# Motion in a Plane

## Class 11 NCERT Physics

### NCERT

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.

##### Solution :

$\textbf{Scalar}:$ Volume, mass, speed, density, number of moles, angular frequency$\\$ $\textbf{Vector}:$ Acceleration, velocity, displacement, angular velocity$\\$ A scalar quantity is specified by its magnitude only. It does not have any direction associated with it.$\\$ Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.$\\$ A vector quantity is specified by its magnitude as well as the direction associated with it.$\\$ Acceleration, velocity, displacement, and angular velocity belong to this category.

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.

##### Solution :

$\textbf{Work}$ and $\textbf{Current}$ are scalar quantities.$\\$ $\text{Work}$ done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.$\\$ $\text{Current}$ is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.

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.

##### Solution :

Let two vectors a and b be represented by the adjacent sides of a parallelogram $OMNP$, as shown in the given figure

Here we can write $|\bar{OM}|=|\bar{a } \quad ...(i)\\ |\bar{MN}|=|\bar{OP}|=|\bar{b}|\quad ...(ii)\\ |\bar{ON}|=|\bar{a}+\bar{b}| \quad ...(iii)$$\\ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in \Delta OMN , we have:\\ ON < ( OM + MN )\\ |\bar{a}+\bar{b}| < |\bar{a}|+|\bar{b}| \quad ....(iv)$$\\$ If the two vectors $\bar{ a}$ and $\bar{b}$ act along a straight line in the same direction, then we can write:$\\$ $|\bar{a}+\bar{b}|=|\bar{a}|+|\bar{b}| \quad ...(v)$$\\ Combining equations (iv) and (v), we get:\\ |\bar{a}+\bar{b}| \leq |\bar{a}|+|\bar{b}|$$\\$ Let two vectors $\bar{a}$ and $\bar{b}$ be represented by the adjacent sides of a parallelogram $OMNP$, as shown in the given figure.$\\$ Here,we have: $\\$ $|\bar{OM}|=|\bar{a } \quad ...(i)\\ |\bar{MN}|=|\bar{OP}|=|\bar{b}|\quad ...(ii)\\ |\bar{ON}|=|\bar{a}+\bar{b}| \quad ...(iii)$$\\ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in \Delta OMN , we have:\\ ON+MN > OM\\ ON+ OM > MN\\ |ON| > |\bar{OM}-\bar{OM}| \quad (\therefore OP=MN)\\ |\bar{a}+\bar{b}| > ||\bar{a}|-|\bar{b}|| \quad (iv)$$\\$ If the two vectors $\bar{a}$ and $\bar{b}$ act along a straight line in the same direction, then we can write: $\\$ $|\bar{a}+\bar{b}|=||\bar{a}|-|\bar{b}||\quad ...(v)$$\\ Combining equations (iv) and (v), we get: \\ |\bar{a}+\bar{b}| \geq ||\bar{a}|-|\bar{b}||$$\\$ Let two vectors a and b be represented by the adjacent sides of a parallelogram $PORS,$ as shown in the given figure.$\\$ $\\$

##### Solution :

$\vec{v} ( t ) =( 3.0\hat{i}-4.0t\hat{ j} ;\vec{a}=-4.0\hat{j}$ $\\$ The position of the particle is given by: $\\$ $\vec{r} = 3.0\hat{i}-2.0t^ 2\hat{ j}+4.0k$ $\\$ Velocity, $\vec{v}$ of the particle is given as: $\\$ $\vec{v}=\dfrac{d\vec{r}}{dt}=\dfrac{d}{dt}(3.0t \hat{i}-2.0t^2\hat{j}+4.0k)\\ \therefore \vec{v}=3.0 \hat{i}-4.0 t \hat{j}$ $\\$ Acceleration of the particle is given as:$\\$ $\vec{a}=\dfrac{dv}{dt}=\dfrac{d}{dt}(3.0\hat{i}-4.0 t \hat{j})\\ \therefore \vec{a}=-4.0 \hat{j}$ $\\$ We have velocity vector. $\\$ $\vec{v}=3.0\hat{i}-4.0 t \hat{j}$ $\\$ At $t=2.0s:\\ \vec{v}=3.0 \hat{i}-8.0 \hat{j}$ $\\$ The magnitude of velocity is given by:$\\$ $|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{73}=8.54 m/s$ $\\$ Direction, $\theta=\tan^{-1}(\dfrac{v_y}{v_x})\\ =\tan^{-1}(\dfrac{-8}{3})=-\tan^{-1}(2.667)\\ =-69.45^o$ $\\$ The negative sign indicates that the direction of velocity is below the x-axis. 8.54 m/s, 69.45° below the x-axis.

21   A particle starts from the origin at t = 0 s with a velocity of 10.0 j and moves in the x-y plane with a constant acceleration of $(8.0 \hat{i}+2.0 \hat{j})ms^{-2}$ $\\$ (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?

##### Solution :

Velocity of the particle, $\vec{v}=10.0 \hat{j}m/s$ $\\$ Acceleration of the particle $\vec{a}=(8.0 \hat{i}+2.0 \hat{j})$ $\\$ Also, $\\$ But,$\vec{a}=\dfrac{d\vec{v}}{dt}=8.0\hat{i}+2.0 \hat{j}\\ d\vec{v}=(8.0\hat{i}+2.0\hat{j})dt$ $\\$ Integrating both sides:$\\$ $\vec{v}(t)=8.0 \hat{i}+2.0 \hat{j}+\vec{v}$ $\\$ Where, $\\$ $\vec{v}=$ Velocity vector of the particle at t = 0 $\\$ $\vec{v}=$ Velocity vector of the particle at time t $\\$ But= $\vec{v}=\dfrac{d\vec{r}}{dt}\\ d\vec{r}=\vec{v}dt=(8.0t \hat{i}+2.0t \hat{j}+\vec{u})dt$ $\\$ Integrating the equations with the conditions: at t = 0; r = 0 and at t = t; r = r$\\$ $\vec{r}=\vec{u}t+\dfrac{1}{2}8.0 t^2 \hat{i}+\dfrac{1}{2}2.0t^2\hat{j}\\ =\vec{u}t+4.0t^2 \hat{i}+t^2\hat{j}\\ =(10.0 \hat{j})t+4.0t^2\hat{i}+t^2 \hat{j}\\ x\hat{i}+y\hat{j}=4.0t^2\hat{i}+(10t+t^2)\hat{j}$ $\\$ Since the motion of the particle is confined to the x-y plane, on equating the coefficients of $\hat{i}$ and $\hat{j}$ ,$\\$ we get: $\\$ $x=4 t^2\\ t=(\dfrac{x}{4})^{\frac{1}{2}}$ $\\$ And y=10t+t$^2$ $\\$ when x=16m: $\\$ $t=(\dfrac{16}{4})^{\frac{1}{2}}=2s\\ \therefore y=10*2+(2)^2=24m$ $\\$ Velocity of the particle is given by: $\\$ $\vec{v}(t)=8.0t \hat{i}+2.0t \hat{j}+\vec{u}\\ at t=2s\\ \vec{v}(t)=8.0*2 \hat{i}+2.0*2\hat{j}+10\hat{j}\\ =16 \hat{i}+14\hat{j}$ $\\$ $\therefore$ Speed of the particle: $\\$ $|\vec{v}|=\sqrt{(16)^2+(14)^2}\\ =\sqrt{256+196}=\sqrt{452}\\ 21.26 m/s$ $\\$

22   $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors $\hat{i}$+$\hat{j}$, and $\hat{i}$-$\hat{j}$ ? What are the components of a vector a=2$\hat{i}$+3$\hat{j}$along the directions of $\hat{i}$+ $\hat{j}$and$\hat{i}$-$\hat{j}$ ? [You may use graphical method]

##### Solution :

Consider a vector $\vec{P}$ $\\$ $\vec{P} =\hat{ i} +\hat{ j}\\ P , \hat{i }+ P_ y\hat{ j} =\hat{i} +\hat{ j}$ $\\$ On comparing the components on both sides, we get: $\\$ $P_ x = P _y = 1\\ |\vec{P}|=\sqrt{ P^2_ x + P _y ^2 }=\sqrt{ 1^ 2 + 1 ^2 }=\sqrt{ 2} .. ( i )$ $\\$ Hence, the magnitude of the vector$\hat{ i }+\hat{ j}$ is $\sqrt{2}$ $\\$ Let $\theta$ be the angle made by the vector P ,with the x-axis, as shown in the following figure.

$\therefore \tan \theta =(\dfrac{P_y}{P_x})\\ \theta =\tan^{-1}(\dfrac{1}{1})=45^o....(ii)$ $\\$ Hence, the vector i + j makes an angle of 45$^o$ with the x - axis$\\$ Let$\\$ $\vec{Q}=\hat{1}-\hat{j}\\ Q_x\hat{i}-Q_y \hat{j}=\hat{i}-\hat{j}\\ Q_x=Q_y=1\\ |\vec{Q}|=\sqrt{Q^2_x+Q^2_y}=\sqrt{2}....(iii)$ $\\$ Hence, the magnitude of the vector$\hat{i}-\hat{j}$ is $\sqrt{2}$ $\\$ Let $\theta$ be the angle made by the vector $\vec{Q} ,$ with the x- axis, as shown in the following figure.$\\$ $\therefore \tan \theta =(\dfrac{Q_y}{Q_x})\\ \theta=-\tan^{-1}(-\dfrac{1}{1})=-45^o .....(iv)$ $\\$ Hence, the vector $\hat{i} -\hat{ j}$ makes and angle of- 45$^o$ with the axis.$\\$ It is given that:$\\$ On comparing the coefficients of $\hat{i}$ and $\hat{j}$ , we have$\\$ $\\vec{A}=2 \hat{i}+3 \hat{j}\\ A_x\hat{i}+A_y\hat{j}=2 \hat{i}+3\hat{j}$ $\\$ Let $\vec{A}$ make an angle $\theta$ with the x-axis, as shown in the following figure

$\therefore \tan \theta = (\dfrac{A_x}{A_y})\\ \theta =\tan^{-1}(\dfrac{3}{2})\\ =\tan^{-1}(1.5)=56.31^o$ $\\$ Angle between the vectors $(2 \hat{i}+3 \hat{j})$ and $(\hat{i+\hat{j}}),\theta=56.31-45=11.31^o$$\\ \vec{A},along the direction of \vec{P} making an angle \theta '$$\\$ $=(A \cos \theta ) P=(A \cos 11.31)\dfrac{(\hat{i}+\hat{j})}{\sqrt{2}}\\ =\sqrt{13}*\dfrac{0.9806}{\sqrt{2}}(\hat{i}+\hat{j})\\ =2.5(\hat{i}+\hat{j})\\ =\dfrac{25}{10}*\sqrt{2}\\ =\dfrac{5}{\sqrt{2}}.....(v)$ $\\$

Let $\theta$ ' be the angle between the vectors $(2 \hat{i}+3 \hat{j})$ and $(\hat{i}-\hat{j})$ $\\$ $\theta ''=45+56.31=101.31^o$ $\\$ Component of vector $\vec{A}$ , along the direction of $\vec{Q}$ , making an angle $\theta$'' $\\$ $=(A \cos \theta)\vec{Q}=(A \cos \theta )\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}\\ =\sqrt{13 } \cos (901.31^o)\dfrac{(\hat{i}-\hat{j})}{\sqrt{2}}\\ =-\sqrt{\dfrac{13}{2}} \sin 11.30^o(\hat{i}-\hat{j})\\ =-2.550*0.1961(\hat{i}-\hat{j})\\ =-0.5(\hat{i}-\hat{j})\\ =-\dfrac{5}{10}*\sqrt{2}\\ =-\dfrac{1}{\sqrt{2}} ...... (iv)$

23   For any arbitrary motion in space, which of the following relations are true:$\\$ $(a) v_{average}=(\dfrac{1}{2})(v(t_1)+v(t_2))\\ (b) v_{average}=\dfrac{[r(t_2)-r(t_1)]}{(t_2-t_1)}\\ (c) v(t)=v(0)+at\\ (d)r(t)=r(0)+v(0)t+(\dfrac{1}{2})at^2\\ (e)a_{average}=\dfrac{[v(t_2)-v(t_1)]}{(t_2-t_1)}$ $\\$ (The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_ 2$ )

##### Solution :

(a) False. It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.$\\$ (b) True. The arbitrary motion of the particle can be represented by this equation.$\\$ (c) False. The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle e in space.$\\$ (d) False. The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space.$\\$ (e) True. The arbitrary motion of the particle can be represented by this equation.

24   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$\\$

##### Solution :

(a) False. Despite being a scalar quantity, energy is not conserved in inelastic collisions.$\\$ (b) False. Despite being a scalar quantity, temperature can take negative values.$\\$ (c) False. Total path length is a scalar quantity. Yet it has the dimension of length.$\\$ (d) False. A scalar quantity such as gravitational potential can vary from one point to another in space.$\\$ (e) True. The value of a scalar does not vary for observers with different orientations of axes.

25   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 apart is 30 $^o$ , what is the speed of the aircraft?

##### Solution :

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, $OR = 3400 m$$\\ Angle subtended between the positions, \angle POQ = 30^o \\ Time = 10 s\\ In \Delta PRO:\\ \tan 15^o=\dfrac{PR}{OR}\\ PR=OR \tan 15^o\\ =3400*\tan 15^o \\ \Delta PRO is similar to \Delta RQO. \\ \therefore PR=RQ\\ PQ=PR+RQ\\ =2PR=2*3400 \tan 15^o\\ =6800*0.268=1822.4 m \\ \therefore Speed of the aircraft= \dfrac{1822.4}{10}=182.24 m/s 26 A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer. ##### Solution : No. Generally speaking, a vector has no definite locations in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However, a position vector has a definite location in space.\\ Yes. A vector can vary with time. For example, the displacement vector of a particle moving with a certain velocity varies with time.\\ No. Two equal vectors located at different locations in space need not produce the same physical effect. For example, two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.\\ 27 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector? ##### Solution : No. A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.\\ No. Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector. 28 Can you associate vectors with\\ (a) the length of a wire bent into a loop,\\ (b) a plane area,\\ (c) a sphere? Explain. ##### Solution : (a) No. One cannot associate a vector with the length of a wire bent into a loop. Because length of a loop does not have a definite direction.\\ (b) Yes. One can associate an area vector with a plane area. The direction of this vector is represented by a normal drawn outward to the area.\\ (c) No. One cannot associate a vector with the volume of a sphere as it does not have a specific direction. However, a null vector can be associated with the area of a sphere.\\ 29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km a way. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance. ##### Solution : No.\\ Range, R = 3 km\\ Angle of projection, \theta = 30^o$$\\$ Acceleration due to gravity, $g = 9.8 m / s ^2$$\\ Horizontal range for the projection velocity u_ 0 , is given by the relation:\\ R=\dfrac{u_0^2 \sin 2 \theta}{g}\\ 3=\dfrac{u_0^2}{g} \sin 60^o\\ \dfrac{u^2_0}{g}=2\sqrt{3}.....(i) \\ The maximum range ( R_{max} ) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,\\ R_{max}=\dfrac{u^2_0}{g}....(ii) \\ On comparing equations (i) a nd (ii), we get:\\ R_{ max} = 3 \sqrt{3 }= 2 * 1.732 = 3.46 km \\ Hence, the bullet will not hit a target 5 km away 30 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 k m/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^{-1} to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 m s ^{-2} ). ##### Solution : Height of the fighter plane = 1.5 km = 1500 m\\ Speed of the fighter plane, v = 720 km / h = 200 m / s\\ Let \theta be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.\\ Muzzle velocity of the gun, u = 600 m / s\\ Time taken by the shell to hit the plane = t\\ Horizontal distance travelled by the shell = u_ x t \\ Distance travelled by the plane = vt \\ The shell hits the plane. Hence, these two distances must be equal. \\ u_ x t = vt\\ u \sin \theta = v\\ \sin \theta =\dfrac{v}{u}\\ =\dfrac{200}{600}=\dfrac{1}{3}=0.33\\ \theta = \sin^{-1}(0.33)\\ =19.5^o \\ In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.\\ \therefore H=\dfrac{u^2 \sin^2(90-\theta)}{2g}\\ =\dfrac{600^2 \cos^2 \theta}{2g}\\ =\dfrac{360000 * \cos^2 19.5}{2*10}\\ =18000*(0.943)^2\\ =16006.482m\\ \approx 16 km 31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn? ##### Solution : Speed of the cyclist, v = 27 km / h = 7.5 m / s\\ Radius of the circular turn, r = 80 m\\ Centripetal acceleration is given as:\\ a_c=\dfrac{v^2}{r}\\ =\dfrac{(7.5)^2}{80}=0.7 m/s^2 \\ The situation is shown in the given figure: Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s ^2 .$$\\$ This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.$\\$ Since the angle between a c and a r is 90 $^o$ , the resultant acceleration a is given by:$\\$ $a=\sqrt{a^2_c+a^2_r}\\ =\sqrt{(0.7)^2+(0.5)^2}\\ =\sqrt{0.74}=0.85 m/s\\ \tan \theta =\dfrac{a_c}{a_r}$ $\\$ Where $\theta$ is the angle of the resultant with the direction of velocity.$\\$ $\tan \theta =\dfrac{0.7}{0.5}=1.4\\ q=\tan^{-1}(1.4)\\ =54.46^o\\ 0.86 m/s^2; 54.46^o$ $\\$ with the direction of velocity

32   (a) Show that for a projectile the angle between the velocity and the x-axis as a f unction of time is given by $\\$ $\theta (t)=\tan^{-1}(\dfrac{v_{oy}-gt}{v_{ox}})$ $\\$ (b) Show that the projection angle $\theta_ 0$ for a projectile launched from the origin is given by $\\$ $\theta _0=\tan^{-1}(\dfrac{4h_m}{R})$ $\\$ where the symbols have their usual meaning.

##### Solution :

Let $v_{ ox}$ and $v_{ oy}$ respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.$\\$ Let $v_ x$ and $v_y$ respectively be the horizontal and vertical components of velocity at a point P.$\\$

Time taken by the projectile to reach point P = t$\\$ (a) Applying the first equation of motion along the vertical and horizontal directions, we get:$\\$ $v_y=v_{0y}=gt\\ And \ v_x=v_{ox}\\ \therefore \tan \theta = \dfrac{v_y}{v_x}=\dfrac{v_{oy}-gt}{v_ox}\\ =\theta =\tan^{-1}(\dfrac{v_{oy}-gt}{v_{ox}})$ $\\$ (b) Maximum vertical height, $h_m=\dfrac{u^2_0 \sin^2 2 \theta}{2g}....(i)$ $\\$ Horizontal range,$R=\dfrac{u^2_0 \sin^2 2 \theta}{g}......(ii)$ $\\$ Solving equations (i) and (ii), we get:$\\$ $\dfrac{h_m}{R}=\dfrac{\sin^2 \theta}{2 \sin^2 \theta}\\ =\dfrac{\sin \theta * \sin \theta}{2 * 2 \sin \theta \cos \theta}\\ =\dfrac{1 \sin \theta}{4 \cos \theta} =\dfrac{1}{4}\tan \theta\\ \tan \theta = (\dfrac{4 h_m}{R})\\ \theta =\tan^{-1}=(\dfrac{4h_m}{R})s$

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##### Solution :

Mass of the body, $m = 0.40 kg$$\\ Initial speed of the body, u = 10 m/s$$\\$ due north Force acting on the body, $F = -8.0 N$ $\\$ Acceleration produced in the body,$\\$ $a=\dfrac{F}{m}=\dfrac{-8.0}{0.40}=-20m/s^2$ At $t=-5 s$ $\\$ Acceleration, $a' = 0$ and $u = 10 m/s$ $\\$ $s=ut+\dfrac{1}{2}a't^2\\ =10*(-5)=-50m\\ At t =25 s$ $\\$ Acceleration, $a''=-20m/s^2 and \ u=10m/s$

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34   A truck starts from rest and accelerates uniformly at $2.0 ms ^{-2}$ . At $t = 10 s$, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t = 11 s?$(Neglect air resistance.)

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##### Solution :

Mass of the man, $m = 70 kg$ Acceleration, $a = 0$$\\ Using Newton’s second law of motion, we can write the equation of motion as:\\ R - mg = ma \\ Where, ma is the net force acting on the man.\\ (a) As the lift is moving at a uniform speed, acceleration a = 0\\ \therefore R = mg\\ = 70 × 10 = 700 N \\ \therefore Reading on the weighing scale =\dfrac{700}{g}=\dfrac{700}{10}=70kg \\ Mass of the man, m = 70 kg \\ (b) Acceleration, a = 5 m / s ^2 downward\\ Using Newton’s second law of motion, we can write the equation of motion as:\\ = R + mg = ma \ \ R = m ( g – a )\\ =70 (10 – 5) = 70 × 5\\ =350 N$$\\$ $\therefore$ Reading on the weighing scale =$\dfrac{350}{g}=\dfrac{350}{10}=35 kg$ $\\$ (c) Acceleration, $a = 5 m / s^ 2$ upward $\\$ Using Newton’s second law of motion, we can write the equation of motion as:$\\$ Mass of the man,$m = 70 kg$ $\\$ Acceleration,$a = 5 m / s^ 2$ upward$\\$ Using Newton’s second law of motion, we can write the equation of motion as:$\\$ $R – mg = ma\\ R = m(g + a)\\ =70 (10 + 5) = 70 × 15\\ =1050 N$ $\\$ $\therefore$ Reading on the weighing scale$=\dfrac{150}{g}=\dfrac{1050}{10}=105 kg$ $\\$ (d) When the lift moves freely under gravity, acceleration a = g$\\$ Using Newton’s second law of motion, we can write the equation of motion as:$\\$ $R + mg = ma\\ R = m ( g – a )\\ = m ( g – g )= 0$ $\therefore$Reading on the weighing scale =$\dfrac{0}{g}=0kg$ $\\$ The man will be in a state of weightlessness.

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38   Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

##### Solution :

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The equation of motion can be written as:$\\$ $F – T = ma \\ T = F – m _1 a\\ = 600 – 10 * 20 = 400 N$ $\\$ (b) When force F is applied on body B:$\\$

The equation of motion can be written as:$\\$ $F – T = m_ 2 a\\ T = F – m _2 a\\ \therefore T = 600 – 20 * 20 = 200 N$

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43   A batsman deflects a ball by an angle of 45$^o$ without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

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