Motion in a Plane — NCERT Exercises

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 and angular acceleration.

Answer

Scalar quantities : volume, mass, speed, density, number of moles and angular frequency.

Vector quantities : acceleration, velocity, displacement, angular velocity and angular acceleration.

Reason — A scalar quantity is described solely by its magnitude and has no associated direction, while a vector quantity has both magnitude and direction and obeys the laws of vector addition. Volume, mass, speed, density, number of moles and angular frequency are completely specified by a number and a unit alone, so they are scalars. Acceleration, velocity, displacement, angular velocity and angular acceleration cannot be specified without stating a direction, so they are vectors.

Pick out two scalar quantities in the following list :

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

Answer

Work and current are the two scalar quantities.

Reason — Work is the scalar product of force and displacement, $\text W = \vec{\text F} \cdot \vec{\text s}$, and the scalar product of two vectors is always a scalar. Electric current has both magnitude and direction, but it does not obey the laws of vector addition (currents at a junction add algebraically), so it is a scalar. All the remaining quantities in the list have both magnitude and direction and obey the laws of vector addition, so they are vectors.

Pick out the only vector quantity in the following list :

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

Answer

Impulse is the only vector quantity.

Reason — Impulse is the product of force and time interval, $\vec{\text J} = \vec{\text F}\Delta \text t$. Since force is a vector and time is a scalar, the product is a vector having the same direction as the force. All the other quantities listed are completely specified by their magnitude alone, so they are scalars.

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.

Answer

(a) Not always meaningful. Two scalars can be added only if they represent the same physical quantity, that is, if they have the same dimensions. Mass and time cannot be added, but two masses can.

(b) Not meaningful. A scalar has only magnitude while a vector has magnitude as well as direction. Even if the dimensions are the same, the two cannot be added.

(c) Meaningful. On multiplying a vector $\vec{\text A} \space$by a scalar k, a vector $\text k\vec{\text A} \space$is obtained whose magnitude is k times that of $\vec{\text A}$. If k is a pure number the new vector represents the same physical quantity, and if k has a unit the product represents a new physical quantity, as in $\vec{\text p} = \text m\vec{\text v}$.

(d) Meaningful. The product of two scalars is always a scalar, and it represents a new physical quantity, for example power multiplied by time gives work.

(e) Not always meaningful. Two vectors can be added only if they represent the same physical quantity, that is, if they have the same dimensions. Force cannot be added to velocity.

(f) Meaningful. A component of a vector is itself a vector of the same physical quantity, so it can be added to the vector by the laws of vector addition.

Read each of the following statements 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 is either greater than, or equal to, the magnitude of the average velocity of the particle over the same time-interval.

(e) Three vectors not lying in a plane can never add up to give a null vector.

Answer

(a) True. The magnitude of a vector gives only its size and carries no direction, so it is a scalar.

(b) False. A vector is resolved into components which are themselves vectors along the chosen directions, so each component is a vector and not a scalar.

(c) False. The total path length is the actual length of the path travelled, whereas the magnitude of the displacement is the straight line distance between the initial and the final position. The two are equal only when the particle moves along a straight line without reversing its direction; otherwise the path length is greater.

(d) True. The average speed is the total path length divided by the time interval and the magnitude of the average velocity is the magnitude of the displacement divided by the same time interval. Since the path length is always greater than or equal to the magnitude of the displacement, the average speed is greater than or equal to the magnitude of the average velocity.

(e) True. For three vectors to give a null vector they must form a closed triangle, and a triangle always lies in one plane. Hence three vectors which do not lie in a plane can never add up to give a null vector.

Establish the following vector inequalities geometrically or otherwise :

(a) $|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$

(b) $|\vec{A} + \vec{B}| \ge \left| |\vec{A}| - |\vec{B}| \right|$

(c) $|\vec{A} - \vec{B}| \le |\vec{A}| + |\vec{B}|$

(d) $|\vec{A} - \vec{B}| \ge \left| |\vec{A}| - |\vec{B}| \right|$

When does the equality sign above apply?

Answer

Let two vectors $\vec{\text A} \space$and $\vec{\text B} \space$be inclined to each other at an angle θ. By the parallelogram law of vector addition,

$|\vec{\text A} + \vec{\text B}| = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

and replacing θ by (180° − θ) for the difference,

$|\vec{\text A} - \vec{\text B}| = \sqrt{\text A^2 + \text B^2 - 2\text{AB}\cos \theta}$

(a) Since cos θ can at most be 1,

$|\vec{\text A} + \vec{\text B}| \le \sqrt{\text A^2 + \text B^2 + 2\text{AB}} = \sqrt{(\text A + \text B)^2} = \text A + \text B$

Hence $|\vec{\text A} + \vec{\text B}| \le |\vec{\text A}| + |\vec{\text B}|$. The equality holds when cos θ = 1, that is, when θ = 0° and the two vectors are in the same direction.

(b) Since cos θ is at least −1,

$|\vec{\text A} + \vec{\text B}| \ge \sqrt{\text A^2 + \text B^2 - 2\text{AB}} = \sqrt{(\text A - \text B)^2} = |\text A - \text B|$

Hence $|\vec{\text A} + \vec{\text B}| \ge \left| |\vec{\text A}| - |\vec{\text B}| \right|$. The equality holds when cos θ = −1, that is, when θ = 180° and the two vectors are in opposite directions.

(c) For the difference, the largest value occurs when cos θ = −1,

$|\vec{\text A} - \vec{\text B}| \le \sqrt{\text A^2 + \text B^2 + 2\text{AB}} = \text A + \text B$

Hence $|\vec{\text A} - \vec{\text B}| \le |\vec{\text A}| + |\vec{\text B}|$. The equality holds when θ = 180°, that is, when the two vectors are in opposite directions.

(d) The smallest value of the difference occurs when cos θ = 1,

$|\vec{\text A} - \vec{\text B}| \ge \sqrt{\text A^2 + \text B^2 - 2\text{AB}} = |\text A - \text B|$

Hence $|\vec{\text A} - \vec{\text B}| \ge \left| |\vec{\text A}| - |\vec{\text B}| \right|$. The equality holds when θ = 0°, that is, when the two vectors are in the same direction.

Given $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$, which of the following statements are correct?

(a) $\vec{A}$, $\vec{B}$, $\vec{C} \space$and $\vec{D} \space$must each be a null vector.

(b) The magnitude of $(\vec{A} + \vec{C}) \space$equals the magnitude of $(\vec{B} + \vec{D})$.

(c) The magnitude of $\vec{A} \space$can never be greater than the sum of the magnitudes of $\vec{B}$, $\vec{C} \space$and $\vec{D}$.

(d) $(\vec{B} + \vec{C}) \space$must lie in the plane of $\vec{A} \space$and $\vec{D} \space$if $\vec{A} \space$and $\vec{D} \space$are not collinear, and in the line of $\vec{A} \space$and $\vec{D}$, if they are collinear ?

Answer

Statements (b), (c) and (d) are correct and statement (a) is incorrect.

Reason —

(a) The four vectors add up to a null vector because they complete a closed polygon when drawn in succession. This does not require any of them to be a null vector, so the statement is incorrect.

(b) Since $\vec{\text A} + \vec{\text B} + \vec{\text C} + \vec{\text D} = 0$,

$(\vec{\text A} + \vec{\text C}) = -(\vec{\text B} + \vec{\text D})$

Two vectors which are the negative of each other have equal magnitudes, so the statement is correct.

(c) From the given relation, $\vec{\text A} = -(\vec{\text B} + \vec{\text C} + \vec{\text D})$. Since the magnitude of the sum of vectors can never exceed the sum of their magnitudes, the magnitude of $\vec{\text A} \space$can never be greater than the sum of the magnitudes of $\vec{\text B}$, $\vec{\text C} \space$and $\vec{\text D}$. The statement is correct.

(d) From the given relation, $(\vec{\text B} + \vec{\text C}) = -(\vec{\text A} + \vec{\text D})$. The sum $(\vec{\text A} + \vec{\text D}) \space$lies in the plane containing $\vec{\text A} \space$and $\vec{\text D}$, and if $\vec{\text A} \space$and $\vec{\text D} \space$are collinear their sum lies along the same line. Hence $(\vec{\text B} + \vec{\text C}) \space$must lie in that plane or along that line, and the statement is correct.

Three girls A, B and C 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. What is the magnitude of the displacement vector for each girl? For which girl is this equal to the actual length of path skated?

Answer

Given,

• Radius of the circular ice ground, r = 200 m
• P and Q are diametrically opposite points

The displacement vector of each girl is the vector drawn from the initial position P to the final position Q. Since all three girls start from P and reach the same point Q, the displacement vector is the same for each of them, and its magnitude is the diameter PQ,

$\text{PQ} = 2\text r = 2 \times 200 \\[1em] = 400\ \text m$

Hence, the magnitude of the displacement vector is 400 m for each girl.

The magnitude of the displacement is equal to the actual length of the path only when the path is a straight line. Girl B skates along the straight line PQ, so for girl B the length of the path skated is also 400 m, equal to the magnitude of her displacement.

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. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity and (c) average speed of the cyclist?

Answer

Given,

• Radius of the circular park, r = 1 km
• Time taken for the round trip, t = 10 min = $\dfrac{10}{60}$ h = $\dfrac{1}{6}$ h

(a) The cyclist starts from the centre O and finally returns to the same point O. Since the initial and the final positions are the same,

$\text{Net displacement} = 0$

(b) The average velocity is the net displacement divided by the total time taken,

$\vec{\text v}_{av} = \dfrac{\text{Net displacement}}{\text{Total time}} = \dfrac{0}{1/6} \\[1em] = 0$

(c) The total path length covered is

$\text{OP} + \text{arc PQ} + \text{QO}$

Here OP = QO = r = 1 km, and PQ is a quarter of the circumference,

$\text{arc PQ} = \dfrac{1}{4}(2\pi \text r) = \dfrac{1}{4} \times 2 \times 3.14 \times 1 \\[1em] = 1.57\ \text{km}$

Therefore,

$\text{Total path length} = 1 + 1.57 + 1 = 3.57\ \text{km}$

The average speed is

$\text{Average speed} = \dfrac{\text{Total path length}}{\text{Total time}} = \dfrac{3.57}{1/6} \\[1em] = 3.57 \times 6 \\[1em] = 21.4\ \text{km h}^{-1}$

Hence, the net displacement and the average velocity are both zero, and the average speed is 21.4 km h^-1.

On an open ground, a motorist follows a track that turns to his left by an angle of 60° 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.

Answer

Given,

• Length of each straight stretch = 500 m
• Turn through 60° to the left after every stretch

Since the motorist turns through 60° after every 500 m, the exterior angle of the path is 60° and the track is a regular hexagon of side 500 m.

At the third turn : The motorist has covered three sides of the hexagon and reaches the vertex diametrically opposite to the starting point. Hence the magnitude of the displacement is

$2 \times 500 = 1000\ \text m$ ∵ Diagonal of hexagon = 2 × side of hexagon

The total path length is

$3 \times 500 = 1500\ \text m$

At the sixth turn : The motorist completes the hexagon and returns to the starting point. Hence

$\text{Displacement} = 0, \quad \text{Path length} = 6 \times 500 = 3000\ \text m$

At the eighth turn : After six turns the motorist is back at the starting point, so the position after the eighth turn is the same as that after two turns. The displacement is the diagonal joining the ends of two adjacent sides,

$\text{Displacement} = 2 \times 500 \times \cos 30^\circ \\[1em] = 2 \times 500 \times \dfrac{\sqrt{3}}{2} \\[1em] = 500\sqrt{3} = 866\ \text m$

and it makes an angle of 30° with the initial direction of motion. The total path length is

$8 \times 500 = 4000\ \text m$

Comparison : At the third turn the magnitude of the displacement (1000 m) is less than the path length (1500 m); at the sixth turn the displacement is zero while the path length is 3000 m; and at the eighth turn the displacement (866 m) is much less than the path length (4000 m).

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?

Answer

Given,

• Total path length covered = 23 km
• Magnitude of displacement = 10 km
• Time taken, t = 28 min = $\dfrac{28}{60}$ h

(a) The average speed of the taxi is

$\text{Average speed} = \dfrac{\text{Total path length}}{\text{Total time}} = \dfrac{23}{28/60} \\[1em] = \dfrac{23 \times 60}{28} \\[1em] = 49.3\ \text{km h}^{-1}$

(b) The magnitude of the average velocity is

$|\vec{\text v}_{av}| = \dfrac{\text{Magnitude of displacement}}{\text{Total time}} = \dfrac{10}{28/60} \\[1em] = \dfrac{10 \times 60}{28} \\[1em] = 21.4\ \text{km h}^{-1}$

No, the two are not equal. The average speed is greater than the magnitude of the average velocity because the path followed by the taxi is not a straight line, so the path length (23 km) is greater than the magnitude of the displacement (10 km).

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 ms^-1 can go without hitting the ceiling of the hall? (g = 9.8 ms^-2)

Answer

Given,

• Speed of projection, u = 40 m s^-1
• Maximum height available, h = 25 m
• Acceleration due to gravity, g = 9.8 m s^-2

For the ball to just graze the ceiling, the maximum height of the projectile must be 25 m. The maximum height attained is

$\text h = \dfrac{\text u^2 \sin^2 \theta}{2\text g}$

Substituting the values,

$25 = \dfrac{(40)^2 \sin^2 \theta}{2 \times 9.8} \\[1em] \sin^2 \theta = \dfrac{25 \times 2 \times 9.8}{1600} \\[1em] = 0.30625$

Therefore,

$\sin \theta = 0.5534 \quad \text{and} \quad \cos \theta = \sqrt{1 - 0.30625} = 0.8329$

The horizontal range is

$\text R = \dfrac{\text u^2 \sin 2\theta}{\text g} = \dfrac{\text u^2 \times 2 \sin \theta \cos \theta}{\text g}$

Substituting the values,

$\text R = \dfrac{1600 \times 2 \times 0.5534 \times 0.8329}{9.8} \\[1em] = \dfrac{1474.9}{9.8} \\[1em] = 150.5\ \text m$

Hence, the maximum horizontal distance covered without hitting the ceiling is 150.5 m.

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?

Answer

Given,

• Maximum horizontal range, R_max = 100 m

The horizontal range is maximum when the angle of projection is 45°, and then

$\text R_{max} = \dfrac{\text u^2}{\text g}$

Therefore,

$\dfrac{\text u^2}{\text g} = 100 \quad \Rightarrow \quad \text u^2 = 100\ \text g$

To throw the ball to the greatest height, the cricketer must throw it vertically upwards, that is, at θ = 90°. The maximum height then attained is

$\text h = \dfrac{\text u^2 \sin^2 90^\circ}{2\text g} = \dfrac{\text u^2}{2\text g}$

Substituting u² = 100 g,

$\text h = \dfrac{100\ \text g}{2\text g} \\[1em] = 50\ \text m$

Hence, the cricketer can throw the ball to a maximum height of 50 m above the ground.

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 ?

Answer

Given,

• Radius of the circular path, r = 80 cm = 0.80 m
• Number of revolutions = 14 in 25 s

The angular frequency is

$\text n = \dfrac{14}{25}\ \text{rev s}^{-1}$

The angular velocity is

$\omega = 2\pi \text n = 2 \times 3.14 \times \dfrac{14}{25} \\[1em] = 3.52\ \text{rad s}^{-1}$

The centripetal acceleration is

$\text a = \omega^2 \text r$

Substituting the values,

$\text a = (3.52)^2 \times 0.80 \\[1em] = 12.39 \times 0.80 \\[1em] = 9.9\ \text{m s}^{-2}$

Hence, the acceleration of the stone is 9.9 m s^-2, directed along the radius towards the centre of the circular path.

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.

Answer

Given,

• Radius of the horizontal loop, r = 1.00 km = 1000 m
• Speed of the aircraft, v = 900 km h^-1

Converting the speed into m s^-1,

$\text v = 900 \times \dfrac{5}{18} = 250\ \text{m s}^{-1}$

The centripetal acceleration is

$\text a = \dfrac{\text v^2}{\text r} = \dfrac{(250)^2}{1000} \\[1em] = \dfrac{62500}{1000} \\[1em] = 62.5\ \text{m s}^{-2}$

Comparing it with g = 9.8 m s^-2,

$\dfrac{\text a}{\text g} = \dfrac{62.5}{9.8} = 6.38$

Hence, the centripetal acceleration of the aircraft is 6.38 times the acceleration due to gravity.

State, with reason, for each of the following statements, 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.

Answer

(a) False. The net acceleration is directed along the radius towards the centre only in uniform circular motion. If the speed of the particle also changes, a tangential component of acceleration is present, and the net acceleration is then inclined to the radius.

(b) True. The direction of motion of a particle at any point of its path is along the tangent at that point, so the velocity vector is always tangential to the path.

(c) True. In uniform circular motion the centripetal acceleration is constant in magnitude but its direction changes continuously and points towards the centre. Over one complete cycle every direction is paired with an equal and opposite one, so the vector average of the acceleration over one cycle is a null vector.

The position of a particle is given by

$\vec{r} = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k},$

where t is in seconds and the coefficients have the proper units for $\vec{r} \space$to be in metres.

(a) Find $\vec{v} \space$and $\vec{a} \space$of the particle.

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Answer

Given,

• $\vec{\text r} = 3.0\text t\hat{i} - 2.0\text t^2\hat{j} + 4.0\hat{k}$

(a) The velocity vector is obtained by differentiating the position vector with respect to time,

$\vec{\text v} = \dfrac{\text d\vec{\text r}}{\text{dt}} = \dfrac{\text d}{\text{dt}}\left(3.0\text t\hat{i} - 2.0\text t^2\hat{j} + 4.0\hat{k}\right) \\[1em] = 3.0\hat{i} - 4.0\text t\hat{j}\ \text{m s}^{-1}$

The acceleration vector is obtained by differentiating the velocity vector with respect to time,

$\vec{\text a} = \dfrac{\text d\vec{\text v}}{\text{dt}} = \dfrac{\text d}{\text{dt}}\left(3.0\hat{i} - 4.0\text t\hat{j}\right) \\[1em] = -4.0\hat{j}\ \text{m s}^{-2}$

(b) At t = 2.0 s,

$\vec{\text v} = 3.0\hat{i} - 4.0 \times 2.0\hat{j} \\[1em] = 3.0\hat{i} - 8.0\hat{j}\ \text{m s}^{-1}$

The magnitude of the velocity is

$|\vec{\text v}| = \sqrt{(3.0)^2 + (-8.0)^2} \\[1em] = \sqrt{9 + 64} = \sqrt{73} \\[1em] = 8.54\ \text{m s}^{-1}$

If θ is the angle made by the velocity with the X-axis,

$\tan \theta = \dfrac{\text v_y}{\text v_x} = \dfrac{-8.0}{3.0} = -2.667 \\[1em] \theta = \tan^{-1}(-2.667) = -69.5^\circ$

Hence, the velocity at t = 2.0 s is 8.54 m s^-1, directed at 69.5° below the X-axis (that is, clockwise from the positive X-axis).

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

Answer

Given,

• Initial velocity, $\vec{\text u} = 10.0\hat{j}$ m s^-1, so u_x = 0 and u_y = 10.0 m s^-1
• Acceleration, $\vec{\text a} = 8.0\hat{i} + 2.0\hat{j}$ m s^-2, so a_x = 8.0 m s^-2 and a_y = 2.0 m s^-2

(a) Using the equation of motion along the X-direction,

$\text x = \text u_x \text t + \dfrac{1}{2}\text a_x \text t^2 \\[1em] = 0 + \dfrac{1}{2} \times 8.0 \times \text t^2 = 4\text t^2$

Putting x = 16 m,

$16 = 4\text t^2 \quad \Rightarrow \quad \text t = 2\ \text s$

The y-coordinate at this instant is

$\text y = \text u_y \text t + \dfrac{1}{2}\text a_y \text t^2 \\[1em] = 10.0 \times 2 + \dfrac{1}{2} \times 2.0 \times (2)^2 \\[1em] = 20 + 4 = 24\ \text m$

(b) The velocity at t = 2 s is

$\vec{\text v} = \vec{\text u} + \vec{\text a}\text t \\[1em] = 10.0\hat{j} + (8.0\hat{i} + 2.0\hat{j}) \times 2 \\[1em] = 16.0\hat{i} + 14.0\hat{j}\ \text{m s}^{-1}$

The speed is

$|\vec{\text v}| = \sqrt{(16.0)^2 + (14.0)^2} \\[1em] = \sqrt{256 + 196} = \sqrt{452} \\[1em] = 21.3\ \text{m s}^{-1}$

Hence, at t = 2 s the y-coordinate is 24 m and the speed of the particle is 21.3 m s^-1.

$\hat{i}$ 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 $A = 2\hat{i} + 3\hat{j}$, along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ ? [You may use graphical method]

Answer

The magnitude of $(\hat{i} + \hat{j})$ is

$|\hat{i} + \hat{j}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$

and the angle it makes with the X-axis is

$\theta = \tan^{-1}\left(\dfrac{1}{1}\right) = 45^\circ$

The magnitude of $(\hat{i} - \hat{j})$ is

$|\hat{i} - \hat{j}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{2}$

and the angle it makes with the X-axis is

$\theta = \tan^{-1}\left(\dfrac{-1}{1}\right) = -45^\circ$

The component of a vector $\vec{\text A} \space$along a direction is the scalar product of $\vec{\text A} \space$with the unit vector along that direction.

The unit vector along $(\hat{i} + \hat{j})$ is $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$, so the component of $\vec{\text A} = 2\hat{i} + 3\hat{j}$ along it is

$\vec{\text A} \cdot \dfrac{(\hat{i} + \hat{j})}{\sqrt{2}} = \dfrac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} \\[1em] = \dfrac{2 + 3}{\sqrt{2}} = \dfrac{5}{\sqrt{2}} \\[1em] = 3.54$

The unit vector along $(\hat{i} - \hat{j})$ is $\dfrac{\hat{i} - \hat{j}}{\sqrt{2}}$, so the component of $\vec{\text A} \space$along it is

$\vec{\text A} \cdot \dfrac{(\hat{i} - \hat{j})}{\sqrt{2}} = \dfrac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j})}{\sqrt{2}} \\[1em] = \dfrac{2 - 3}{\sqrt{2}} = -\dfrac{1}{\sqrt{2}} \\[1em] = -0.71$

Hence, both $(\hat{i} + \hat{j})$ and $(\hat{i} - \hat{j})$ have magnitude $\sqrt{2}$, directed at 45° and −45° with the X-axis, and the components of $\vec{\text A} \space$along them are 3.54 and −0.71 respectively.

For any arbitrary motion in space, which of the following relations are true ?

(a) $\vec{v}_{av} = \dfrac{\vec{v}(t_1) + \vec{v}(t_2)}{2}$

(b) $\vec{v}_{av} = \dfrac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}$

(c) $\vec{v}(t) = \vec{v}(0) + \vec{a}t$

(d) $\vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + \dfrac{1}{2}\vec{a}t^2$

(e) $\vec{a}_{av} = \dfrac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$,

$\vec{v}_{av}$ and $\vec{a}_{av}$ stands for average of the quantities over the time-interval $t_1$ to $t_2$.

Answer

Relations (b) and (e) are true; relations (a), (c) and (d) are false.

Reason —

(a) False. This relation holds only when the acceleration is uniform. For arbitrary motion the velocity may change in any manner, so the average velocity is not the arithmetic mean of the initial and the final velocities.

(b) True. The average velocity is by definition the displacement divided by the corresponding time interval, and this definition holds for every kind of motion.

(c) False. This is an equation of motion derived for uniform acceleration, so it cannot be applied to arbitrary motion in which the acceleration itself changes.

(d) False. This equation is also derived for uniform acceleration and therefore does not hold for arbitrary motion.

(e) True. The average acceleration is by definition the change in velocity divided by the corresponding time interval, and this definition holds for every kind of motion.

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.

Answer

(a) False. A scalar need not be conserved. For example, the kinetic energy of a body is a scalar, but it is not conserved in an inelastic collision.

(b) False. A scalar may be negative. For example, temperature on the Celsius scale and gravitational potential energy can both take negative values.

(c) False. A scalar may have dimensions. For example, mass, work and density are scalars having definite dimensions, while only pure numbers such as the refractive index are dimensionless.

(d) False. A scalar may vary from point to point in space. For example, the temperature or the pressure of the air in a room is different at different points.

(e) True. A scalar has only magnitude and no direction, so its value does not depend on the orientation of the coordinate axes chosen by the observer. Hence all observers, however their axes are oriented, obtain the same value of a scalar.

An aircraft is flying at a height of 3400 m above the ground with uniform speed. If the angle subtended at a ground observation point by the aircraft positions 10 s apart is 30°, what is the speed of the aircraft?

Answer

Given,

• Height of the aircraft above the ground, h = 3400 m
• Angle subtended at the observation point, 2θ = 30°, so θ = 15°
• Time interval, t = 10 s

Let the aircraft move from position A to position B in 10 s, and let O be the observation point on the ground. The perpendicular from O meets AB at its mid-point C, so that

$\angle \text{AOC} = \angle \text{BOC} = 15^\circ$

In the right-angled triangle AOC,

$\tan 15^\circ = \dfrac{\text{AC}}{\text{OC}} \quad \Rightarrow \quad \text{AC} = \text{OC} \times \tan 15^\circ$

Substituting OC = 3400 m and tan 15° = 0.2679,

$\text{AC} = 3400 \times 0.2679 = 910.9\ \text m$

The total distance travelled by the aircraft is

$\text{AB} = 2 \times \text{AC} = 2 \times 910.9 = 1821.8\ \text m$

Therefore, the speed of the aircraft is

$\text v = \dfrac{\text{AB}}{\text t} = \dfrac{1821.8}{10} \\[1em] = 182.2\ \text{m s}^{-1}$

Hence, the speed of the aircraft is 182.2 m s^-1.