3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 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 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. 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 R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(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 (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v 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. 3.13(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 : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 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 -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  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° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) 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. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (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 Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t 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 r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆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. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 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. 3.2 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. 3.3 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. 3.4 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. 3.5 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 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. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (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 ? 3.8 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. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 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. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 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. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? Reprint 2025-26 48 PHYSICS 3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 3.16 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 3.17 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 2.0 j m s-2. (a) At what time is ( ) the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t1 to t2) 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.

9.20 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm, u2 = – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.23 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The Reprint 2025-26 Answers effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.