# Physics NEET: Forces and Laws of Motion

## Course Content

Total learning: 3 lessons / 4 quizzes

### Introduction

The concepts of position, displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line is already discussed in previous chapter. 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 in detail to enable us to use vectors for the use of defining velocity and acceleration 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 special significance in daily-life situations. We shall discuss uniform circular motion in some detail.
The equations developed for motion in a plane can be can be made as a base to easily extend to the case of three dimensions.

### Scalars and Vectors

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 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 rule for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided just as ordinary numbers (Addition and subtraction of scalars makes sense only for quantities with same units. However, you can multiply and divide scalars of different units). 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 temperature on a particular day are 35.60 C and 24.20 C respectively. Then, the difference between the two temperatures is 11.4°C. Similarly, if a uniform solid cube of aluminum of side 10 cm has 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. 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 . Thus, both v and represent the velocity vector. The magnitude of a vector is often called its absolute value, indicated by I v I =u. 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.

#### 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’, respectively [Fig. 19(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 P (at time t’). 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. 19 b, given the initial and final positions as P and Q, the displacement vector is the same PQ for different 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 emphasized in the previous chapter also while discussing motion along a straight line.

#### 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. (In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves unchanged. Such vectors are called free vectors. However in some physical applications, location or line of application of a vector is important. Such vectors are called localized vectors).

Fig. 20(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. 20(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’.

### 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:

l λA l = l λ A l 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 l A l as shown in Fig. 21(a).

Multiplying a vector A with a negative number λ gives a vector λA whose direction is opposite to the direction of A and whose magnitude is -λ times l A l.

Multiplying a given vector A by a given numbers, say -1 and -1.5, gives vectors as shown in Fig. 21(b). The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λA is the product of the dimensions λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.

### Addition and Subtraction of Vectors – Graphical Method

As mentioned in section for 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. 22(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. 22(b). Then we join tail of 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, 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 shown Fig 22(c), the same vector R is obtained. Thus vector addition is commutative:

A + B = B + A ……………………………………………………………………………..(Eq. 1)

The addition of vectors also obeys the associative law as illustrated in Fig. 22(d). The result of adding vectors A and B first and then adding vector C is same as the result of adding B and C first and then adding B and C first and then adding vector A:are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented

(A + B) + C = A + (B+C) ………………………………………………………………………………………(Eq. 2)

What is the result of adding two equal and opposite vector? Consider two vectors A and –A shown in Fig. 21(b). Their sum is A + (-A). Since the magnitudes of 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 …………………………………………………………………………(Eq. 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 ……………………………………………………………………………………………………………(Eq. 4) Fig. 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.

What is the physical meaning of a zero vector? Consider the position and displacement vectors in a plane as shown in Fig. 19(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 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) ………………………………………………………………………………………………..(Eq. 5)

It is shown in Fig. 23. 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. To add these vectors, we bring their tales to common origin O as shown in Fig. 24(a).Then we draw a line from the head of 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. 24(b)]. In Fig. 24(c), the triangle law is used to obtain the resultant of A and B and we see that the two methods are equivalent. Fig. 6

(a) Two vectors A and B with their tails brought to a common origin (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method.

Example 1
Rain is falling vertically with a speed of 35 m s-1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella? Fig. 7

The velocity of the rain and the wind are represented by the vectors vr and vw in Fig. 7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is

R = √(vr2 + vw2) = √(352 + 122) m s-1 = 37 m s-1

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

tan θ = vw /vr = 12/35 = 0.343

Or, θ = tan-1(0.343) = 190

Therefore, the boy should his umbrella in the vertical plane at an angle of about 190 with the vertical towards the east.

### 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. 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 thee tail and head of the vector A. Then, through O, draw a straight line parallel to b. Let them intersect at Q. Then, we have,

A = OP = OQ + QP …………………………………………………………………………………………..(Eq. 6)

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

OQ = λ a and QP = μ b ……………………………………………………………………………………….(Eq. 7)

Where λ and m are real numbers.

Therefore, A = λ a + μ b ………………………………………………………………………………………(Eq. 8) Fig. 8

(a) Two non-co linear vectors a and b. (b) Resolving a vector A in terms of vectors a and b.

We say that A has been resolved in to two component vectors λ a and μ b along a and b respectively. Using this method one can resolve a given vector in to two component vectors along a set of two vectors-all 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 the we discuss now.

#### Unit Vectors

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 i, j and k respectively, as shown in Fig. 27.

Since these are unit vectors, we have

|i | = | j | = | k | = 1 …………………………………………………………………………………………(Eq. 9)

These unit vectors are perpendicular to each other. Since we are dealing with motion in two dimensions we require use of only two unit vectors. If we multiply a unit vector, say n by a scalar, the result is a vector.

λ = λ n. In general, a vector A can be written as

A = | A | n ……………………………………………………………………………………………………..(Eq. 10)

where n is a unit converter along A.

We can now resolve a vector A in terms of component vectors that lie along unit vectors i and j. Consider a vector A that lies in x-y plane as shown in Fig. 27(b) and get vectors A1 and A2 such that A1 + A2 = A. Since A1 is parallel to i and A2 parallel to j, we have:

A1 = Axi, A2 = Ayj …………………………………………………………………………. .(Eq. 11)

where Ax and Ay are real numbers.

Thus, A = AXi + Ayj ………………………………………………………………………………(Eq. 12)

This is represented in Fig. 27(c). The quantities Ax and Ay are called x and y components of the vector A. Note that AXi is a vector and so is Ayj. Using simple trigonometry, we can express Ax and Ay in terms of the magnitude of BA and the angle θ it makes with x-axis:

Ax = A cosθ

Ay = A sinθ …………………………………………………………………………………………….(Eq. 13)

As is clear from Eq. (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:

• Its magnitude A and the direction θ it makes with the x-axis; or
• Its components Ax and Ay

If A and θ are given, Ax and Ay can be obtained using Eq. (13). If Ax and Ay are given, A and θ can be obtained as follows:

Ax2 + Ay2 = A2 cos2 θ + A2 sin2 θ

= A2

Or, A = √( Ax2 + Ay2) …………………………………………………………………………………………(Eq. 14)

and tan θ =( Ay/ Ax ), θ = tan-1 (Ay/Ax) ……………………………………………………………………(Eq. 15) So far we have considered a vector lying in x-y plane. The same procedure can be used to resolve a general vector A in to three components along x, y and z axes in three dimensions. If α, β and γ are the angles (Note that angles α, β and γ are angles in space. They are between pairs of lines, which are not coplanar) between A and the x, y and z axes, respectively.

Fig. 9 d, we have Ax = A cos α, Ay = A cos β, Az = A cos γ ………………………………………………………………….(Eq. 16a)

In general, we have

A = Axi + Ayj + Azk ……………………………………………………………………………………………(Eq. 16b)

The magnitude of vector A is

A = √(A2x + A2y + A2z) ………………………………………………………………………………………(Eq. 16c)

A position vector r can be expressed as

r = x i + y j + z k ……………………………………………………………………………………………….(Eq. 17)

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

### Vector Addition – Analytical Method

Although the graphical method of adding vectors helps us in visualizing 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 :

A = Axi + Ayj ………………………………………………………………………………………………………(Eq. 18)

B = Bxi + Byj Let R be their sum. We have

R = A +B

= (Axi + Ayj) + (Bxi + Byj) ………………………………………………………………………………..(Eq. 19a)

Since vectors obey the cumulative and associative laws, we can arrange and regroup the vectors in Eq. (19a) as convenient to us:

R = (Ax+ Bx)i + (Ay+ By)j ………………………………………………………………………………….(Eq. 19b)

Since R = Rxi + Ryj ………………………………………………………………………………………….(Eq. 20)

We have, Rx = Ax + Bx , Ry = Ay + By …………………………………………………………………(Eq .21)

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

In three dimensions, we have

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

R = A +B = Rxi + Ryj + Rzk

With Rx = Ax + Bx ,

Ry = Ay + By

Rz = Az + Bz …………………………………………………………………………………………………(Eq. 22)

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

a = axi + ayj + azk

b = bxi + byj + bzk

c = cxi + cyj + czk ………………………………………………………………………………………(Eq.23a)

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

Tx = ax + bx -cx

Ty = ay + by –cy

Tz = az + bz –cz ……………………………………………………………………………………………(Eq. 23b)

Example 2
Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle θ between them. Let OP and OQ represent the two vectors A and B making an angle θ (Fig. 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+ 2A B cos θ

R = √( A2 +B2+ 2A B cos θ) ………………………… ………………………………………………(Eq. 24a)

In ΔOSN, SN = OS sinα = R sinα and

In ΔPSN, SN = PS sin θ = B sin θ

Therefore, R sinα = B sin θ

Or, (R/sin θ) = (B/ sinα) ………………………………………………………………………………..(Eq. 24b)

Similarly,

PM = A sinα = B sin β

Or, (A/ sin β) = (B/ sinα) …………………………………………………………………………(Eq. 24c)

Combining Eqs. (24b) and (24c), we get

(R/sin θ) = (A/ sin β) = (B/ sin α) ……………………………………………………………..(Eq. 24d)

Using Eq. (24d), we get:

sinα = (B/R) sin θ …………………………………………………………………………………………(Eq.24e)

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

or, tan α = SN/(OP+PN) = B sinθ/(A+B cos θ) ………………………………………………(Eq. 24f)

Equation (24a) gives the magnitude of the resultant and Eqs. (24e) and (24f) its direction. Equation (24a) is known as the Laws of cosines and Eq. (24d) as the Law of sines.

Example 3
A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 600 east of south. Find the resultant velocity of the boat.

The vector vb representing the velocity of the motor boat and the vector vc representing the water current are shown in Fig. 30 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

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

R = √(vb2 + vc2 + 2vbvc cos1200

=√(252 + 1 02 + 2 × 25 × 10(-1/2) ≅ 22 km/h

To obtain the direction, we apply the Law of sines

(R/sin θ) = vc / sinφ or sinφ = (vc / R) sin θ

= (10 x sin 120)/21.8 = (10√3)/(2 x 21.8) ≅0.397

≅ 23.40

### Motion in a Plane

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

#### 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. 12) is given by

r = xi + yj where x and y are components of r along x and y axes or simply they are the coordinates of the object. 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. 12(b)]. Then, the displacement is:

Δr = r’ – r …………………………………………………………………………………………………….(Eq. 25)

and is directed from P to P’.

We can write Eq. (25) in a component form:

Δr = (x’i + y’j) – (xi + yj)→

= iΔx + jΔy

Where Δx = x’-x, Δy = y’-y, ………………………………………………………………………………(Eq. 26)

#### Velocity

The average velocity (vav) = (Δr/Δt) = (Dxi + Dyj)/Δt = i (Δx/Δt) + j (Δy/Δt) ….(Eq. 27)

Or, vav = vx i + vy j

Since vav = Δr/Δt, the direction of the average velocity is the same as that of Δr (Fig. 12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero:

v = limΔt→0r/Δt) = dr/dt …………………………………………………………………………..(Eq. 28)

The meaning of the limiting process can be easily understood with the help of Fig. 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 Δt1, Δt2 and Δt3. Δr1, Δr2 and Δr3 are the displacements of the object in times Δt1, Δt2 and Δt3 respectively. The direction of the average velocity vav is shown in figures (a), (b) and (c) for three decreasing values of Δt. i.e. Δt1, Δt2 and Δt3. (Δt1˃ Δt2 ˃Δt3). As Δt→o, Dr→o and is along the tangent to the path (Fig. 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:

v = dr/dt

= limΔt→0 [(Δxt)i +yt)j]…………………………………………………….(Eq. 29)

= i limΔt→0 (Δxt) + j limΔt→0yt)

Or, v = i (dx/dt) + j (dy/dt) = vx i + vyj

Where vx = (dx/dt).vy = dy/dt …………………………………………………………………………(Eq. 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

v = √ (vx2 + vy2) ………………………………………………………………………………………….(Eq. 30b)

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

tan θ = vy/vx , θ = tan-1(vy/vx) ………………………………………………………………………(Eq. 30c)

vx, vy and angle θ are shown in Fig . 33 for a velocity vector v.

#### Acceleration

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

aav = vt) = Δ(vx i + vyj)/ Δt = (Δvxt)i + (Δvyt)j …………………………..(Eq.31a)

Or, aav = ax i + ayj ……………………………………………………………………………………….(Eq. 31b)

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

a = limΔt→0 (Δv/Δt) …………………………………………………………………………….(Eq.32a)

Since Δv = Δvx i +Δ vyj , we have

a = i limΔt→0vxt) + j limΔt→0vyt)

Or, a = ax i + ayj ………………………………………………………………………………….(4.32b)

where, ax = dvx / dt, ay = dvy / dt ………………………………………………………….(Eq.32c)

(In terms of x and y, ax and ay can be expressed as ax = d/dt(dx/dt) = d2x/dt2, ay = d/dt(dy/dt) = d2y/dt2)

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 Fig. 34(a) to (d). P represents the position of the object at time t and P1, P2, P3 positions after time Δt1, Δt2, Δt3 respectively (Δt1˃ Δt2 ˃Δt3) are also shown in Fig. 34(a), (b) and (c). In each case of Δt, Δv is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of Δv. We see that as Δt decreases the direction of Δv changes and consequently, the direction of the acceleration changes. Finally, in the limit Δt→0 Fig. 15(d), the average acceleration becomes the instantaneous acceleration and has the direction as shown. Fig.15

The average acceleration for three time intervals (a) Δt1, (b) Δt2, and (c) Δt3, (Δt1> Δt2> Δt3). (d) In the limit Δt →0, the average acceleration becomes the acceleration.

Note that in one dimension, 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 00 and 1800 between them.

Example 4
The position of a particle is given by

R = 3.0ti + 2.0t2j + 5.0k

Where t is in seconds and the coefficients have the proper units for r to be in meters. Find v(t) and at t = 1.0 s.

v(t) = dr/dt = d/dt (3.0 t i + 2.0t2j + 5.0 k)

= 3.0i + 4.0tj

a(t) = dv/dt = +4.0j

a = 4.0 m s-2 along y-direction

At t = 1.0 s, v = 3.0i + 4.0 j

Its magnitude is v = √(32 +42 )= 5.0 m s-1 and direction is θ = tan-1(vy/vx) = tan-1(4/3) @ 530 with x-axis.

### 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 v0 at time t =0 and v at time t.

Then, by definition

a = (v-v0)/(t-0) = (v-v0)/t

Or, v = v0 + at …………………………………………………………………………………………(Eq.33a)

In terms of components:

vx = vax + axt

vy = vay + ayt ………………………………………………………………………………………….(Eq.33b)

Let us now find how the position r changes with time. We follow the method used in the one dimensional case. Let r0 and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be v0 and v. Then, over this time interval t, the average velocity is (v0 +v)/2. The displacement is the average velocity multiplied by the time interval:

r – r0 =[ (v + v0/2)]/t = [{(v0 + at) + v0}/2]/t

= v0t + ½ at2

Or, r = r0 + v0t + ½ at2 ……………………………………………………………………………(Eq. 34a)

It can be easily verified that the derivative of Eq. (34a), i.e. dr/dt gives Eq. (33a) and it also satisfies the condition that at t=0, r = r0. Equation (34a) can be written in component form as

x = x0 + vaxt + ½ axt2

y = y0 + vayt + ½ ayt2 ……………………………………………………………………………….(Eq. 34b)

One immediate interpretation of Eq. (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 analyzing 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 Projectile motion.

Example 5
A particle starts from origin t t=0 with a velocity 5.0 i m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i + 2.0j) m/s2. (a) What is the y coordinate of the particle at the instant its x coordinate is 84 m? (b) What is the speed of the particle at this time?

The position of the particle is given by

r(t) = v0t + ½ at2

= 5.0i t + (1/2)(3.0i + 2.0j) t2

= (5.0 t + 1.5 t2)i + 1.0t2j

Therefore, x(t) = 5.0t + 1.5t2

y(t) = +1.0t2

Given x(t) = 84m, t = ?

5.0 t + 1.5 t2 = 84 ⇒t = 6 s

At t = 6 s, y = 1.0 (6)2 = 36.0 m

Now, the velocity v = dr/dt = (5.0 +3.0t)i + 2.0 t j

At t = 6 s, v = 23.0i + 12.0 j

Speed= IvI = √(232 + 122) ≅ 26 m s-1

### Relative Velocity in two Dimensions

The concept of relative velocity, introduced in section for motion in 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 = vAvB ………………………………………………………………………………………(Eq .35a)

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

vBA = vBvA

Therefore, vAB = – vBA ………………………………………………………………………………(Eq. 35b)

and, | vAB | = | vBA | …………………………………………………………………………………(Eq. 35c)

Example 6
Rain is falling vertically with a speed of 35 m s-1. A woman rides a bicycle with a speed of 12 m s-1 in east to west direction. What is the direction in which she should hold her umbrella?

In Fig. 16 vr represents the velocity of rain and vb, the velocity of bicycle, the woman is riding. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the velocity of the bicycle she is riding. That is vrb = vrvb Fig.16

This relative velocity vector as shown in Fig. 16 makes an angle θ with the vertical. It is given by

tan θ = vb/vr = 12/35 = 0.343

Or, θ = 190

Therefore, the woman should hold her umbrella at an angle of about 190 with the vertical towards west.

Note carefully the difference between this Example and the Example 1. In Example 1, the boy experiences the resultant (vector sum) of two velocities while in this example, the woman experiences the velocity of rain relative to the bicycle (the vector difference of the two velocities).

### 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 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 v0 that makes an angle θ0 with the x-axis as shown in Fig. 17.

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

a = –gj

Or, ax = 0, ay = -g …………………………………………………………………………………………..(Eq. 36)

The components of initial velocity v0 are:

vax = v0 cos θ0,

vax = v0 sin θ0, ………………………………………………………………………………………………(Eq. 37) Fig 17

Motion of an object projected with velocity v0 at angle θ0.

If we take the initial position to be the origin of the reference frame as shown in Fig. 17, we have:

x0 = 0, y0 = 0

Then, Eq. (37b) becomes:

x = vax t = (v0 cos θ0) t

and y = (v0 sin θ0) t –(1/2)gt2 …………………………………………………………………………..(Eq. 38)

The components of velocity at time t can be obtained using Eq. (33b):

vx = vax = v0 cos θ0

vy = vay = v0 sin θ0 –gt ……………………………………………………………………………………(Eq. 39)

Equation (38) gives the x and y coordinates of the position of a projectile at time t in terms of two parameters – initial speed v0 and projection angle θ0. 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. 37. Note that at the point of maximum height, vy =0 and therefore,

θ = tan(vy/vx) = 0

#### 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 Eq. (4.38). We obtain:

y = (tan θ0) x – [g/{2(v0 cos θ0)2}]x2 ………………………………………………………………..(Eq. 40)

Now, since g, θ0 and v0 are constants, Eq. (4.40) is of the form y = ax + bx2, 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. 18). Fig. 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. 39):

vy = v0 sin θ0 gtm= 0

Or, tm = v0 sin θ0/g ……………………………………………………………………………….(Eq. 41a)

The total time Tf during which the projectile is in flight can be obtained by putting y = 0 in (Eq. 38), we get:

Tf = 2(v0 sin θ0) /g ………………………………………………………………………………….(Eq. 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. 38):

ym = hm = (v0 sin θ0)( v0 sin θ0/g) – g/2(v0 sin θ0/g)2

Or, hm =(v0 sin θ0)2/2g ………………………………………………………………………….(Eq. 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 = (v0 cos θ0)(Tf)

= (v0 cos θ0)(2 v0 sin θ0)/g

R =( v02sin0)/g ……………………………………………………………………………………(Eq. 43a)

Equation (43a) shows that for a given projection velocity v0, R is maximum when sin 2θ0 is maximum. i.e., when θ0 =450.

The maximum horizontal range is, therefore,

Rm = v02/g ………………………………………………………………………………………………..(Eq. 43b)

Example 7
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 450 by equal amounts, the ranges are equal”. Prove this statement.

For a projectile launched with velocity v0 at an angle θ0, the range is given by

R = v02 sin 2θ0/g

Now, for angles, (450 + α) and (450 α), 0 is (900 + 2α) and (900 -2α), respectively. The values of sin (900 + 2α) and sin (900 – 2α) are same, equal to that of cos 2α. Therefore, ranges are equal to elevations which exceed or fall short of 450 by equal amounts α.

Example 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 the 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).

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) = x0 + voxt

y(t) = y0 + voyt + (1/2)ayt2

Here, x0 = y0 = 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 –gt so that when the stone hits the ground:

vox = 15 m s-1.

voy = 0 – 9.8 ₓ 10 = -98 m s-1.

Therefore, the speed of the stone is

√(vx 2+ vy2) = (15 2+ 982) = 99 m s-1

Example 9
A cricket ball is thrown at a speed of 28 m s-1 in a direction 300 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.

(a) The maximum height is given by

hm = (v0 sin θ0)2/2g = (28 sin 300)2/(2 ₓ 9.8) m

= (14 ₓ 14)/(2 ₓ 9.8) = 10.0 m

(b) The time taken to return to the same level is Tj = (2 v0 sin θ0)/g = (2 × 28 × sin 300)/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 = (v02 sin 2θ0)/g = ((28 × 28 × sin 600)/9.8 = 69 m

### Uniform Circular Motion

When an object follows 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. 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. Fig. 19

Velocity and acceleration of an object in uniform circular motion. The time interval Δt decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle.

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 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 shown in Fig 19(a1). Δv is obtained in Fig 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, Δv is perpendicular to Δr. Since average acceleration is along Δv (ā = Δv/Δt), the average acceleration ā is perpendicular to Δr. If we place Δv on the line that bisects the angle between r and r’, we see that it is directed towards the center of circle. Fig. 19(b) shows the same quantities for smaller time interval, Δv and hence ā is again directed towards center. In Fig. 19(c), Δt→ 0 and the average acceleration becomes the instantaneous acceleration. It is directed towards the center (In the limit Δt→ 0, Δr becomes perpendicular to r. In this limit Δv→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the center at each point of the circular path.). Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the center of the circle. Let us now find the magnitude of the acceleration.

The magnitude of a, by definition, given by |a| = limΔ t→ 0 (|Δv |/ Δt). Let the angle between position vectors r and r’ be Δθ . Since the velocity vectors v and v’ are always perpendicular to the position vectors, the angle between them is also Δθ . Therefore, the triangle CPP’ formed by the position vectors and the triangle GHI formed by the velocity velocity vectors v, v’ and Δv are similar (Fig. 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:

v |/ v = |Δr |/ R

Or, |Δv | = v (|Δr |/ R)

Therefore,

|a| = lim Δ t→ 0 (|Δv |/ Δt) = lim Δ t→ 0 (|v (|Δr |/R Δt) = (v/R) lim Δ t→ 0 (|Δr |/ Δt)

If Δt is small, Δθ will also be small and then are PP’ can be approximately taken to be |Δr |:

r | ≅ t

(r |/Δt) ≅ v

Or, lim Δt→ 0 (r |/Δt) = v

Therefore the centripetal acceleration ac is:

ac = (v/R)v = v2/R ……………………………………………………………………………………(Eq. 44)

Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude v2/R and is always directed towards the center. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christian Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term which means ‘center-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes- pointing always towards center. Therefore, a centripetal acceleration is not a constant vector.

We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the object moves from P to P’ in time Δt (=t’ – t), the line CP (Fig. 19) turns through an angle Δθ as shown in the figure is called angular distance. We define the angular speed ω(Greek letter omega) as the rate of change of angular displacement:

w = Δθ/Δt ……………………………………………………………………………………………..(Eq. 45)

Now, if the distance travelled by the object during the time Δt is Δs, i.e. PP’ is Δs, then:

v = Δs/Δt but Δs = RΔθ. Therefore:

v = R (Δθ/Δt) = R ω ………………………………………………………………………………..(Eq. 46)

We can express centripetal acceleration ac in terms of angular speed:

ac = v2/R = ω2R2/R = ω2R

ac = ω2R ……………………………………………………………………………………………………(Eq. 47)

The time taken by an object to make a revolution is known as its time period T and the number of revolution made in one second is called its frequency v (=1/T). However, during this time the distance moved by the object is s = 2πR.

Therefore, v = 2πR/T = 2πRv …………………………………………………………………. (Eq. 48)

In terms of frequency v, we have,

w = 2πv

v=Rv

ac = 4π2v2/R ………………………………………………………………………………………………..(Eq. 49)

Example 10
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) what is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is the magnitude? 