Chapter Four

Motion in a Plane

4.1 Introduction

In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line. We found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two directions are possible. But in order to describe motion of an object in two dimensions (a plane) or three dimensions (space), we need to use vectors to describe the above-mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to add, subtract and multiply vectors ? What is the result of multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and acceleration in a plane. We then discuss motion of an object in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail the projectile motion. Circular motion is a familiar class of motion that has a special significance in daily-life situations. We shall discuss uniform circular motion in some detail.

The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions.

4.2 Scalars and vectors

In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers*. For example, if the length and breadth of a rectangle are 1.0 m and 0.5 m respectively, then its perimeter is the sum of the lengths of the four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = 3.0 m. The length of each side is a scalar and the perimeter is also a scalar. Take another example: the maximum and minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of 2.7 kg, then its volume is 10–3 m3 (a scalar) and its density is 2.7×103 kg m–3 (a scalar).

A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition. So, a vector is specified by giving its magnitude by a number and its direction. Some physical quantities that are represented by vectors are displacement, velocity, acceleration and force.

To represent a vector, we use a bold face type in this book. Thus, a velocity vector can be represented by a symbol v. Since bold face is difficult to produce, when written by hand, a vector is often represented by an arrow placed over a letter, say . Thus, both v and represent the velocity vector. The magnitude of a vector is often called its absolute value, indicated by |v| = v. Thus, a vector is represented by a bold face, e.g. by A, a, p, q, r, ... x, y, with respective magnitudes denoted by light face A, a, p, q, r, ... x, y.

4.2.1 Position and Displacement Vectors

To describe the position of an object moving in a plane, we need to choose a convenient point, say O as origin. Let P and P′ be the positions of the object at time t and t′, respectively [Fig. 4.1(a)]. We join O and P by a straight line. Then, OP is the position vector of the object at time t. An arrow is marked at the head of this line. It is represented by a symbol r, i.e. OP = r. Point P′ is represented by another position vector, OP′ denoted by r′. The length of the vector r represents the magnitude of the vector and its direction is the direction in which P lies as seen from O. If the object moves from P to P′, the vector PP′ (with tail at P and tip at P′) is called the displacement vector corresponding to motion from point P (at time t) to point P′ (at time t′).

Fig. 4.1 (a) Position and displacement vectors. (b) Displacement vector PQ and different courses of motion.

It is important to note that displacement vector is the straight line joining the initial and final positions and does not depend on the actual path undertaken by the object between the two positions. For example, in Fig. 4.1(b), given the initial and final positions as P and Q, the displacement vector is the same PQ for different paths of journey, say PABCQ, PDQ, and PBEFQ. Therefore, the magnitude of displacement is either less or equal to the path length of an object between two points. This fact was emphasised in the previous chapter also while discussing motion along a straight line.

4.2.2 Equality of Vectors

Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.**

Fig. 4.2 (a) Two equal vectors A and B. (b) Two vectors A� and B� are unequal though they are of the same length.

Figure 4.2(a) shows two equal vectors A and B. We can easily check their equality. Shift B parallel to itself until its tail Q coincides with that of A, i.e. Q coincides with O. Then, since their tips S and P also coincide, the two vectors are said to be equal. In general, equality is indicated as A = B. Note that in Fig. 4.2(b), vectors A′ and B′ have the same magnitude but they are not equal because they have different directions. Even if we shift B′ parallel to itself so that its tail Q′ coincides with the tail O′ of A′, the tip S′ of B′ does not coincide with the tip P′ of A′. 

4.3 Multiplication of vectors by real numbers

Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A :

|λ A| = λ |A | if λ > 0.

For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. 4.3(a).

Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|.

Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig 4.3(b). 

Fig. 4.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. (b) Vector A and resultant vectors after multiplying it by a negative number –1  and –1.5.

The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.

4.4 Addition and subtraction of vectors — graphical method

As mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently, the parallelogram law of addition. We shall now describe this law of addition using the graphical method. Let us consider two vectors A and B that lie in a plane as shown in Fig. 4.4(a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors. To find the sum A + B, we place vector B so that its tail is at the head of the vector A, as in Fig. 4.4(b). Then, we join the tail of A to the head of B. This line OQ represents a vector R, that is, the sum of the vectors A and B. Since, in this procedure of vector addition, vectors are arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition. If we find the resultant of B + A as in Fig. 4.4(c), the same vector R is obtained. Thus, vector addition is commutative:

Fig. 4.4 (a) Vectors A and B. (b) Vectors A and B added graphically. (c) Vectors B and A added graphically. (d) Illustrating the associative law of vector addition.

A + B = B + A                                                         (4.1)

The addition of vectors also obeys the associative law as illustrated in Fig. 4.4(d).  The result of adding vectors A and B first and then adding vector C is the same as the result of adding B and C first and then adding vector A :

(A + B) + C = A + (B + C)                                      (4.2)

What is the result of adding two equal and opposite vectors ? Consider two vectors A and –A shown in Fig. 4.3(b). Their sum is A + (–A). Since the magnitudes of the two vectors are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented by 0 called a null vector or a zero vector :

A – A = 0 |0|= 0                                                      (4.3)

Since the magnitude of a null vector is zero, its direction cannot be specified.

The null vector also results when we multiply a vector A by the number zero. The main properties of 0 are :

A + 0 = A

λ 0 = 0

0 A = 0                                                                   (4.4)

What is the physical meaning of a zero vector? Consider the position and displacement vectors in a plane as shown in Fig. 4.1(a). Now suppose that an object which is at P at time t, moves to P′ and then comes back to P. Then, what is its displacement? Since the initial and final positions coincide, the displacement is a “null vector”. 

Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors A and –B :

A – B = A + (–B)                                                     (4.5)

Fig. 4.5 (a) Two vectors A and B, – B is also shown. (b) Subtracting vector B from vector A – the result is R2. For comparison, addition of vectors A and B, i.e. R1 is also shown.

 

It is shown in Fig 4.5. The vector –B is added to vector A to get R2 = (A – B). The vector R1 = A + B is also shown in the same figure for comparison.  We can also use the parallelogram method to find the sum of two vectors. Suppose we have two vectors A and B. To add these vectors, we bring their tails to a common origin O as shown in Fig. 4.6(a). Then we draw a line from the head of A parallel to B and another line from the head of B parallel to A to complete a parallelogram OQSP. Now we join the point of the intersection of these two lines to the origin O. The resultant vector R is directed from the common origin O along the diagonal (OS) of the parallelogram [Fig. 4.6(b)]. In Fig.4.6(c), the triangle law is used to obtain the resultant of A and B and we see that the two methods yield the same result. Thus, the two methods are equivalent. 

Fig. 4.7

The direction θ that R makes with the vertical is given by

Or,

Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19o with the vertical towards the east. 

4.5 Resolution of vectors

Let a and b be any two non-zero vectors in a plane with different directions and let A be another vector in the same plane(Fig. 4.8). A can be expressed as a sum of two vectors — one obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have

A = OP = OQ + QP                                               (4.6)

But since OQ is parallel to a, and QP is parallel to b, we can write :

OQ = λ a, and QP = µ b                                        (4.7)

where λ and µ are real numbers.

Therefore, A = λ a + µ b                                         (4.8)

Fig. 4.8 (a) Two non-colinear vectors a and b.  (b) Resolving a vector A in terms of vectors a and b.

We say that A has been resolved into two component vectors λ a and µ b along a and b respectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors – all the three lie in the same plane.  It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. Unit vectors along the x-, y- and z-axes of a rectangular coordinate system are denoted by

and
, respectively, as shown in Fig. 4.9(a).

Since these are unit vectors, we have

(4.9)

These unit vectors are perpendicular to each other. In this text, they are printed in bold face with a cap (^) to distinguish them from other vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of only two unit vectors. If we multiply a unit vector, say by a scalar, the result is a vector

. In general, a vector A can be written as

(4.10)

where is a unit vector along A. 

We can now resolve a vector A in terms of component vectors that lie along unit vectors and

Consider a vector A that lies in x-y plane as shown in Fig. 4.9(b). We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 4.9(b), and get vectors A1 and A2 such that A1 + A2 = A. Since A1 is parallel to

and A2 is parallel to

, we have : 

(4.11)

where Ax and Ay are real numbers.

(4.12)

This is represented in Fig. 4.9(c). The quantities Ax and Ay are called x-, and y- components of the vector A. Note that Ax is itself not a vector, but 

is a vector, and so is

. Using simple trigonometry, we can express Ax and Ay in terms of the magnitude of A and the angle θ it makes with the x-axis :

Ax = A cos θ

Ay = A sin θ                                                                                 (4.13)

As is clear from Eq. (4.13), a component of a vector can be positive, negative or zero depending on the value of θ.

Now, we have two ways to specify a vector A in a plane. It can be specified by :

(i) its magnitude A and the direction θ it makes with the x-axis; or

(ii) its components Ax and Ay

If A and θ are given, Ax and Ay can be obtained using Eq. (4.13). If Ax and Ay are given, A and θ can be obtained as follows :

= A2

Or,

And

$\tan \theta = \frac{A_y}{A_x}, \theta ={\tan }^{-1} \frac{A_y}{A_x}$

So far we have considered a vector lying in an x-y plane. The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions. If α, β, and γ are the angles* between A and the x-, y-, and z-axes, respectively [Fig. 4.9(d)], we have

(d)

In general, we have

The magnitude of vector A is

A position vector r can be expressed as

(4.17)

where x, y, and z are the components of r along x-, y-, z-axes, respectively.

4.6 Vector addition – analytical method

Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components. Consider two vectors A and B in x-y plane with components Ax, Ay and Bx, By :

(4.18)

Let R be their sum. We have

R = A + B

(4.19a)

Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (4.19a) as convenient to us :

Since

we have,

Thus, each component of the resultant vector R is the sum of the corresponding components of A and B.

In three dimensions, we have

with

This method can be extended to addition and subtraction of any number of vectors. For example, if vectors a, b and c are given as

(4.23a)

then,  a vector T = a + b – c has components :

Answer Let OP and OQ represent the two vectors A and B making an angle θ (Fig. 4.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector R :

R = A + B

SN is normal to OP and PM is normal to OS.

From the geometry of the figure,

OS2 = ON2 + SN2

but ON = OP + PN = A + B cos θ SN = B sin θ

OS2 = (A + B cos θ)2 + (B sin θ)2

or, R2 = A2 + B2 + 2AB cos θ

In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ

Therefore, R sin α = B sin θ

or,

Similarly,

PM = A sin α = B sin β

or,

Combining Eqs. (4.24b) and (4.24c), we get

Using Eq. (4.24d), we get:

where R is given by Eq. (4.24a).

or,

Equation (4.24a) gives the magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction.  Equation (4.24a) is known as the Law of cosines and Eq. (4.24d) as the Law of sines. 

Answer The vector vb representing the velocity of the motorboat and the vector vc representing the water current are shown in Fig. 4.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

Fig. 4.11

We can obtain the magnitude of R using the Law of cosine :

To obtain the direction, we apply the Law of sines

or,

4.7 Motion in a plane

In this section we shall see how to describe motion in two dimensions using vectors.

4.7.1 Position Vector and Displacement

The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. 4.12) is given by

where x and y are components of r along x-, and y- axes or simply they are the coordinates of the object.

(a)

(b)

Fig. 4.12 (a) Position vector r. (b) Displacement ∆r and average velocity v of a particle.

Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′ [Fig. 4.12(b)]. Then, the displacement is : 

∆r = r′ – r                                                              (4.25)

and is directed from P to P′. 

We can write Eq. (4.25) in a component form:

 

where ∆x = x ′ – x, ∆y = y′ – y                              (4.26)

 

Velocity

The average velocity of an object is the ratio of the displacement and the corresponding time interval :

Since

, the direction of the average velocity is the same as that of ∆r (Fig. 4.12). The velocity(instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero :

 

(4.28)

Fig. 4.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction ofis parallel to the line tangent to the path.

The meaning of the limiting process can be easily understood with the help of Fig 4.13(a) to (d).  In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and P3 represent the positions of the object after times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the displacements of the object in times ∆t1, ∆t2, and ∆t3, respectively. The direction of the average velocity

is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 4.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.

We can express v in a component form :

(4.29)

Or,

. where

(4.30a)

So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find vx and vy.

The magnitude of v is then 

 

 

 

(4.30b)

and the direction of v is given by the angle θ :

(4.30c)

vx, vy and angle θ are shown in Fig. 4.14 for a velocity vector v at point p.

 

Acceleration

The average acceleration a of an object for a time interval ∆t moving in x-y plane is the change in velocity divided by the time interval :

(4.31a)

Or,

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero :

(4.32a)

Since

we have

Or,

(4.32b)

where, 

(4.32c)*

As in the case of velocity, we can understand graphically the limiting process used in defining acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 4.15(a) to (d). P represents the position of the object at time t and P1, P2, P3 positions after time ∆t1, ∆t2, ∆t3, respectively (∆t1> ∆t2>∆t3). The velocity vectors at points P, P1, P2, P3 are also shown in Figs. 4.15 (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of ∆v. We see that as ∆t decreases, the direction of ∆v changes and consequently, the direction of the acceleration changes. Finally, in the limit ∆t g0 [Fig. 4.15(d)], the average acceleration becomes the instantaneous acceleration and has the direction as shown.

Fig. 4.15 The average acceleration for three-time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. 

Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them.

Example 4.4 The position of a particle is given by

where t is in seconds and the coefficients have the proper units for r to be in meters. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at
t = 1.0 s.

Answer

a = 4.0 m s–2 along y-direction

At t = 1.0 s,

It’s magnitude is

and direction is

with x-axis.

4.8 Motion in a plane with constant acceleration

Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be v0 at time t = 0 and v at time t.

Then, by definition

Or,

(4.33a)

In terms of components :

(4.33b)

Let us now find how the position r changes with time. We follow the method used in the one-dimensional case. Let ro and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then, over this time interval t, the average velocity is (vo + v)/2. The displacement is the average velocity multiplied by the time interval :

(4.34a)

Or,

It can be easily verified that the derivative of Eq. (4.34a), i.e. 

gives Eq.(4.33a) and it also satisfies the condition that at t=0, r = ro. Equation (4.34a) can be written in component form as

(4.34b)

One immediate interpretation of Eq.(4.34b) is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical situations, as we shall see in section 4.10 for projectile motion.

Answer From Eq. (4.34a) for r0 = 0, the position of the particle is given by

Therefore,

Given x (t) = 84 m, t = ?

5.0 t + 1.5 t2 = 84 ⇒ t = 6 s

At t = 6 s, y = 1.0 (6)2 = 36.0 m 

At t = 6 s,

v = 23.0 î +12.0 ĵ 

speed

.

4.9 Relative velocity in two dimensions

The concept of relative velocity, introduced in section 3.7 for motion along a straight line, can be easily extended to include motion in a plane or in three dimensions. Suppose that two objects A and B are moving with velocities vA and vB (each with respect to some common frame of reference, say ground

.). Then, velocity of object A relative to that of B is :

vAB = vA – vB                                                         (4.35a)

and similarly, the velocity of object B relative to that of A is :

vBA = vB – vA 

Therefore, vAB = – vBA                                           (4.35b)

and,

This relative velocity vector as shown in Fig. 4.16 makes an angle θ with the vertical. It is given by

4.10 Projectile motion

As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems (1632).

In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle θo with the x-axis as shown in Fig. 4.17.

After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:

 a = -g  ĵ 

Or, ax = 0, ay = – g                                                       (4.36)

The components of initial velocity vo are :

vox = vo cos θo

voy= vo sin θo                                                              (4.37)

If we take the initial position to be the origin of the reference frame as shown in Fig. 4.17, we have :

xo = 0, yo = 0

Then, Eq.(4.34b) becomes :

x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2                                  (4.38)

The components of velocity at time t can be obtained using Eq.(4.33b) :

vx = vox = vo cos θo

vy = vo sin θo – g t                                                                                             (4.39)

Equation (4.38) gives the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed vo and projection angle θo. Notice that the choice of mutually perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in a simplification.  
One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y- component changes, like an object in free fall in vertical direction. This is shown graphically at few instants in Fig. 4.18. Note that at the point of maximum height, vy= 0 and therefore,

Equation of path of a projectile

What is the shape of the path followed by the projectile? This can be seen by eliminating the time between the expressions for x and y as given in Eq. (4.38). We obtain:

Now, since g, θo and vo are constants, Eq. (4.40) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 4.18). 

Fig. 4.18 The path of a projectile is a parabola.

Time of maximum height

How much time does the projectile take to reach the maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (4.39):

vy = vo sinθo – g tm = 0

Or, tm = vo sinθo /g                                                                         (4.41a)

The total time Tf during which the projectile is in flight can be obtained by putting y = 0 in Eq. (4.38). We get :

Tf = 2 (vo sin θo )/g                                                                          (4.41b)

Tf is known as the time of flight of the projectile. We note that Tf = 2 tm , which is expected because of the symmetry of the parabolic path.

Maximum height of a projectile

The maximum height hm reached by the
projectile can be calculated by substituting t = tm in Eq. (4.38) :

Or,

(4.42)

Horizontal range of a projectile

The horizontal distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight Tf . Therefore, the range R is 

R = (vo cos θo) (Tf )

=(vo cos θo) (2 vo sin θo)/g

Or,

(4.43a)

Equation (4.43a) shows that for a given projection velocity vo , R is maximum when sin 2θ0is maximum, i.e., when θ0 = 450.

The maximum horizontal range is, therefore,

(4.43b)

 

 

Example 4.7 Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.

Answer For a projectile launched with velocity vo at an angle θo , the range is given by

 

Now, for angles, (45° + α) and ( 45° – α), 2θo is (90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.  

 

Example 4.8 A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s-1. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ).

Answer We choose the origin of the x-,and y-axis at the edge of the cliff and t = 0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x-, and y- components of the motion can be treated independently. The equations of motion are :

x (t) = xo + vox t

y (t) = yo + voy t +(1/2) ay t2

Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2,

vox = 15 m s-1.

The stone hits the ground when y(t) = – 490 m. 

– 490 m = –(1/2)(9.8) t2.

This gives t =10 s.

The velocity components are vx = vox and vy = voy – g t

so that when the stone hits the ground :

vox = 15 m s–1

voy = 0 – 9.8 × 10 = – 98 m s–1

Therefore, the speed of the stone is

 

Example 4.9 A cricket ball is thrown at a speed of 28 m s–1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.

Answer (a) The maximum height is given by

 

(b) The time taken to return to the same level is

Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 = 28/9.8 s = 2.9 s

(c) The distance from the thrower to the point where the ball returns to the same level is

R 

4.11 Uniform circular motion

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R as shown in Fig. 4.19. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration.

Let r and r′ be the position vectors and v and v′ the velocities of the object when it is at point P and P′ as shown in Fig. 4.19(a). By definition, velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors v and v′ are as shown in Fig. 4.19(a1). ∆v is obtained in Fig. 4.19 (a2) using the triangle law of vector addition. Since the path is circular, v is perpendicular to r and so is v′ to r′. Therefore, ∆v is perpendicular to ∆r. Since average acceleration is along ∆v

, the average acceleration  is perpendicular to ∆r. If we place ∆v on the line that bisects the angle between r and r′, we see that it is directed towards the centre of the circle. Figure 4.19(b) shows the same quantities for smaller time interval. ∆v and hence

is again directed towards the centre. In Fig. 4.19(c), ∆t→0 and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre*. Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle. Let us now find the magnitude of the acceleration.

The magnitude of a is, by definition, given by

* In the limit ∆t→0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path.

Let the angle between position vectors r and r′ be ∆θ. Since the velocity vectors v and v′ are always perpendicular to the position vectors, the angle between them is also ∆θ . Therefore, the triangle CPP′ formed by the position vectors and the triangle GHI formed by the velocity vectors v, v′ and ∆v are similar (Fig. 4.19a). Therefore, the ratio of the base-length to side-length for one of the triangles is equal to that of the other triangle. That is :

Or,

Therefore,

If ∆t is small, ∆θ will also be small and then arc PP′ can be approximately taken to be|∆r|:

Or,

Therefore, the centripetal acceleration ac is : 

Exercises

Question 4.1: State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

4.2 Pick out the two scalar quantities in the following list :

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

4.3 Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

4.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :

(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.

 4.5 Read each statement below carefully and state with reasons if it is true or false :

(a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

 4.6 Establish the following vector inequalities geometrically or otherwise :

(a) |a+b| ≤ | a| + |b|

(b) |a+b| ≥ ||a| −|b||

(c) |a−b| ≤ |a| + |b|

(d) |a−b| ≥||a| − |b||

When does the equality sign above apply?

 4.7 Given a + b + c + d = 0, which of the following statements are correct :

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of ( b + d),

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

 

NEETprep Answer

4.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

 

 

NEETprep Answer

4.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Fig. 4.21

NEETprep Answer

4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth, and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

NEETprep Answer

Exercises

Exercises

EXEMPLAR QUESTIONS

4.1 The angle between $\mathrm{A}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\mathrm{B}=\mathrm{i}-\mathrm{j}$ is
(a) 45° (b) 90° (c) –45° (d) 180°

4.2. Which one of the following statements is true?

a) a scalar quantity is the one that is conserved in a process

b) a scalar quantity is the one that can never take negative values

c) a scalar quantity is the one that does not vary from one point to another in space

d) a scalar quantity has the same value for observers with different orientations of the axes

4.3. Figure shows the orientation of two vectors u and v in the XY plane.

$\text{ If }\mathrm{u}=\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}\text{ and }\mathrm{v}=\mathrm{p}\hat{\mathrm{i}}+\mathrm{q}\hat{\mathrm{j}}$

which of the following is correct?

a) a and p are positive while b and q are negative

b) a, p, and b are positive while q is negative

c) a, q, and b are positive while p is negative

d) a, b, p, and q are all positive

4.4. The component of a vector r along the X-axis will have maximum value if

a) r is along positive Y-axis

b) r is along positive X-axis

c) r makes an angle of 45^o with the X-axis

d) r is along the negative Y-axis

4.5. The horizontal range of a projectile fired at an angle of 15^o is 50 m. If it is fired with the same speed at an angle of 45^o, its range will be

a) 60 m

b) 71 m

c) 100 m

d) 141 m

4.6. Consider the quantities of pressure, power, energy, impulse gravitational potential, electric charge, temperature, area. Out of these, the only vector quantities are

a) impulse, pressure, and area

b) impulse and area

c) area and gravitational potential

d) impulse and pressure

4.7. In a two-dimensional motion, instantaneous speed ${\mathrm{v}}_0$ is a positive constant. Then which of the following are necessarily true?

a) the average velocity is not zero at any time

b) average acceleration must always vanish

c) displacements in equal time intervals are equal

d) equal path lengths are traversed in equal intervals

4.8. In a two-dimensional motion, instantaneous speed ${\mathrm{v}}_0$ is a positive constant. Then which of the following are necessarily true?

a) the acceleration of the particle is zero

b) the acceleration of the particle is bounded

c) the acceleration of the particle is necessarily in the plane of motion

d) the particle must be undergoing a uniform circular motion

4.9. Three vectors A, B, and C add up to zero. Find which is false

a) vector (A×B)C is not zero unless vectors B, C are parallel

d) vector (A×B).C is not zero unless vectors B, C are parallel

c) if vectors A, B, C define a plane, (A×B)C is in that plane

d) (A×B).C = |A||B||C| $\to$ C² = A² + B²

4.10. It is found that |A + B| = |A|. This necessarily implies

a) B = 0

b) A, B are antiparallel

c) A, B are perpendicular

d) A.B ≤ 0

EXEMPLAR QUESTIONS

4.11. Two particles are projected in the air with speed v₀, at angles θ₁ and θ₂ to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

a) angle of the project: q₁ > q₂

b) time of flight: T₁ > T₂

c) horizontal range: R₁ > R₂

d) total energy: U₁ > U₂

4.12. A particle slides down a frictionless parabolic track starting from rest at point A. Point B is at the vertex of the parabola and point C is at a height less than that of point A. After C, the particle moves freely in the air as a projectile. If the particle reaches the highest point at P, then

a) KE at P = KE at B

b) height at P = height at A

c) total energy at P = total energy at A

d) time of travel from A to B = time of travel from B to P

4.13. The following are four different relations about displacement, velocity, and acceleration for the motion of a particle in general. Choose the incorrect one (s):

a) v_av = 1/2 [v(t₁) + v(t₂)]

b) v_av = r(t₂)-r(t₁)/t₂-t₂

c) r = 1/2 [v(t₂)-v(t₁)](t₂-t₁)

d) a_av = v(t₂)-v(t₁)/t₂-t₁

4.14. For a particle performing uniform circular motion, choose the correct statement from the following:

a) magnitude of particle velocity (speed) remains constant

b) particle velocity remains directed perpendicular to the radius vector

c) direction of acceleration keeps changing as particle moves

d) angular momentum is constant in magnitude but direction keeps changing

4.15. For two vectors A and B, |A+B=|A - B| is always true when

a) |A| = |B| $\ne$ 0

b) $\mathrm{A} \perp \mathrm{B}$

c) |A| = |B| $\ne$ 0 and A and B are parallel or antiparallel

d) when either |A| or |B| is zero

4.16. A cyclist starts from center O of a circular park of radius 1 km and moves along the path OPRQO as shown in the figure. If he maintains a constant speed of 10 m/s, what is his acceleration at point R in magnitude and direction?

4.17. A particle is projected in the air at some angle to the horizontal, moves along parabola as shown in the figure, where x and y indicate horizontal and vertical directions respectively. Show in the diagram, direction of velocity and acceleration at points A, B, and C.

4.18. A ball is thrown from a rooftop at an angle of 45^o above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

a) greatest speed

b) smallest speed

c) greatest acceleration?

4.19. A football is kicked into the air vertically upwards. What is its

a) acceleration

b) velocity at the highest point

4.20. A, B, and C are three non-collinear, non-co-planar vectors. What can you say about the direction of A(B × C)?

EXEMPLAR QUESTIONS

4.21. A boy traveling in an open car moving on a labeled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give an explanation to support your diagram.

4.22. A boy throws a ball in the air at 60^o to the horizontal along a road with a speed of 10 m/s. Another boy sitting in passing by car observes the ball. Sketch the motion of the ball as observed by the bot in the car, if the car has a speed of 18 km/h. Give an explanation to support your diagram.

4.23. In dealing with the motion of a projectile in the air, we ignore the effect of air resistance on the motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

4.24. A fighter plane is flying horizontally at an altitude of 1.5 km with a speed of 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target?

4.25. a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s²?

b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 10¹¹m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s²?

4.26 Given below in column I are the relations between vectors a, b, and c, and in column II are the orientations of a, b, and c in the XY plane. Match the relation in column I to correct orientations in column II.

Column I	Column II
(a) A + b = c
(b) A - c = b
(c) b - a = c
(d) a + b + c = 0

4.27. If |A| = 2 and |B| = 4, then match the relations in column I with the angle between θ between A and B in column II.

   Column I      Column II

(a) AB = 0     (i) $\mathrm{\theta } = 0°$

(b) AB + 8     (ii) $\mathrm{\theta } = 90°$

(c) AB = 4     (iii) $\mathrm{\theta } = 180°$

(d) AB = -8   (iv) $\mathrm{\theta } = 60°$

4.28. If |A| = 2 and |B| = 4, then match the relations in column I with the angle θ between A and B in column II.

         Column I                  Column II

(a) |A x B| = 0            (i) $\mathrm{\theta } = 30°$

(b) |A x B| = 8           (ii) $\mathrm{\theta } =$45°   

(c) |A x B| = 4           (iii) $\mathrm{\theta } = 90°$

(d) |A x B| = 4$\sqrt{2}$     (iv) $\mathrm{\theta } = 0°$

4.29. A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of the hill and can be moved on the ground at a speed of 2 m/s so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g = 10 m/s².

4.30. A gun can fire shells with maximum speed v0 and the maximum horizontal range that can be achieved is $R=\frac{v_0^2}{g}$. If a target farther away by distance ∆x has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least $h=\mathrm{\Delta }x[1+\frac{\mathrm{\Delta }x}{R}]$

This problem can be approached in two different ways

(i) Refer to the diagram, target T is at a horizontal distance $x=R+\mathrm{\Delta }x$ and between the point of projection y = - h.

(ii) From point P in the diagram projection at speed vo at an angle 0 below horizontal with height h and horizontal range ($∆$rA)

EXEMPLAR QUESTIONS

4.31. A particle is projected in the air at an angle β to a surface which itself is inclined at an angle α to the horizontal.

a) find an expression of range on the plane surface

b) time of flight

c) β at which range will be maximum

4.32. A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit the second time.

4.33. A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45^o to the vertical. What is the speed of the rain? In what direction does rainfall as observed by a ground-based observer?

4.34. A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s.

a) if swimmer starts swimming due north, what will be his resultant velocity?

b) if he wants to start from point A on south bank and reach opposite point B on north bank,

i) which direction should he swim?

ii) what will be his resultant speed?

c) from two different cases as mentioned in a) and b) above, in which case will he reach opposite bank in shorter time?

4.35. A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle θ to the horizontal, find

a) the effective angle to the horizontal at which the ball is projected in the air as seen by a spectator

b) what will be the time of flight?

c) what is the distance from the point of projection at which the ball will land?

d) find θ at which he should throw the ball that would maximize the horizontal range as found c)

e) how does θ for maximum range change if $u>u_0,u=u_0,u<v_0$?

f) how does θ in e) compare with that for u = 0$(i.e.,{45}^{\circ })$?

4.36. Motion in two dimensions, in a plane, can be studied by expressing position, velocity, and acceleration as a vector in Cartesian coordinates $A=A_x\hat{i}+A_y\hat{j}$ where i and j are unit vectors along x and y directions, respectively and Ax and Ay are corresponding components of A. Motion can also be studied by expressing vectors in circular polar coordinates as $A=A_r\hat{r}+A_{\theta }\hat{\theta }$ where $\hat{r}=\frac{r}{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$ $\text{ and }\hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j}$ are unit vectors along the direction in which r and θ are increasing.

a) express $\hat{i}\text{ and }\hat{j}$ in terms of $\hat{\mathrm{r}}\text{ and }\hat{\mathrm{\theta }}.$.

b) show that both $\hat{r}\text{ and }\hat{\theta }$ are unit vectors and are perpendicular to each other

c) show that  $\frac{d}{dt}\left(\hat{r}\right)=\omega \hat{\theta },\text{ where }\omega =\frac{d\theta }{dt}\text{ and }\frac{d}{dt}\left(\hat{\theta }\right)=-\theta \hat{r}$

d) for a particle moving along a spiral given by $r=a\theta \hat{r}$ where a = 1 find dimensions of ‘a’

e) find velocity and acceleration in polar vector representation for a particle moving along spiral described in d) above

4.37. A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m × 50 m is filled with sand. Outside this square, he can walk at a speed of 1 m/s. In the central square, he walks only at a speed of v m/s. What is the smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?