x1 = 2 m – 4 m = –2 m Case III Suppose that at time t1, the particle is at A, and at a later time t2, it is at A. The position is at x1 = 2 m and x2 = 2 m. The displacement of the particle is d = x2 – x 1 = 2 m – 2 m = 0 From the example, we see that the displacement can be positive, negative or zero. In Fig. 4, the positive direction is from left to the right. If the displacement of a particle is positive, its final position is to the right of the initial position. If the displacement is negative, the final position is to the left of the initial position. If the displacement is zero, the final position is the same as the initial position. Thus, the sign of d gives the direction of the final position as seen from the initial position of the particle. For motion of a particle along a straight line, a number along with a unit gives the displacement. The numerical value of this number gives us how far the final position is from the initial position of the particle. The sign tells us the direction. IMPORTANT NOTES Distance is the actual path (length) covered by a moving particle (or object) in a given interval of time while displacement is the change in position, i.e., a vector joining initial to final position. 1. Distance is a scalar while displacement is a vector both having same SI unit m. 2. The magnitude of displacement is equal to minimum possible distance between two positions; so Distance > |Displacement| 3. For motion between two points displacement is single-valued while distance depends on actual path and so can have many values. 4. For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position.

4

Kinematics in One Dimension 5. For a moving particle distance can never be negative or zero while displacement can be. (Zero displacement means that the body after motion has come back to initial position.) Distance > 0 but Displacement > = or < 0 6. In general, magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction.

5 UNIFORM AND NON-UNIFORM MOTION Classification of motion Uniform Motion If a body covers equal distances in any equal time intervals, then its motion is uniform. Non-uniform Motion If a body covers unequal distances in any equal time intervals, then its motion is non-uniform. Average Speed versus Instantaneous Speed Since a moving object often changes its speed during its motion, it is common to distinguish between the average speed and the instantaneous speed. The distinction is as follows. Instantaneous Speed - speed at any given instant in time. Average Speed - average of all instantaneous speeds; found simply by the following formula : Average Speed =

Total distance Total time

Average speed in different circumstances (a) Bodies covering different distances with different speeds : Average speed, vav

=

=

total distances covered d = total time taken t

s1  s 2  s 3 = t1  t 2  t 3 ......

s1  s 2  s3  ......  s1 s 2     .......  v1 v 2 

Special case : If s1 = s2 = s ; equal distances with different speeds, vav =

2s 1 1 s   v1 v 2

  

2 1 1 2 v1 v 2   or  a 1  2 [Harmonic mean] v1  v 2

(b) Bodies moving with different speeds in different time intervals : t = t1 + t2 + t3 + ......... Average speed = vav =

v1t 1  v 2 t 2  ..... t1  t 2

Special case : If t1 = t2 = t3 ...= tn = t ; equal time intervals with different speeds, vav =

(v1  v 2  ...v n )t v1  v 2  ....  [Arithmetic Mean] nt n

Illustratin 2. A car travels from city A to city B with a constant speed of 40 km/hr and returns back to city A with a constant speed of 60 km/hr. Find its average speed during its entire journey. Solution. Equal distances with different speeds

vav 

2v1v 2 v1  v2

v av 

2  40  60 40  60

5

Kinematics in One Dimension

v av 

2  40  60 100

vav  48 km / hr.

The instantaneous speed : of an object is equal to the distance travelled by it in a short time interval . Uniform speed : If an object travels equal distances in equal time intervals, it is said to move with uniform speed or constant speed. If an object moves with a uniform speed, its speed at any instant is the same as its average speed in any time interval. If it covers a distance S in a time interval t, its speed at any instant is given by the equation. v=

s or s = v t

Unit of speed : The SI unit of speed is metre/second. This is written in short as m/s. If the distance is measured in kilometers and time in hours, the speed will be in kilometer / hour, or km / h. Velocity : The velocity of an object is its displacement per unit time. Velocity has magnitude as well as direction. It is a vector quantity. So velocity is "direction-aware." When evaluating the velocity of an object, we have to keep track of direction. It would not be enough to say that an object has a velocity of 55 km/hr. For instance, we can describe an object's velocity as being 55km/hr, east. The direction of the velocity is simply the same as the direction, which an object is moving. It would not matter whether the object is speeding up or slowing down. If the object is moving rightwards, then its velocity is described as being rightwards. If an object is moving downwards, then its velocity is described as being downwards. The Velocity of an object is a quantity that gives the speed of the object as well as its direction of motion. The velocity of an object changes if the speed or the direction of motion of an object is changed. The unit of velocity is the same as that of speed. The SI unit of velocity is thus metre per second, written as m/s. We often use the unit km/h for convenience. Definition : Speed is the distance covered by a body in unit time. Velocity is the displacement of a body in unit time. Measuring Tecnique : Speed = Distance / Time ; Velocity = Displacement / Time Scalar or Vector Nature : Speed is a scalar. Velocity is vector. Units : Speed and velocity have the same units. C.G.S - cm/s M.K.S - m/s SI - m/s

Velocity of an object moving along a straight line For an object moving along a straight line, there are only two possible directions of motion. In such a case, magnitude of its velocity may be represented by the speed of the object. A positive sign is used before the magnitude of velocity if it is moving in the positive direction of the motion. If an object is moving in the negative direction, its velocity may be represented by the speed of the object and a negative sign before it. The resulting number gives the speed as well as the direction of the motion, and hence, represents velocity.

Types of velocity Uniform velocity : When a body covers equal displacements in equal intervals of time in a specified direction, however short the time intervals may be, the body is said to be moving with a uniform velcity. It means the velocity of an object does not change as times passes. This means that the object is moving along a straight line, without changing direction, with a fixed speed. The body will have a uniform velocity only if (i) it covers equal distances in equal intervals of time, i.e., the magnitude does not change and (ii) its direction remains the same. If the velocity is v, the displacement s in a time interval t is given by s = vt Variable velocity : When a body covers unequal displacements in equal intervals of time in a secified direction or equal displacements in equal intervals of time, but its direction changes, then the body is said to be moving with a variable velocity. If a car covers 100 m in 1 minute from O to A and 200 m in 1 minute from A to B and so on, it is covering unequal distances in equal intervals of time. Thus we can say that the car is moving with variable velocity or non-uniform velocity. Average velocity : The displacement of a body per unit time, when the body is actually moving with variable velocity, is called average velocity.

6

Kinematics in One Dimension Average velocity =

Change in Position Displacement = Total time Time

vav =

s t

Instantaneous Velocity : A velocity at one specific instant of time or at one specific point in the path. Such a velocity is called instantaneous velocity, and it needs to be defined carefully. The instantaneous velocity at the first point can then be defined as the value that the average velocity approaches when the second point is taken closer and closer to the first. v = lim

Δt  0

Δs ds  Δt dt

Regarding average speed and average velocity it is worth noting that : 1. Average speed is a scalar while average velocity is a vector both having same units (m/s). 2. Average speed or velocity depends on time interval over which it is defined. 3. For a given time interval average velocity is single valued while average speed can have many values depending on path followed. 4. If after motion the body comes back to its initial positin, Average velocity = 0 Average speed > 0 and finite 5. For a moving body average speed can never be negative or zero (unless t =  ) while average velocity can be i.e., Average speed > 0 while Average velocity > = or < 0. 6. In general, average speed is not equal to magnitude of average velocity. However, it can be so if the motion is along a straight line without change in direction. Regarding speed and velocity, it is worth noting that : 1. Velocity is a vector while speed is a scalar, both having same units (m/s). 2. If during motion velocity remains constant throughout a given interval of time the motion is said to be uniform and for uniform motion, Instantaneous velocity = average velocity However, converse may or may not be true, i.e., if average velocity = instantaneous velocity, the motion may or may not be uniform. 3. If velocity is constant, speed = (|velocity|) will also be constant. However, may converse may or may not be true, i.e., if speed = constant, velocity may or may not be constant as velocity has direction in addition to magnitude which may or may not change, e.g., in case of straight line motion with uniform velocity, then |v| = constant and so speed = constant While in case of uniform circular motion, Speed = constant but velocity is not constant due to change in direction. 4. Velocity can be positive or negative as it is a vector but speed can never be negative. Acceleration of an Object moving along a Straight Line : The final mathematical quantity discussed is acceleration. An often-confused quantity is acceleration. Acceleration has to do with changing how fast an object is moving. If an object is not changing its velocity, then the object is not accelerating. Anytime an object's velocity is changing, that object is said to be accelerating; it has acceleration. Acceleration is defined as the change in velocity per unit time. It is a vector quantity. It is represented by the symbol 'a'. Mathematically a=

v-u t

Where a = acceleration, v = final velocity,

u = initial velocity t = time

7

Kinematics in One Dimension The SI unit of acceleration is m/s2. If the acceleration of a body is 'a' m/s2, it means the velocity of the body is increasing by 'a' units for every second.

Types of Acceleration : Uniform Acceleration or constant acceleration : If the velocity of the body changes equal amounts in equal intervals of time, then it is said to be in uniform acceleration. This is reffered to as a constant acceleration since the velocity is changing by a constant amount each second. An object with a constant acceleration should not be confused with an object with a constant velocity. If an object is changing its velocity - whether by a constant amount or a varying amount - then it is an accelerating object. And an object with a constant velocity is not accelerating. If the velocity of a uniformly accelerating body changes from u to v, then its acceleration is given by a=

v u t

Rearranging the above formula, we will get the following equation : v = u + at The Direction of the Acceleration Vector Since acceleration is a vector quantity, it will always have a direction associated with it. The direction of the acceleration vector depends on two things : ˜ Whether the object is speeding up or slowing down ˜ Whether the obect is moving in the +ve or –ve direction.

The general Rule of thumb is : If an object is slowing down, then its acceleraton is in the opposite direction of its motion. Deceleration : If the speed of a particle decreases wth time, we say that it is decelerating, or it has deceleration or retardation. The acceleration of the body is opposite to that of velocity, and then the body decelerates. Non-uniform acceleration : If the velocity of the body changes unequal amounts in equal intervals of time, then it is said to be in non-uniform acceleration. Average acceleration : The average acceleration of any object over a given interval of time can be calculated using the equation a=

v change in velocity = time interval t

Illustration 3. A bird flies north at 20 m/s for 15 s. It rests for 5 s and then flies south at 25 m/s for 10 s. For the whole trip find, (a) the average speed (b) the average velocity (c) the average acceleration Solution. Distance travelled towards north = AC = 20 m/s  15 s = 300 m. Distance travelled towards south = CB = 25 m/s  10 s = 250 m. C

path B

A

Average Speed =

distance 300  250 = m/s = 18.34 m/s. time 15  5  10

Average velocity =

displacemn et 300  250 = = 1.67 m/s time 15  5  10

8

Kinematics in One Dimension Average Acceleration, aav =

v v f  v i (-25) - (+20)  = m/s2 = – 1.5 m/s2. 30 t t

Instantaneous acceleration : Instantaneous acceleration is given by the following formula : a = Lt

t 0

v dv = t dt

Regarding the acceleration it is worth noting that 1. It is a vector with SI units (m/s2) 2. If acceleration is zero, velocity will be constant and the motion will be uniform. However, if acceleration is constant, acceleration is uniform but motion is non-uniform and if acceleration is not constant both motion and acceleration are non-uniform. 3. Acceleration can be positive or negative.

6. EQUATIONS OF KINEMATICS AND PROBLEM-SOLVING : The Equations of Kinematics s = ut +

1 2 at 2

v2 - u2 = 2as v = u + at s=

u v ´t 2

There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol S stands for the displacement of the object The symbol t stands for the time for which the object moved. The symbol 'a' stands for the acceleration of the object. And the symbol u stands for the initial velocity value of the object. The symbol v stands for final velocity value of the object. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to determine unknown information about an object's motion if other information is known. Illustration 4. Derive a formula for finding the displacement of a body in the nth second. Solution. Let the displacement of a body in nth second be Sn. The displacement for n seconds = S1. The displacement for (n-1) seconds = S2. For an initial velocity u and constant acceleration a, the formula for finding the displacement is S = ut +

S1 = un +

1 2 at 2 1 2 an 2

S2 = u(n-1) +

1 a (n-1)2 2

But S1 – S2 = displacement of body in the nth second i.e., Sn. Sn = S1 – S2 = un +

1 2 1 an – [u(n-1) + a (n-1)2] 2 2

9

Kinematics in One Dimension 2

2

 Sn = un + 0.5an – [un – u + 0.5a (n +1–2n)]  Sn = un + 0.5an2 – un + u – 0.5 an2 – 0.5a + an  Sn = u - 0.5a + an = u + a (n – 0.5) 

Sn = u + a (n –

1 ) 2

Therefore the displacement of the body in the nth second is, Sn = u + a (n –

1 ) 2

Special Cases of Equations of Kinematics 1. If a body starts from rest, then the initial velocity, u = 0 and the equations of kinematics are : v = at ;

S=

1 2 at ; 2

v2 = 2aS

2. If a body comes to rest, then v = 0 and acceleration becomes negative. Then the equations of kinematics are : u – at = 0 ;

S = ut –

1 2 at ; u2 = 2aS 2

3. If a body has uniform velocity then change in velocity is zero, i.e., a = 0, and the equation of kinematics are : v=u; S = ut Illustration 5. A particle starting from rest, covers a distance s, moving along a straight line and comes to rest. It has a uniform aceleration and retardation, in its two parts of journey. If v be the maximum velocity attained by it, show that the total time of motion is 2s/v. A

t1

v

S1

t2

B

S2

Solution. Let s1 and s2 be the displacement of the particle in the first and second parts of the journey respectively, so that s1 + s2 = s (figure.) If t1 and t2 be the corresponding times, then u+ν s=  t  2 

Using

We have

s1 = (s1 + s2) =

(0 + v)t1 (ν + 0)t 2 and s2 = 2 2

ν (t + t ) 2 1 2

2s ν

i.e.,

t1 + t2 =

Total time =

.......

[ s1 + s2 = s]

2s ν

Illustration 6. A train stopping at two stations d distance apart takes t time on the journey from one station to the other. Assuming that its motion is first that of uniform acceleration  and then that of uniform retardation  , show that

1 1 t2     2d

Solution. Let A and B be the stations d distance apart. Further let the train accelerate from A to C, over a distance s1 and then decelerate from C to B, over a distance s2. Also, let v be the maximum velocity attained. If t1 be the time from A to

10

Kinematics in One Dimension C, then

 uv   using s   t  2   

v 0+v s1 =   t1 = 2 t1 2  

.............. (1)

0 t 2  2 

and t2 be the time from C to B, then s2 = 

s2 =

v t2 2

............... (2)

Adding equaitons (1) and (2), we get s1 + s2

d

=

=

v (t + t ) 2 1 2

v (t) 2

..............(3)

Again for the motion from A to C, v =  t1

..............(4) [ u  0]

and for motion from C to B,  = b t2

................(5) [ u  0]

 1 1   =t +t =t   1 2

From equaitons (4) and (5),

v 

Hence, from equation (3), v =

2d t

 2d   1 1   t  α + β  = t   

1 1 t2     2d

7. EQUATIONS OF KINEMATICS AND FREE FALL The conceptual characteristics of freely falling object are described as follows : An object in free fall experiences an acceleration of -9.8 m/s2. (The  sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematics equations is  9.8 m/s2 for any freely falling object. If an object is merely dropped (as opposed to being thrown) from an elevated height to the ground below, then the initial velocity of the object is 0 m/s. If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematics equations; for example, the final velocity (v) after travelling to the peak would be assigned a value of 0 m/s. If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity, which it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of + 30 m/s will have a downward velocity of  30 m/s when it returns to the same height. We can use equations of motion for freely falling bodies. If a body is falling from a height 'h' with an acceleration due to gravity 'g' then the equations of motion becomes : v = u + gt h = ut +

1 2 gt 2

v2  u2 = 2gh If the initial velocity of freely falling bodies is zero i.e. u = 0 Then, the equations of motion become : v = gt

11

Kinematics in One Dimension

1 2 gt 2

h=

v2 = 2gh If a body is projected vertically upwards, then what are the equations of motion? We are considering the acceleration due to gravity of a freely falling body as positive, we take a negative value for a body projected upwards, and the equations of motion are v = u  gt h = ut 

1 2 gt 2

v2  u2 = –2gh When a body is projected upwards with a velocity 'u', it is raised to a certain height and then it starts falling down. Time of flight : s  ut 

1 2 gt 2

 s  0

t=T 

T

2u g

Time of descent = time of ascent =

u . g

8. MOTION ON AN INCLINED PLANE : (i) Suppose a body of mass m is allowed to move down the plane with an initial velocity zero. i.e., from the position of rest. Let the plane is inclined at an angle  w.r.t. horizontal direction and the body is at a height h at initial instant (t = 0) (Fig)

R h  mg sin s

mg cos mg

 fig 9

Hence in such a case, u = 0, a = g sin  = constant and equations of motion are : v = (g sin  )t s=

1 (gsin  )t2 and v2 =2 (g sin  )s 2

where s is the distance travelled by the body along the plane in time t, when it reaches at the bottom. (ii)

If s is given, then : t2 =

2s 2 g sin  or t  s.

If t' is the time taken to travel

1 s th of the total distance, then : t'2  4 4

2

t  t' 1 Hence,    or t' =  t 4 2

12

Kinematics in One Dimension i.e., in first halftime the body moves

1 3 th of the total distance while in next half it moves th of the total distance 4 4

on an inclined plane. (iii) Time taken to move down an inclined plane s=

or

t=

h h , = sin  or s = s sin 

As

(iv)

1 (g sin  ) t2 2

Because, v2 = 2g sin  . s and s =

Hence

v2 = 2g sin 

(2s / g sin )

Hence t =

1 sin 

2h g

h sin 

h = 2 gh sin 

or

v=

[ 2gh ]

Thus, speed is independent of  and depend on height h. NOTE : (i) If two bodies allowed to move down from same height along two inclined planes having different gradients  1 and  2 (  1 >  2) (Fig.) then final velocity acquired by both will be same i.e., v1 = v2 =

[ 2gh ]

h 1

2 Fig.10

(ii)

t

1 sin 

2h 1  t g sin 

Because  1 >  2, hence t1 < t2 (v) If we compare the behaviour of two identical bodies moving under gravity in different conditions i.e., one falling vertically downwards and other moving along an inclined plane from same height h, then in case of first body :

vF =

[ 2gh ] , aF = g

and

tF =

2h g

but in the case of second body,

vs =

1 2h [ 2gh ] , as = g sin  and ts = sin  g

Hence, vF = vs, aF > as and tF < ts. Illustration 7. A body starting from rest slide on an inclined plane of length s as shown in fig. the time taken to cover half the distance. Sol. When a body slides on an inclined plane, acceleration of the body a = g sin  = constant. So, equation of motion can be applied. From equaiton of motion, here s = 0 + 1/2 (g sin  ) t2

[as u = 0]

13

........... (1)

Kinematics in One Dimension R

S mg cos

h

mg 

2

t

s (s / 2 )  t '     i.e., If s' = , s 2 t

t' =

2

 0.7 t

i.e., in about 0.7 times the time of descent the body covers half of the total distance. Illustration 8. A mass A is released from the top of a frictionless inclined plane 18 m long and reached the bottom 3 s. later. At the instant when A is released, a second mass B is projected upwards along the plane from the bottom with a certain initial velocity. The mass B travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with A. The two masses do not collide. Find the acceleration and initial velocity of B. How far up the inclined plane does B travel? Sol. Acceleration down the plane, a = g sin  = constant for two masses. A starts from rest, 18 = 0  3 +

1 a  32. 2

i.e., a = 4 m/s2. ................. (1) Now if B is projected up the plane with a velocity u, it will go up the plane till its velocity becomes zero i.e., (v = 0) as for it acceleration will be down the plane. So from equation of motion, time taken to go up the plane. u=0 v=0 C

18m

B

0 = u – at1 i.e., t1 = (u / a) ............ (2) and from equation of motion distance moved up the plane 0 = u2 – 2as, i.e., s = (u2 / 2a) ..............(3) Now the body B will slide down the plane and so time taken by it to slide a distance s down the plane starting from rest at C, from equation of motion, s = 0 + 1/2 at2 2. i.e.,

t2 =

2  u2  u  u2      s   a  2a  a  2a 

..............(4)

But it is given that body B reaches to ground with A, i.e.,

t1 + t2 = 3

or

u=

or

2u = 3 [as from Eqns. (2) and (4), t1 = t2 = u/a] a

3 3 a=  4 = 6 m / s. 2 2

[as from Eqns. (1) a = 4]

So the acceleration of B is 4 m/s2 down the plane while its velocity is 6 m/s up the plane. (The distance travelled by B up the plane from Eqns. (3) i.e., s = (u2 / 2a) = 4.5 m) Illustration 9. A body is projected from the bottom of a smooth inclined plane with a velocity of 10 m/s. If it be just sufficient to carry it to the top in 2s, find the inclination and height of the plane. [g = 10 m/s2] Sol. Let AB be the inclined plane, with angle of inclination  and height h. (figure)

14

Kinematics in One Dimension Now, for the motion along the plane up, t=25 B

h u=10 m/s 

A

C

Using

u = 10; a = –10 sin  ; v = 0; t = 2s v = u + at

we have

0 = 10 – (10 sin  ) (2)  sin  =

10   = 30o 20

Now, using u2 = 2gh, we have h =

102 =5m 2  10

Illustration 10. A particle starts from rest from the top of a smooth inclined plane of a given base. Find the inclination of the plane to the horizontal, for the time of fall to be least. Sol. Let 'x' be the base of the inclined plane and '  ' any angle of inclination; Length of the incline AB = x sec  . (figure) B

S

A

 x

C

Now, for the motion along BA, s = x sec  ; a = g sin  ; u = 0; t = ? Using,

s = ut +

t=

1 2 1 at ; x sec  = g sin  t2 2 2

2x = g sin  cos 

4x ( 2 sin  cos  = sin 2  ) g sin 2 

For minimum time of fall, sin 2θ should be maximum, i.e., sin 2  1

2  900

;

  450

15

Kinematics in One Dimension

PROBLEM - SOLVING TECHNIQUE Motion of a particle in a straight line To determine the motion of a particle on an axis : 1. Select the origin of the x axis and the direction in which the axis is positive. Usually, it is best to place the particle at the origin at time t = 0; then, x0 = 0. A sketch showing the axis is helpful. It is also helpful to show the position of the particle at some later time on the same sketch. 2. Take the positive direction of the velocity and the acceleration in the same direction as the axis. 3. State the problem in your own words and then translate into symbols and equations. For example, the problem might ask "When does the particle arrive at x = 10m?" You might paraphrase this is as "What is the value of time t when x = 10m?" Or, the problem may read "Where is the particle when its velocity is v = 10m/s?" which you then paraphrase as "What is the value of x when v = 10m/s?". 4. Make a list of given quantities, such as the initial position x0, the initial velocity v0, and the initial acceleration a0. 5. Identify the unknowns, and label them with symbols. Use kinematic Eqs. to relate the unknowns to the knowns. Solve for the unknowns in terms of the known values. Sometimes you will have to solve simultaneous equations for the unknowns. 6. Examine your results carefully to see whether they make sense.

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