NCERT Solutions Class 11 Physics Chapter 3 — Motion in a Plane

NCERT Class 11 Physics — Motion in a Plane, Chapter 3 Exercises. All 22 questions solved.

The exercises at the end of Chapter 3 fall into three groups. Questions 3.1 to 3.7, 3.19 and 3.21 are about vectors themselves — which quantities are vectors, what can be added to what, and how a vector is split into components. Questions 3.8 to 3.11 separate displacement from path length, and average velocity from average speed. Questions 3.12 to 3.18, 3.20 and 3.22 are motion in a plane proper: projectiles, uniform circular motion, and a particle whose position is given as a function of time. The working runs on three sets of results:

• Components — a vector $\vec A$ at angle $\theta$ to the $x$-axis is $A_x\hat{\imath} + A_y\hat{\jmath}$ with $A_x = A\cos\theta$, $A_y = A\sin\theta$; along any unit vector $\hat n$ its component is $\vec A\cdot\hat n$.
• Projectile launched from the ground with speed $u$ at angle $\theta$ — maximum height $H = \dfrac{u^2\sin^2\theta}{2g}$ and horizontal range $R = \dfrac{u^2\sin2\theta}{g}$, with $g = 9.8\ \text{m s}^{-2}$.
• Uniform circular motion — centripetal acceleration $a = \dfrac{v^2}{r} = \omega^2 r$, directed towards the centre, with $\omega = 2\pi\nu$.

Key insight. A vector in a plane is two independent pieces of information, and the questions are easiest when you keep them apart. Displacement depends only on where the motion starts and ends, never on the route, which settles 3.8 to 3.11 before any arithmetic. In a projectile the horizontal motion (constant velocity) and the vertical motion (constant acceleration $g$) run side by side without affecting each other, which is what 3.12, 3.13 and 3.18 use. And in circular motion the velocity keeps changing direction even when its size does not, so there is an acceleration towards the centre (3.14, 3.15), yet over a whole cycle the velocity comes back to where it started (3.16).

Question 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.

Solution. A quantity is a vector if it needs a direction to be specified completely and it combines with others of its kind by the triangle law; otherwise it is a scalar. Volume, mass, density and number of moles are fully described by a number and a unit. Speed is the magnitude of velocity, with no direction attached. Angular frequency, $\omega = 2\pi\nu$, says how many radians are swept per second, not about which axis, so it too is a scalar. Acceleration, velocity and displacement each have a direction, and angular velocity points along the axis of rotation.

Question 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.

Solution. Work is $W = \vec F\cdot\vec d$, a scalar (dot) product, so it is a number with no direction. Current is the trickier one: it flows one way along a wire, but two currents meeting at a junction add as ordinary numbers, not by the triangle law, so current is a scalar. Everything else in the list has a direction and adds as a vector.

Question 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.

Solution. Impulse is force multiplied by time, $\vec F\,\Delta t$, which equals the change in momentum, $\Delta\vec p$. A vector multiplied by a scalar is a vector, so impulse has the direction of the force. Pressure may look like a candidate, but at a point it acts equally in all directions, and the force it produces on a surface takes its direction from the surface, so pressure itself is a scalar. The rest are plainly scalars.

Question 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.

The word that matters throughout is any. Adding physical quantities makes sense only when they have the same dimensions and are the same kind of quantity; multiplying them always makes sense, because the product is simply a new quantity with new dimensions.

Question 3.4 (a)

Solution. Two scalars can be added only if they have the same dimensions. $2\ \text{kg} + 3\ \text{kg}$ is meaningful; $2\ \text{kg} + 3\ \text{s}$ is not. Since the question says any two scalars, the operation is not meaningful in general.

Question 3.4 (b)

Solution. A scalar has no direction, and a vector has one. There is no rule that says which direction the sum would have, so a scalar can never be added to a vector, even when the dimensions match.

Question 3.4 (c)

Solution. Multiplying a vector by a scalar gives another vector: its magnitude is scaled and its direction kept (or reversed, for a negative scalar). Mass times velocity giving momentum, and charge times electric field giving force, are everyday examples.

Question 3.4 (d)

Solution. Any two scalars can be multiplied, whatever their dimensions; the product is a new scalar. Power times time gives energy, and density times volume gives mass.

Question 3.4 (e)

Solution. Like scalars, vectors can be added only if they have the same dimensions: a displacement can be added to a displacement, but not to a velocity. So adding any two vectors is not meaningful.

Question 3.4 (f)

Solution. In this chapter the components of $\vec A$ are defined as the numbers $A_x$ and $A_y$ in $\vec A = A_x\hat{\imath} + A_y\hat{\jmath}$, and the book stresses that $A_x$ is itself not a vector. Adding the number $A_x$ to the vector $\vec A$ is then the same kind of operation as (b), and is not meaningful. (If “component” is read as the vector component $A_x\hat{\imath}$ instead, the sum $\vec A + A_x\hat{\imath}$ is a legitimate vector, which is why some solutions call this operation meaningful; with the book’s own definition of a component, it is not.)

Question 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.

Question 3.5 (a)

Solution. The magnitude of a vector is a number with a unit and no direction, and it is the same whichever way the axes are drawn — the length of an arrow does not change when the axes are rotated.

Question 3.5 (b)

Solution. A scalar has the same value for every choice of axes. A component does not: $A_x = A\cos\theta$ changes as the axes are turned, and it can be positive, negative or zero for the same vector. So a component is not a scalar. Nor is it a vector: it is the vector component $A_x\hat{\imath}$ that carries the direction.

Question 3.5 (c)

Solution. Displacement is the straight line from the starting point to the end point, while path length measures the route actually taken. They are equal only for motion along a straight line without turning back. A particle going once round a circle covers a path of $2\pi r$ with zero displacement.

Question 3.5 (d)

Solution. The path length is never less than the magnitude of the displacement, since the straight line is the shortest route between two points. Dividing both by the same time interval keeps the inequality, so the average speed is greater than or equal to the magnitude of the average velocity, with equality only for straight-line motion in one direction.

Question 3.5 (e)

Solution. Take the three vectors to be $\vec a$, $\vec b$ and $\vec c$. The sum $\vec a + \vec b$ lies in the plane of $\vec a$ and $\vec b$. For the total to be zero, $\vec c$ would have to be $-(\vec a + \vec b)$, which also lies in that plane. So three vectors that add up to zero must be coplanar, and three that are not coplanar cannot add up to zero.

Question 3.6

Establish the following vector inequalities geometrically or otherwise: (a) $|\vec a + \vec b| \le |\vec a| + |\vec b|$ (b) $|\vec a + \vec b| \ge \big||\vec a| – |\vec b|\big|$ (c) $|\vec a – \vec b| \le |\vec a| + |\vec b|$ (d) $|\vec a – \vec b| \ge \big||\vec a| – |\vec b|\big|$. When does the equality sign above apply?

Solution. Place the tail of $\vec b$ at the head of $\vec a$. The vector from the tail of $\vec a$ to the head of $\vec b$ is $\vec a + \vec b$, so the three form a triangle with sides of length $|\vec a|$, $|\vec b|$ and $|\vec a + \vec b|$. For $\vec a – \vec b$, write it as $\vec a + (-\vec b)$ and do the same with $-\vec b$, which has the same length as $\vec b$; that gives a triangle with sides $|\vec a|$, $|\vec b|$ and $|\vec a – \vec b|$. Everything then follows from two facts about any triangle: one side is never longer than the sum of the other two, and never shorter than their difference. Where the vectors are parallel or antiparallel the triangle collapses into a straight line, and that is exactly where the equality signs apply.

Question 3.6 (a)

Solution. In the triangle for $\vec a + \vec b$, the side $|\vec a + \vec b|$ cannot exceed the sum of the other two sides, so $|\vec a + \vec b| \le |\vec a| + |\vec b|$. The two sides add up exactly only when the triangle is flattened with $\vec b$ continuing in the same direction as $\vec a$.

Question 3.6 (b)

Solution. In the same triangle, one side is never shorter than the difference of the other two, so $|\vec a + \vec b| \ge \big||\vec a| – |\vec b|\big|$. The shortest possible $\vec a + \vec b$ occurs when $\vec b$ doubles back along $\vec a$, so equality needs the two to point in opposite directions.

Question 3.6 (c)

Solution. In the triangle for $\vec a + (-\vec b)$, the side $|\vec a – \vec b|$ is at most $|\vec a| + |-\vec b| = |\vec a| + |\vec b|$. Equality needs $-\vec b$ to point the same way as $\vec a$, that is, $\vec b$ opposite to $\vec a$.

Question 3.6 (d)

Solution. In the same triangle, $|\vec a – \vec b| \ge \big||\vec a| – |\vec b|\big|$. Equality needs $-\vec b$ to point against $\vec a$, that is, $\vec b$ in the same direction as $\vec a$.

Note on the book’s answer. The key says only that equality holds “for collinear vectors”. That is right but incomplete: (a) and (d) need the vectors to point the same way, and (b) and (c) need them to point opposite ways. All four also hold with equality when either vector is zero.

Question 3.7

Given $\vec a + \vec b + \vec c + \vec d = \vec 0$, which of the following statements are correct: (a) $\vec a$, $\vec b$, $\vec c$ and $\vec d$ must each be a null vector, (b) the magnitude of $(\vec a + \vec c)$ equals the magnitude of $(\vec b + \vec d)$, (c) the magnitude of $\vec a$ can never be greater than the sum of the magnitudes of $\vec b$, $\vec c$ and $\vec d$, (d) $\vec b + \vec c$ must lie in the plane of $\vec a$ and $\vec d$ if $\vec a$ and $\vec d$ are not collinear, and in the line of $\vec a$ and $\vec d$, if they are collinear?

Solution. Each statement is tested by rearranging the given equation.

(a) False. Four non-zero vectors can add up to zero: the four sides of any quadrilateral, taken head to tail round it, do exactly that.

(b) True. Rearranging, $\vec a + \vec c = -(\vec b + \vec d)$. A vector and its negative have the same magnitude.

(c) True. Rearranging, $\vec a = -(\vec b + \vec c + \vec d)$, so $|\vec a| = |\vec b + \vec c + \vec d| \le |\vec b| + |\vec c| + |\vec d|$ by the triangle inequality of Question 3.6 applied twice.

(d) True. Rearranging, $\vec b + \vec c = -(\vec a + \vec d)$. If $\vec a$ and $\vec d$ are not collinear, $\vec a + \vec d$ lies in their plane, and so does its negative. If they are collinear, $\vec a + \vec d$ lies along their common line, and so does $\vec b + \vec c$.

Question 3.8

Three girls skating on a circular ice ground of radius $200\ \text{m}$ 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 which girl is this equal to the actual length of path skated?

Solution. Displacement is the straight line from the starting point to the finishing point, and it pays no attention to the route. All three girls start at $P$ and finish at $Q$, so they all have the same displacement: the diameter $PQ$, of length $2\times200 = 400\ \text{m}$. Girl $B$ skates straight along that diameter, so her path length is also $400\ \text{m}$. Girl $A$ and the third girl (labelled $C$ here; the book leaves her path unlabelled) take curved routes that are longer than the straight line.

Question 3.9

A cyclist starts from the centre $O$ of a circular park of radius $1\ \text{km}$, 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 $10\ \text{min}$, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

In the figure $P$ and $Q$ are a quarter of the circle apart ($\angle POQ = 90^\circ$), so the ride along the circumference is a quarter of the circumference.

Question 3.9 (a)

Solution. The cyclist ends where he started, at $O$. Displacement depends only on the start and end points, so it is zero, however far he rode.

Question 3.9 (b)

Solution. Average velocity is displacement divided by time. The displacement is zero, so the average velocity is zero too.

Question 3.9 (c)

Solution. Average speed uses the whole path length instead. The path is $OP$, then the quarter arc $PQ$, then $QO$:

$$\text{path} = 1 + \frac{2\pi(1)}{4} + 1 = 2 + \frac{\pi}{2} = 3.571\ \text{km}$$

The time is $10\ \text{min} = \tfrac16\ \text{h}$, so

$$\text{average speed} = \frac{3.571\ \text{km}}{1/6\ \text{h}} = 21.4\ \text{km h}^{-1}$$

Question 3.10

On an open ground, a motorist follows a track that turns to his left by an angle of $60^\circ$ after every $500\ \text{m}$. 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.

Solution. Turning left by $60^\circ$ after every $500\ \text{m}$ means six legs complete a full turn of $360^\circ$, so the track is a regular hexagon of side $500\ \text{m}$. Measure every direction from the first leg, the direction in which the motorist leaves the starting turn.

Third turn. The motorist has driven three sides, $1500\ \text{m}$, and stands at the corner diametrically opposite the start. Across a regular hexagon that distance is twice the side, $1000\ \text{m}$. Adding the three legs as components confirms it: $500(\cos0^\circ + \cos60^\circ + \cos120^\circ) = 500$ and $500(\sin0^\circ + \sin60^\circ + \sin120^\circ) = 866$, so the displacement is $\sqrt{500^2 + 866^2} = 1000\ \text{m}$ at $\tan^{-1}(866/500) = 60^\circ$ to the first leg.

Sixth turn. After six sides he is back where he began, so the displacement is zero, while the path is $3000\ \text{m}$.

Eighth turn. The first six legs cancel, leaving the effect of two legs, one at $0^\circ$ and one at $60^\circ$: $(500 + 250,\ 0 + 433) = (750,\ 433)$, of magnitude $\sqrt{750^2 + 433^2} = 866\ \text{m}$ at $\tan^{-1}(433/750) = 30^\circ$ to the first leg. The path is $8\times500 = 4000\ \text{m}$.

In each case the path length is greater than the magnitude of the displacement, because the track keeps changing direction.

Question 3.11

A passenger arriving in a new town wishes to go from the station to a hotel located $10\ \text{km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\ \text{km}$ long and reaches the hotel in $28\ \text{min}$. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

The time is $28\ \text{min} = \tfrac{28}{60}\ \text{h}$.

Question 3.11 (a)

Solution. Average speed divides the distance actually travelled, $23\ \text{km}$, by the time:

$$\text{average speed} = \frac{23}{28/60} = 49.3\ \text{km h}^{-1}$$

Question 3.11 (b)

Solution. Average velocity divides the displacement by the time, and the displacement is the straight-line $10\ \text{km}$ from station to hotel, whatever route the cab took:

$$|\text{average velocity}| = \frac{10}{28/60} = 21.4\ \text{km h}^{-1}$$

The two are not equal. They would be only if the cab had driven straight from the station to the hotel, so that the path length equalled the displacement.

Question 3.12

The ceiling of a long hall is $25\ \text{m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\ \text{m s}^{-1}$ can go without hitting the ceiling of the hall?

Solution. The range $R = u^2\sin2\theta/g$ grows as the angle of projection rises towards $45^\circ$, but so does the height. At $45^\circ$ the ball would climb to $u^2/4g = 1600/39.2 = 40.8\ \text{m}$ and hit the ceiling. So the best the thrower can do is choose the steepest angle whose highest point just touches the $25\ \text{m}$ ceiling — any steeper and the ball hits it, any shallower and the range is smaller:

$$H = \frac{u^2\sin^2\theta}{2g} = 25 \quad\Rightarrow\quad \sin^2\theta = \frac{2\times9.8\times25}{40^2} = 0.306,\qquad \sin\theta = 0.553,\qquad \theta = 33.6^\circ$$

The range at this angle is

$$R = \frac{u^2\sin2\theta}{g} = \frac{1600\times\sin67.2^\circ}{9.8} = \frac{1600\times0.922}{9.8} = 150.5\ \text{m}$$

The hall is described as long, so the ball does not reach a wall first.

Question 3.13

A cricketer can throw a ball to a maximum horizontal distance of $100\ \text{m}$. How much high above the ground can the cricketer throw the same ball?

Solution. The maximum range comes from throwing at $45^\circ$, where $\sin2\theta = 1$ and $R_{\max} = u^2/g$. That fixes the speed he can give the ball:

$$\frac{u^2}{g} = 100\ \text{m} \quad\Rightarrow\quad u^2 = 980\ \text{m}^2\,\text{s}^{-2}$$

To send the ball as high as possible he throws it straight up with the same speed. It rises until its velocity is zero, so $0 = u^2 – 2gH$ and

$$H = \frac{u^2}{2g} = \frac{R_{\max}}{2} = 50\ \text{m}$$

Question 3.14

A stone tied to the end of a string $80\ \text{cm}$ long is whirled in a horizontal circle with a constant speed. If the stone makes $14$ revolutions in $25\ \text{s}$, what is the magnitude and direction of acceleration of the stone?

Solution. The speed is constant, so there is no acceleration along the path; the only acceleration is the centripetal one that keeps turning the velocity. The frequency is $\nu = 14/25\ \text{s}^{-1}$, so the angular speed is

$$\omega = 2\pi\nu = 2\pi\times\frac{14}{25} = 3.52\ \text{rad s}^{-1}$$

With $r = 0.80\ \text{m}$:

$$a = \omega^2 r = (3.52)^2\times0.80 = 9.9\ \text{m s}^{-2}$$

It points along the string, from the stone towards the centre of the circle, and so changes direction continuously as the stone goes round.

Question 3.15

An aircraft executes a horizontal loop of radius $1.00\ \text{km}$ with a steady speed of $900\ \text{km/h}$. Compare its centripetal acceleration with the acceleration due to gravity.

Solution. Convert to SI units first: $900\ \text{km/h} = 900\times\dfrac{1000}{3600} = 250\ \text{m s}^{-1}$ and $r = 1000\ \text{m}$. Then

$$a = \frac{v^2}{r} = \frac{250^2}{1000} = 62.5\ \text{m s}^{-2},\qquad \frac{a}{g} = \frac{62.5}{9.8} = 6.4$$

Question 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.

Question 3.16 (a)

Solution. The acceleration points exactly at the centre only if the speed is constant. If the particle speeds up or slows down, it also has a tangential acceleration along its path, and the net acceleration is the vector sum of the two, which points off the radius.

Question 3.16 (b)

Solution. Velocity is the limit of $\Delta\vec r/\Delta t$ as $\Delta t \to 0$. As the interval shrinks, the chord $\Delta\vec r$ between two nearby points on the path turns into the tangent at the point, so the velocity always lies along the tangent, whatever the shape of the path.

Question 3.16 (c)

Solution. Average acceleration is the change in velocity divided by the time taken. After one full cycle the particle is back at its starting point moving with exactly the same velocity, so $\Delta\vec v = \vec 0$ and the average acceleration over the cycle is a null vector — even though the instantaneous acceleration is never zero.

Question 3.17

The position of a particle is given by

$$\vec r = 3.0t\,\hat{\imath} – 2.0t^2\,\hat{\jmath} + 4.0\,\hat k\ \text{m}$$

where $t$ is in seconds and the coefficients have the proper units for $\vec r$ to be in metres. (a) Find the $\vec v$ and $\vec a$ of the particle. (b) What is the magnitude and direction of velocity of the particle at $t = 2.0\ \text{s}$?

Question 3.17 (a)

Solution. Velocity is the rate of change of position, and acceleration the rate of change of velocity, so each is found by differentiating component by component. The $\hat k$ term is constant and drops out:

$$\vec v = \frac{d\vec r}{dt} = (3.0\,\hat{\imath} – 4.0t\,\hat{\jmath})\ \text{m s}^{-1},\qquad \vec a = \frac{d\vec v}{dt} = -4.0\,\hat{\jmath}\ \text{m s}^{-2}$$

The acceleration is constant: the particle moves uniformly along $x$ while accelerating steadily in the $-y$ direction, like a projectile, in the plane $z = 4.0\ \text{m}$.

Question 3.17 (b)

Solution. At $t = 2.0\ \text{s}$, $\vec v = 3.0\,\hat{\imath} – 8.0\,\hat{\jmath}\ \text{m s}^{-1}$. Its magnitude is

$$v = \sqrt{3.0^2 + 8.0^2} = \sqrt{73} = 8.54\ \text{m s}^{-1}$$

and it makes an angle $\theta = \tan^{-1}(8.0/3.0) = 69.4^\circ \approx 70^\circ$ with the $x$-axis. The $y$-component is negative, so the velocity points below the $x$-axis, towards $-y$.

Note on the book’s answer. The key gives “$70^\circ$ with $x$-axis”, which is the right size of angle; the direction is on the $-y$ side of the axis, i.e. $-70^\circ$.

Question 3.18

A particle starts from the origin at $t = 0\ \text{s}$ with a velocity of $10.0\,\hat{\jmath}\ \text{m/s}$ and moves in the $x$-$y$ plane with a constant acceleration of $(8.0\,\hat{\imath} + 2.0\,\hat{\jmath})\ \text{m s}^{-2}$. (a) At what time is the $x$-coordinate of the particle $16\ \text{m}$? What is the $y$-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

The acceleration is constant, so the equations of uniformly accelerated motion hold separately along $x$ and along $y$: $x = u_xt + \tfrac12a_xt^2$ and so on, with $u_x = 0$, $u_y = 10.0\ \text{m/s}$, $a_x = 8.0\ \text{m s}^{-2}$ and $a_y = 2.0\ \text{m s}^{-2}$.

Question 3.18 (a)

Solution. Along $x$ the particle starts from rest:

$$x = \tfrac12(8.0)t^2 = 4.0t^2 = 16 \quad\Rightarrow\quad t = 2.0\ \text{s}$$

At that same instant, along $y$:

$$y = u_yt + \tfrac12a_yt^2 = 10.0(2.0) + \tfrac12(2.0)(2.0)^2 = 20 + 4 = 24\ \text{m}$$

Question 3.18 (b)

Solution. The velocity components at $t = 2.0\ \text{s}$ are $v_x = a_xt = 16\ \text{m/s}$ and $v_y = u_y + a_yt = 10.0 + 4.0 = 14\ \text{m/s}$. The speed is the magnitude of the velocity:

$$v = \sqrt{16^2 + 14^2} = \sqrt{452} = 21.26\ \text{m s}^{-1}$$

Question 3.19

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

Solution. The two vectors. $\hat{\imath} + \hat{\jmath}$ has components $(1, 1)$, so its magnitude is $\sqrt{1^2 + 1^2} = \sqrt2$ and it makes $\tan^{-1}(1/1) = 45^\circ$ with the $x$-axis. $\hat{\imath} – \hat{\jmath}$ has components $(1, -1)$: the same magnitude $\sqrt2$, at $\tan^{-1}(-1/1) = -45^\circ$, that is, $45^\circ$ below the $x$-axis.

The components of $\vec A$. The component of a vector along a direction is its projection on that direction, found by dotting it with the unit vector along it. So first divide each vector by its magnitude, $\sqrt2$:

$$\hat n_1 = \frac{\hat{\imath} + \hat{\jmath}}{\sqrt2},\qquad A_1 = \vec A\cdot\hat n_1 = \frac{2 + 3}{\sqrt2} = \frac{5}{\sqrt2}$$

$$\hat n_2 = \frac{\hat{\imath} – \hat{\jmath}}{\sqrt2},\qquad A_2 = \vec A\cdot\hat n_2 = \frac{2 – 3}{\sqrt2} = -\frac{1}{\sqrt2}$$

The negative sign means the projection falls on the far side of $O$ from $\hat{\imath} – \hat{\jmath}$, as the figure shows. The two directions are perpendicular, so the components must rebuild $\vec A$, and they do: $\tfrac{5}{\sqrt2}\hat n_1 – \tfrac{1}{\sqrt2}\hat n_2 = \tfrac52(\hat{\imath} + \hat{\jmath}) – \tfrac12(\hat{\imath} – \hat{\jmath}) = 2\,\hat{\imath} + 3\,\hat{\jmath}$.

Question 3.20

For any arbitrary motion in space, which of the following relations are true: (a) $\vec v_{\text{average}} = \tfrac12\big(\vec v(t_1) + \vec v(t_2)\big)$ (b) $\vec v_{\text{average}} = \big[\vec r(t_2) – \vec r(t_1)\big]/(t_2 – t_1)$ (c) $\vec v(t) = \vec v(0) + \vec a\,t$ (d) $\vec r(t) = \vec r(0) + \vec v(0)\,t + \tfrac12\vec a\,t^2$ (e) $\vec a_{\text{average}} = \big[\vec v(t_2) – \vec v(t_1)\big]/(t_2 – t_1)$. (The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_2$.)

Solution. The test is whether a relation is a definition, which holds for every motion, or a result derived on the assumption that the acceleration is constant, which does not.

(b) and (e) are true. They are the definitions of average velocity (displacement over time) and average acceleration (change in velocity over time), and they need no assumption about how the motion happens in between.

(a), (c) and (d) are false in general. All three come from integrating a constant acceleration. For (a), the average of the initial and final velocities equals the average velocity only when the velocity changes uniformly with time. In (c) and (d), $\vec a$ is a single fixed vector, which an arbitrary motion does not have — a particle in uniform circular motion, for one, has an acceleration that keeps turning.

Question 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.

Question 3.21 (a)

Solution. Being a scalar says nothing about being conserved. Kinetic energy is a scalar, yet in an inelastic collision some of it is lost to heat and sound; the temperature of a cooling cup of tea is a scalar that falls steadily.

Question 3.21 (b)

Solution. Many scalars can be negative: a temperature of $-10\ ^\circ\text{C}$, the gravitational potential energy of a body bound to the Earth, or the charge on an electron.

Question 3.21 (c)

Solution. Mass, speed, density and energy are all scalars, and every one of them has dimensions. Dimensionless scalars exist (the refractive index, a coefficient of friction), but they are the exception.

Question 3.21 (d)

Solution. A scalar can be different at different places. Temperature varies from point to point in a room, and gravitational potential changes with height above the Earth. A scalar that varies in space is called a scalar field.

Question 3.21 (e)

Solution. This is exactly what makes a quantity a scalar. Rotating the axes changes the components of a vector, but it cannot change the mass of a body, the length of a rod or the temperature of a gas; every observer, however the axes are turned, measures the same value.

Question 3.22

An aircraft is flying at a height of $3400\ \text{m}$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0\ \text{s}$ apart is $30^\circ$, what is the speed of the aircraft?

Solution. The question does not say where the observer stands, and the answer depends on it. The intended reading, and the one that gives the book’s answer, is that the observer is directly below the midpoint $M$ of the two positions $A$ and $B$, so the $30^\circ$ is split into $15^\circ$ on either side of the vertical $OM = 3400\ \text{m}$. In the right-angled triangle $OMB$,

$$MB = OM\tan15^\circ = 3400\times0.2679 = 911\ \text{m}$$

so the aircraft covers $AB = 2\times911 = 1822\ \text{m}$ in $10.0\ \text{s}$. It flies level at constant speed, so

$$v = \frac{1822}{10.0} = 182\ \text{m s}^{-1}$$

Common mistakes

• Question 3.4(f) and 3.5(b): treating a component as either a vector or a scalar. The book defines the component $A_x$ as a number that is not a vector, which is why (f) is not meaningful. But a component is not a scalar either, because it changes when the axes are rotated, which is why 3.5(b) is false. Students who have learnt “vectors and scalars are the only two kinds” find both answers puzzling.
• Question 3.9: giving a non-zero average velocity. The cyclist rides $3.57\ \text{km}$ but ends where he started. Average velocity uses displacement, which is zero; only average speed uses the distance ridden.
• Question 3.10: giving the sixth-turn displacement as $3\ \text{km}$, or losing the angle at the eighth turn. Path length and displacement part company as soon as the track bends. Adding the $500\ \text{m}$ legs as components, each turned a further $60^\circ$, keeps the directions honest.
• Question 3.12: using $\theta = 45^\circ$ because it gives the maximum range. At $45^\circ$ the ball would rise to $40.8\ \text{m}$ and hit the ceiling. The ceiling, not the $45^\circ$ rule, sets the angle here: $33.6^\circ$, found from $H = 25\ \text{m}$.
• Question 3.13: giving $100\ \text{m}$, or the $25\ \text{m}$ height of the $45^\circ$ throw. The greatest height comes from throwing straight up, and it is $u^2/2g$, exactly half the maximum range $u^2/g$.
• Question 3.14: forgetting the $2\pi$. The number of revolutions per second is $\nu = 0.56\ \text{s}^{-1}$, but the formula needs radians per second, $\omega = 2\pi\nu$. Leaving out the $2\pi$ gives an acceleration $4\pi^2 \approx 39$ times too small; forgetting to turn $80\ \text{cm}$ into $0.80\ \text{m}$ makes it $100$ times too large.
• Question 3.17(b): putting the velocity $70^\circ$ above the $x$-axis. At $t = 2.0\ \text{s}$ the $y$-component is $-8.0\ \text{m s}^{-1}$, so the velocity points below the axis. Always check the sign of each component before stating a direction.
• Question 3.20: applying $\vec v = \vec v(0) + \vec a t$ to every motion. The equations of motion were derived for constant acceleration. Only the two definitions, (b) and (e), hold for arbitrary motion.

Practise next

• Chapter 4, Laws of Motion — the accelerations found here, including the centripetal one, now get a cause: the net force.
• Chapter 2, Motion in a Straight Line — the equations of uniformly accelerated motion that this chapter applies separately along each axis.