Motion in a Plane — Practice & Self Evaluation

A vector quantity is represented as:

1. a physical quantity with direction only
2. a physical quantity with magnitude only
3. a quantity with both magnitude and direction, following vector addition rules
4. a physical quantity with both magnitude and direction.

Answer

a quantity with both magnitude and direction, following vector addition rules

Reason — The quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities. Merely having a magnitude and a direction is not enough, and this is why option 4 is incomplete. For example, electric current has both magnitude and direction but it is a scalar quantity, since it does not obey the laws of vector addition (the triangle or parallelogram law).

The result of subtracting two vectors is:

1. scalar
2. zero
3. another vector
4. undefined.

Answer

another vector

Reason — The subtraction of a vector $\vec{\text B} \space$from a vector $\vec{\text A} \space$means the addition of $(-\vec{\text B}) \space$to $\vec{\text A}$, that is, $\vec{\text A} - \vec{\text B} = \vec{\text A} + (-\vec{\text B})$. Since the sum of two vectors is again a vector, the difference of two vectors is also a vector. It becomes a zero vector only in the special case when the two vectors are equal.

A unit vector has a magnitude of:

1. zero
2. one
3. infinite
4. none of these.

Answer

one

Reason — A unit vector is a vector that has a magnitude of exactly one unit. It is obtained by dividing a vector by its own magnitude, $\hat{\text A} = \dfrac{\vec{\text A}}{|\vec{\text A}|}$, and it retains the direction of the original vector while scaling its magnitude to one. It is used to indicate direction without regard to magnitude.

Which one is not a zero vector?

1. Position vector of origin
2. Acceleration vector of a body moving with constant velocity
3. Sum of two equal vectors
4. Difference of two equal vectors.

Answer

Sum of two equal vectors

Reason — The position vector of the origin itself is a zero vector, since its initial and terminal points coincide. The acceleration vector of a body moving with constant velocity is a zero vector, since the velocity does not change. The difference of two equal vectors is a zero vector, because $\vec{\text A} - \vec{\text B} = \vec{0} \space$if $\vec{\text A} = \vec{\text B}$. But the sum of two equal vectors is $\vec{\text A} + \vec{\text A} = 2\vec{\text A}$, which has twice the magnitude of $\vec{\text A} \space$and is therefore not a zero vector.

A free vector can be moved parallel to itself :

1. anywhere
2. only in one direction
3. to the origin
4. along a curve.

Answer

anywhere

Reason — A free vector is a vector that can be moved anywhere in space without changing its properties, that is, its magnitude and direction remain unchanged. It is not tied to a specific point. For example, a wind blowing with a velocity of 10 m s^-1 to the north is the same free vector at every location in a field.

Which of the following is a bound vector?

1. Force
2. Displacement
3. Velocity
4. All of these.

Answer

Force

Reason — Bound vectors are associated with a specific point in space, that is, they are functions of the cartesian coordinates of the point of application. A force acts at a definite point of a body and its effect changes if it is shifted to another point, so it is a bound vector. Displacement and velocity are not tied to a particular point and can be drawn anywhere without changing their meaning.

The multiplication of a vector by a scalar result in:

1. a scalar
2. a vector
3. a zero vector
4. a unit vector.

Answer

a vector

Reason — On multiplying a vector $\vec{\text A} \space$by a scalar k, a vector $\vec{\text R} = \text k\vec{\text A}$is obtained whose magnitude is k times the magnitude of $\vec{\text A} \space$and whose direction is the same as that of $\vec{\text A}$(opposite if k is negative). A vector scaled by a scalar therefore remains a vector. For example, multiplying velocity $\vec{\text v} \space$by mass m gives the vector quantity momentum $\vec{\text p} = \text m\vec{\text v}$.

When a vector is multiplied by zero, the result is:

1. the same vector
2. a zero vector
3. a unit vector
4. infinite vector.

Answer

a zero vector

Reason — One of the properties of the zero vector is that the multiplication of a finite vector $\vec{\text A} \space$by zero is equal to the zero vector, that is, $0\vec{\text A} = \vec{0}$. The resulting vector has zero magnitude and no specific direction, and its initial and terminal points are coincident.

In projectile motion, the path followed by the projectile is:

1. circular
2. straight line
3. parabolic
4. elliptical.

Answer

parabolic

Reason — For a body projected with velocity u at an angle θ₀ with the horizontal, the equation of the trajectory is

$\text y = \text x \tan \theta_0 - \dfrac{\text{gx}^2}{2\text u^2 \cos^2 \theta_0}$

This equation is quadratic in x and linear in y, so it represents a parabola. Hence the path of a projectile, called its trajectory, is parabolic.

The horizontal component of a projectile's velocity:

1. increases
2. decreases
3. remains constant
4. becomes zero at the highest point.

Answer

remains constant

Reason — The horizontal and the vertical motions of a projectile take place independent of each other. No force acts on the projectile along the horizontal direction (air resistance being neglected), and the force of gravity is vertical, so it changes only the vertical component of velocity. Hence the horizontal component u_x = u cos θ remains unchanged throughout the flight, including at the highest point.

The angle of projection for maximum range is:

1. 30°
2. 45°
3. 60°
4. 90°.

Answer

45°

Reason — The horizontal range of a projectile is

$\text R = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$

For a given speed of projection u, the range is maximum when sin 2θ₀ is maximum, that is, when sin 2θ₀ = 1 or 2θ₀ = 90°, giving θ₀ = 45°. This is why a long-jumper takes his jump at an angle of about 45°.

The range of a projectile depends on:

1. only the angle of projection
2. only the initial velocity
3. both initial velocity and angle of projection
4. only gravity.

Answer

both initial velocity and angle of projection

Reason — The horizontal range is given by $\text R = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$. A higher initial velocity allows the projectile to cover a greater distance, and the angle of projection decides the trajectory and hence how long the projectile stays in the air and how far it travels horizontally. The range also depends on g, but not on g alone.

The maximum height of a projectile is directly proportional to:

1. initial velocity
2. horizontal velocity
3. square of the initial velocity
4. the angle of projection.

Answer

square of the initial velocity

Reason — The maximum height attained by a projectile is

$\text h = \dfrac{\text u^2 \sin^2 \theta_0}{2\text g}$

For a fixed angle of projection, h ∝ u². The greatest height for a given speed is obtained by putting θ₀ = 90°, which gives $\text h_{max} = \dfrac{\text u^2}{2\text g}$, again directly proportional to the square of the velocity of projection.

The projectile launched from a height follows a:

1. straight path
2. hyperbolic path
3. parabolic path
4. circular path.

Answer

parabolic path

Reason — When a projectile is launched from a height, its initial velocity is resolved into a horizontal component and a vertical component. The horizontal motion is uniform because no horizontal force acts on it, while the vertical motion is uniformly accelerated due to gravity. The combination of a uniform horizontal motion and a uniformly accelerated vertical motion produces a parabolic trajectory, whatever the launch height may be.

If two vectors are perpendicular to each other, their dot product is:

1. positive
2. negative
3. zero
4. infinite.

Answer

zero

Reason — The scalar product of two vectors is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$. If the two vectors are perpendicular, θ = 90° and cos 90° = 0, so

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 90^\circ = 0$

Thus the scalar product of two mutually perpendicular vectors is zero even though neither of them is a null vector. For example, the work done is zero when the force is perpendicular to the displacement.

The velocity of a car A is $5\hat{i} + 11\hat{j}$. The velocity of another car B is $11\hat{i} + 5\hat{j}$. What is the relative velocity of car B with respect to car A?

1. $6\hat{i} - 6\hat{j}$
2. $5\hat{i} + 11\hat{j}$
3. $6\hat{i} + 6\hat{j}$
4. $5\hat{i} + 5\hat{j}$.

Answer

$6\hat{i} - 6\hat{j}$

Reason — The relative velocity of a body B with respect to a body A is

$\vec{\text v}_{BA} = \vec{\text v}_B - \vec{\text v}_A$

Substituting the values,

$\vec{\text v}_{BA} = (11\hat{i} + 5\hat{j}) - (5\hat{i} + 11\hat{j}) \\[1em] = (11 - 5)\hat{i} + (5 - 11)\hat{j} \\[1em] = 6\hat{i} - 6\hat{j}$

Note: The textbook’s printed answer key gives option 2, which represents only the velocity of car A. However, book's own solution hint correctly calculates the relative velocity as $6\hat{i} - 6\hat{j}$. Therefore, the solution given above is correct.

A plane is flying at 200 m/s due north. A wind is blowing due east at 50 m/s. What is the magnitude of the plane's velocity relative to the ground?

1. 250 m/s
2. 150 m/s
3. 206 m/s
4. 220 m/s.

Answer

206 m/s

Reason — The velocity of the plane relative to the ground is the vector sum of the velocity of the plane and the velocity of the wind. Since the two are mutually perpendicular,

$\text v = \sqrt{(200)^2 + (50)^2} \\[1em] = \sqrt{40000 + 2500} = \sqrt{42500} \\[1em] = 206\ \text{m s}^{-1}$

A car is negotiating a curve of radius 50 m at a speed of 20 m/s. What is the minimum coefficient of friction required to prevent sliding on a flat road?

1. 0.1
2. 0.2
3. 0.4
4. 0.8.

Answer

0.8

Reason — On a flat curved road the necessary centripetal force is provided entirely by the force of friction between the tyres and the road. Hence

$\mu \text{mg} = \dfrac{\text{mv}^2}{\text r} \quad \Rightarrow \quad \mu = \dfrac{\text v^2}{\text{rg}}$

Substituting the values,

$\mu = \dfrac{(20)^2}{50 \times 9.8} \\[1em] = \dfrac{400}{490} \\[1em] = 0.82 \approx 0.8$

A motorcycle rounds a curve of radius 30 m at a speed of 15 m/s. What is the centripetal acceleration?

1. 0.5 m/s²
2. 2.5 m/s²
3. 7.5 m/s²
4. 10 m/s².

Answer

7.5 m/s²

Reason — The centripetal acceleration of a body moving on a circular path of radius r with speed v is

$\text a = \dfrac{\text v^2}{\text r}$

Substituting the values,

$\text a = \dfrac{(15)^2}{30} = \dfrac{225}{30} \\[1em] = 7.5\ \text{m s}^{-2}$

Note: The textbook’s printed answer key gives option 2, which is $2.5\ \text{m s}^{-2}$. However, the book's own solution hint correctly calculates the acceleration as $7.5\ \text{m s}^{-2}$. Therefore, the solution given above is correct.

If the magnitude of vector A is 5 units and vector B is 10 units, and the angle between them is 60°, the magnitude of the resultant vector is:

1. 5 units
2. 10 units
3. 125 units
4. 13 units.

Answer

13 units

Reason — By the parallelogram law of vector addition, the magnitude of the resultant is

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

Substituting the values,

$\text R = \sqrt{(5)^2 + (10)^2 + 2 \times 5 \times 10 \times \cos 60^\circ} \\[1em] = \sqrt{25 + 100 + 100 \times \dfrac{1}{2}} \\[1em] = \sqrt{175} = 13.2 \approx 13\ \text{units}$

A vector has components $3\hat{i} + 4\hat{j}$. The magnitude of this vector is:

1. 5
2. 7
3. 4
4. 3

Answer

5

Reason — If a vector is written in terms of its rectangular components as $\vec{\text A} = \text A_x\hat{i} + \text A_y\hat{j}$, then its magnitude is

$\text A = \sqrt{\text A_x^2 + \text A_y^2}$

Substituting the values,

$\text A = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} \\[1em] = \sqrt{25} = 5$

If vector $\vec{A} \space$is $2\hat{i} + 3\hat{j}$ and vector $\vec{B} \space$is $4\hat{i} - \hat{j}$, the dot product of $\vec{A} \space$and $\vec{B} \space$is:

1. 5
2. 8
3. 6
4. 11

Answer

5

Reason — The scalar product of two vectors is equal to the sum of the products of their corresponding x-, y- and z-components,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Substituting the values,

$\vec{\text A} \cdot \vec{\text B} = (2)(4) + (3)(-1) \\[1em] = 8 - 3 = 5$

The time of flight of a projectile depends on:

1. only horizontal velocity
2. only vertical velocity
3. both horizontal and vertical components
4. none of the above.

Answer

only vertical velocity

Reason — The time taken by a projectile to reach the highest point is obtained from v = u − gt by putting v = 0 and u = u sin θ₀, which gives $\text t = \dfrac{\text u \sin \theta_0}{\text g}$. Hence the time of flight is

$\text T = 2\text t = \dfrac{2\text u \sin \theta_0}{\text g}$

Only the vertical component u sin θ₀ appears in this expression, so the time of flight depends on the vertical component of velocity alone.

At the highest point of the trajectory, the vertical and horizontal components of the velocity of a projectile will be, respectively:

1. $(v_0 \sin θ, 0)$
2. $(v \sin θ, 0)$
3. $(0, v_0 \cos θ)$
4. $(v_0 \sin θ, v_0 \cos θ)$.

Answer

$(0, v_0 \cos θ)$

Reason — At the highest point of the trajectory the projectile stops rising, so the vertical component of its velocity becomes zero. The horizontal component is unaffected by gravity and remains unaltered at v₀ cos θ throughout the flight. Hence at the peak the vertical and horizontal components are 0 and v₀ cos θ respectively.

Why does a cyclist lean inward while negotiating a turn on a flat curved road?

1. To balance the gravitational force acting on him.
2. To balance the centrifugal force acting outward.
3. To increase his speed around the curve.
4. To reduce friction between the tyres and the road.

Answer

To balance the centrifugal force acting outward.

Reason — When a cyclist takes a turn, an outward centrifugal force acts on him due to his inertia. By leaning inward he counteracts this force, so that the combined effect of the gravitational force and the normal reaction points towards the centre of the curve and supplies the necessary centripetal force. This keeps him in balance while negotiating the turn.

Why does a car moving on a flat curved road require friction between the tyres and the road?

1. To balance the gravitational force.
2. To provide centripetal force for circular motion.
3. To increase the car's speed on curves.
4. To prevent the car from tipping over.

Answer

To provide centripetal force for circular motion.

Reason — A body moving on a circular path needs a net inward force directed towards the centre. On a flat curved road this centripetal force is provided entirely by the friction between the tyres and the road. In the absence of this force the car would continue to move along a straight line because of its inertia, and would not be able to follow the curve.

Why does an object moving in a circular path experience a centrifugal force in a non-inertial frame of reference?

1. Because of the object's inertia resisting the centripetal force.
2. Because there is an actual force acting outward on the object.
3. Because the object is accelerating toward the center.
4. Because the gravitational force is balanced by centrifugal force.

Answer

Because of the object's inertia resisting the centripetal force.

Reason — In a non-inertial frame, such as the rotating frame of an observer moving with the object, an outward force appears to act on the object. This centrifugal force is not a real force; it arises from the inertia of the object, which resists the centripetal acceleration directed towards the centre. It is therefore called a pseudo force and has no existence in an inertial frame.

The magnitude of centripetal force depends on:

1. velocity and mass
2. velocity and radius
3. mass and radius
4. All of these.

Answer

All of these.

Reason — The centripetal force required to keep a body of mass m moving with speed v on a circular path of radius r is

$\text F = \dfrac{\text{mv}^2}{\text r}$

The expression contains the mass, the velocity as well as the radius, so the magnitude of the centripetal force depends on all three of them.

The main purpose of banking a road is to:

1. increase the speed limit
2. decrease fuel consumption
3. provide additional centripetal force
4. make the road more aesthetically pleasing.

Answer

provide additional centripetal force

Reason — When a road is banked, the outer edge is raised above the inner edge so that the normal reaction becomes inclined to the vertical. Its horizontal component then acts towards the centre of the circular path and supplies an additional centripetal force. The vehicle can therefore take the turn safely without depending on friction alone.

A vehicle moving in a circular path on a flat road requires:

1. only friction to maintain circular motion
2. only the banking angle to maintain circular motion
3. a combination of friction and banking
4. no forces at all.

Answer

only friction to maintain circular motion

Reason — A flat road has no banking, so the normal reaction is vertical and has no component towards the centre of the circular path. The whole of the centripetal force needed for the circular motion must therefore be supplied by the friction between the tyres and the road.

When a four-wheeler moves on a curved banked road at the designed speed, the role of friction is:

1. to provide centripetal force
2. to balance the forces completely
3. to assist only when speed is less than designed speed
4. to help in acceleration.

Answer

to balance the forces completely

Reason — At the designed (optimum) speed the horizontal component of the normal reaction by itself provides exactly the centripetal force required for the circular motion. Friction is then not needed for turning, and it only serves to balance the remaining forces, coming into play to correct any excess whenever the speed departs from the designed value.

If a vehicle goes too fast on a banked curve, it will:

1. slide inward
2. slide outward
3. maintain its path
4. stop completely.

Answer

slide outward

Reason — If the speed exceeds the designed speed for the given banking angle, the centripetal force required becomes greater than that supplied by the horizontal component of the normal reaction. The friction available is then not enough to counteract the outward centrifugal effect, and the vehicle slides outward, away from the centre of the curve.

In the case of a vehicle on a flat road, if the speed is doubled, the required frictional force will:

1. remain the same
2. increase by four times
3. increase by twice
4. decrease by half.

Answer

increase by four times

Reason — On a flat road the friction supplies the whole centripetal force,

$\text F = \dfrac{\text{mv}^2}{\text r}$

so F ∝ v² for a given mass and radius. If the speed is doubled, the required force becomes

$\text F' = \dfrac{\text m(2\text v)^2}{\text r} = 4 \times \dfrac{\text{mv}^2}{\text r} = 4\text F$

that is, four times the original frictional force.

If a vehicle moves on a banked road at a speed lower than the design speed, the frictional force will:

1. act up the slope
2. act down the slope
3. be zero
4. act perpendicular to the slope.

Answer

act up the slope

Reason — At a speed lower than the designed speed, the horizontal component of the normal reaction is greater than the centripetal force actually required, so the vehicle tends to slide down the banked surface towards the inner edge. Friction opposes this tendency and therefore acts up the slope, keeping the vehicle moving on its circular path.

When a vehicle is travelling on a curved path, which of the following statements is true?

1. Only gravity acts on the vehicle.
2. Centripetal force is directed outward.
3. The vehicle must have a net inward force to maintain circular motion.
4. The vehicle will never slide if it has enough speed.

Answer

The vehicle must have a net inward force to maintain circular motion.

Reason — For circular motion a net inward force, that is, a centripetal force directed towards the centre of the curve, is always required. It can be provided by friction, by the banking of the road, or by both together. Without such an inward force the vehicle would move along a straight line because of its inertia.

The magnitude of cross product of two parallel vectors is:

1. zero
2. maximum
3. minimum
4. equal to the dot product.

Answer

zero

Reason — The magnitude of the vector product of two vectors is $|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta$. When the two vectors are parallel the angle between them is 0° (or 180°), and sin 0° = 0, so

$|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin 0^\circ = 0$

Thus the vector product of two parallel vectors is a null vector.

A body projected at an angle θ has the same range as when projected at (90° − θ). This property is known as:

1. symmetry of motion
2. complementary angles
3. principle of invariance
4. none of the above.

Answer

complementary angles

Reason — Substituting (90° − θ₀) in place of θ₀ in the expression for the range,

$\text R = \dfrac{\text u^2 \sin 2(90^\circ - \theta_0)}{\text g} = \dfrac{\text u^2 \sin (180^\circ - 2\theta_0)}{\text g} \\[1em] = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$

which is the same as the range for θ₀. Hence the horizontal range is the same for two complementary angles of projection, and a football kicked at 30° or at 60° strikes the ground at the same place.

A vector lies in the XY-plane. Which of its components is necessarily zero?

1. X-component
2. Y-component
3. Z-component
4. None of these.

Answer

Z-component

Reason — A vector lying in the XY-plane can be written in terms of its rectangular components as $\vec{\text A} = \text A_x\hat{i} + \text A_y\hat{j}$. Since the vector has no extension along the Z-direction, its component along the Z-axis is necessarily zero, that is, A_z = 0. The X- and Y-components may have any values.

If a projectile's horizontal range is R and the maximum height is H, the ratio of R to H for an angle of 45° is:

1. 2
2. 4
3. 1
4. ∞.

Answer

4

Reason — For a projectile,

$\text R = \dfrac{\text u^2 \sin 2\theta}{\text g} \quad \text{and} \quad \text H = \dfrac{\text u^2 \sin^2 \theta}{2\text g}$

Therefore,

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

Putting θ = 45°,

$\dfrac{\text R}{\text H} = \dfrac{2 \times \sin 90^\circ}{(\sin 45^\circ)^2} = \dfrac{2}{1/2} \\[1em] = 4$

A vector A makes an angle of 45° with both X and Y-axes. The angle it makes with the Z-axis is:

1. 90°
2. 45°
3. 60°
4. None of these.

Answer

90°

Reason — If α, β and γ are the angles made by a vector with the X-, Y- and Z-axes, the direction cosines satisfy

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

Putting α = β = 45°,

$\dfrac{1}{2} + \dfrac{1}{2} + \cos^2 \gamma = 1 \quad \Rightarrow \quad \cos \gamma = 0 \quad \Rightarrow \quad \gamma = 90^\circ$

Thus the vector lies in the XY-plane and is perpendicular to the Z-axis.

The unit vector perpendicular to both $\vec{A} = \hat{i} + 2\hat{j}$ and $\vec{B} = 3\hat{i} + \hat{j}$ is given by:

1. $\hat{i} + \hat{j}$
2. $\hat{k}$
3. $\hat{i} - \hat{j}$
4. None of these.

Answer

$\hat{k}$

Reason — The vector product $\vec{\text A} \times \vec{\text B} \space$gives a vector perpendicular to the plane containing $\vec{\text A} \space$and $\vec{\text B}$. Here both vectors lie in the XY-plane, so their cross product must be along the Z-axis,

$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 3 & 1 & 0 \end{vmatrix} \\[1em] = \hat{k}(1 \times 1 - 2 \times 3) = -5\hat{k}$

The unit vector perpendicular to both is therefore $\hat{k}$ (directed along the Z-axis).

The angle of projection for which horizontal range and maximum height are equal is:

1. 30°
2. 45°
3. 60°
4. 76°.

Answer

76°

Reason — Equating the horizontal range and the maximum height,

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

so that

$2 \sin 2\theta = \sin^2 \theta \\[1em] 2 \times 2 \sin \theta \cos \theta = \sin^2 \theta \\[1em] \tan \theta = 4$

Since tan 76° = 4, the angle of projection is θ ≈ 76°.

A boat moving with a velocity of 10 m/s with respect to the still water crosses a river flowing at 5 m/s. If the boat needs to move directly across the river, in which direction should it head?

1. At an angle of 45° downstream
2. At an angle of 30° upstream
3. At an angle of 60° upstream
4. At an angle of 30° downstream.

Answer

At an angle of 60° upstream

Reason — For the boat to move directly across the river, the component of its velocity along the flow must exactly cancel the velocity of the river. If θ is the angle made by the heading of the boat with the line joining the two banks (that is, with the direction straight across),

$\text v_{river} = \text v_{boat} \sin \theta \\[1em] 5 = 10 \sin \theta \quad \Rightarrow \quad \sin \theta = 0.5 \quad \Rightarrow \quad \theta = 30^\circ$

So the boat must head 30° upstream of the line straight across, which is the same as making an angle of 60° with the upstream bank. Measured from the bank, the required heading is therefore 60° upstream.

The horizontal displacement of a projectile after 2 s is 40 m. What is the horizontal velocity?

1. 10 m/s
2. 20 m/s
3. 40 m/s
4. 30 m/s.

Answer

20 m/s

Reason — The horizontal component of the velocity of a projectile remains constant, so the horizontal motion is uniform. Hence

$\text x = \text u_x \text t \quad \Rightarrow \quad \text u_x = \dfrac{\text x}{\text t}$

Substituting the values,

$\text u_x = \dfrac{40}{2} = 20\ \text{m s}^{-1}$

A river flows east to west at a speed 5 m/min. A man capable of swimming 10 m/min in still water is standing on the south bank of the river. He wants to swim across the river in the shortest time possible. Which of the following path should he choose?

1. Due North
2. Due North-East
3. Due North-East with double the speed of the river
4. None of the above.

Answer

Due North

Reason — The time taken to cross the river depends only on the component of the swimmer's velocity perpendicular to the banks. This component is greatest when the man swims directly due north, since then his full speed of 10 m min^-1 is used in crossing. The current will carry him some distance to the west, but the drift does not affect the time of crossing, so the time taken is the shortest.

Note: The textbook’s printed answer key gives option 2. However, the book's own solution hint correctly states that the man should swim directly towards the north. Therefore, the solution given above is correct.

Assertion (A): Minimum number of non-equal vectors in a plane to give zero resultant is three.

Reason (R): If $\vec{A} + \vec{B} + \vec{C} = 0$ then they must lie in one plane.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a zero resultant it must be possible to represent the given vectors by the sides of a closed polygon taken in the same order. Since two vectors of different magnitudes cannot be added to get a zero resultant, and the minimum number of sides of a closed polygon is three, at least three unequal vectors are needed in a plane to give a zero resultant.

Reason (R) is also correct: If $\vec{\text A} + \vec{\text B} + \vec{\text C} = 0$, the three vectors form a closed triangle, and a triangle always lies in one plane. Hence the three vectors must be coplanar.

However, the Reason only states that three such vectors are coplanar. It does not explain why the minimum number required is three.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion (A): $\vec{A} \times \vec{B} \space$is perpendicular to both $\vec{A} + \vec{B} \space$as well as $\vec{A} - \vec{B}$.

Reason (R): $\vec{A} + \vec{B} \space$as well as $\vec{A} - \vec{B} \space$lie in the plane containing $\vec{A} \space$and $\vec{B} \space$but $\vec{A} \times \vec{B} \space$lies perpendicular to the plane containing $\vec{A} \space$and $\vec{B}$.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The vector product $\vec{\text A} \times \vec{\text {B }}$is perpendicular to both the sum and the difference of $\vec{\text {A }}$and $\vec{\text B}$.

Reason (R) is also correct: The vector product of two vectors has the direction perpendicular to the plane containing the two vectors, its sense being given by the right-hand screw rule. Both $\vec{\text A} + \vec{\text {B }}$and $\vec{\text A} - \vec{\text {B }}$are obtained by the laws of vector addition from $\vec{\text {A }}$and $\vec{\text B}$, so they lie in the same plane as $\vec{\text {A }}$and $\vec{\text B}$. A vector perpendicular to this plane is therefore perpendicular to every vector lying in it.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Angle between $\hat{i} + \hat{j}$ and $\hat{i}$ is 45°.

Reason (R): $\hat{i} + \hat{j}$ is equally inclined to both $\hat{i}$ and $\hat{j}$ and angle between $\hat{i}$ and $\hat{j}$ is 90°.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Taking the scalar product,

$\cos \theta = \dfrac{(\hat{i} + \hat{j}) \cdot \hat{i}}{|\hat{i} + \hat{j}||\hat{i}|} = \dfrac{1}{\sqrt{2} \times 1} = \dfrac{1}{\sqrt{2}}$

so that θ = 45°.

Reason (R) is also correct: Since $\hat{i}$ and $\hat{j}$ are unit vectors of equal magnitude, their sum lies along the diagonal of the square formed by them and is therefore equally inclined to both. As the angle between $\hat{i}$ and $\hat{j}$ is 90°, this diagonal bisects it, giving an angle of $\dfrac{90^\circ}{2} = 45^\circ$ with each. This is exactly why the angle in the Assertion is 45°.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A vector $\vec{C} \space$is perpendicular to both vectors $\vec{A} \space$and $\vec{B}$, so it will be parallel to $\vec{A} \times \vec{B}$.

Reason (R): Cross product of two vectors is perpendicular to the plane containing those vectors.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A vector which is perpendicular to both $\vec{\text {A }}$and $\vec{\text {B }}$must lie along the normal to the plane containing them, and so must be parallel (or antiparallel) to $\vec{\text A} \times \vec{\text B}$.

Reason (R) is also correct: By definition, the vector product of two vectors has the direction perpendicular to the plane containing the two vectors, its sense being decided by the right-hand screw rule. Since there is only one direction normal to a given plane, any vector perpendicular to both $\vec{\text {A }}$and $\vec{\text {B }}$must be along that normal, which is the direction of $\vec{\text A} \times \vec{\text B}$.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): In javelin throw, the athlete throws the projectile at an angle slightly more than 45°.

Reason (R): The maximum range does not depend upon angle of projection.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: In javelin throw and discus throw the projectile is released from a place above the surface of the earth, and for such a projection the maximum range is obtained for an angle of projection slightly less than 45°, not more than 45°. Hence the player throws the projectile at an angle slightly less than 45° to the horizontal.

Reason (R) is also false: The horizontal range $\text R = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$ clearly depends on the angle of projection, and it becomes maximum only for a particular angle of projection (45° for projection from the ground). Hence the maximum range does depend upon the angle of projection.

Therefore, both assertion and reason are false.

Assertion (A): The projectile has only vertical component of velocity at the highest point of trajectory.

Reason (R): At the highest point only one component of velocity is present.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is false but reason is true.
4. If both assertion and reason are false.

Answer

If assertion is false but reason is true.

Explanation

Assertion (A) is false: At the highest point of the trajectory it is the vertical component of velocity that becomes zero, while the horizontal component u cos θ₀ remains unaltered. Hence the projectile has only the horizontal component of velocity there, not the vertical component.

Reason (R) is correct: At the highest point only one component of velocity, namely the horizontal component, is present, since the vertical component has been reduced to zero by gravity.

Therefore, assertion is false but reason is true.

Select scalar and vector quantities from the following:

(i) mass, displacement, pressure, work, acceleration.

(ii) momentum, energy, impulse, density, temperature, speed.

(iii) moment of a force, surface tension, momentum and temperature.

(iv) is pressure a vector or a scalar quantity?

Answer

(i) Scalars : mass, pressure, work. Vectors : displacement, acceleration.

(ii) Scalars : energy, density, temperature, speed. Vectors : momentum, impulse.

(iii) Scalars : surface tension, temperature. Vectors : moment of a force, momentum.

(iv) Pressure is a scalar quantity, since it is completely specified by its magnitude alone and does not obey the laws of vector addition.

The magnitude of vector $\vec{A} \space$is 2.5 m and it is directed towards north. What will be the magnitudes and directions of the followings?

(a) $-\vec{A}$

(b) $\vec{A}/2.0$

(c) $-2.5\vec{A}$

(d) $4.0\vec{A}$

Answer

If the direction of a vector is reversed, the sign of the vector is reversed, and on multiplying a vector by a scalar k the magnitude becomes k times while the direction remains the same (opposite if k is negative).

(a) $-\vec{\text A}$: magnitude 2.5 m, directed towards south.

(b) $\dfrac{\vec{\text A}}{2.0}$ : magnitude 1.25 m, directed towards north.

(c) $-2.5\vec{\text A}$: magnitude 6.25 m, directed towards south.

(d) $4.0\vec{\text A}$: magnitude 10 m, directed towards north.

Can a scalar quantity be added to a vector quantity? Is their product possible ?

Answer

No, a scalar quantity cannot be added to a vector quantity, because a scalar has only magnitude whereas a vector has magnitude as well as direction. Even if the two have the same dimensions, they cannot be added.

Yes, their product is possible. On multiplying a vector $\vec{\text {A }}$by a scalar k, a vector $\text k\vec{\text {A }}$is obtained whose magnitude is k times that of $\vec{\text A}$. For example, multiplying the velocity $\vec{\text {v }}$by the mass m gives the momentum $\vec{\text p} = \text m\vec{\text v}$.

Is the magnitude of $(\vec{A} + \vec{B}) \space$same as that of $(\vec{B} + \vec{A})$ ?

Answer

Yes. The addition of vectors is commutative, that is,

$\vec{\text A} + \vec{\text B} = \vec{\text B} + \vec{\text A}$

The two sums are represented by vectors which are parallel and of equal length, so their magnitudes as well as their directions are the same.

Is the magnitude of $(\vec{A} - \vec{B}) \space$same as that of $(\vec{B} - \vec{A})$ ?

Answer

Yes, the magnitudes are the same, but the directions are opposite. Since

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

the two difference vectors are parallel and equal in length, but they point in opposite directions.

When is the sum of two vectors maximum and when minimum ?

Answer

The magnitude of the sum of two vectors is

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

The sum is maximum when the two vectors are in the same direction, that is, θ = 0°, and then R = A + B.

The sum is minimum when the two vectors are in opposite directions, that is, θ = 180°, and then R = A − B (or B − A, whichever is positive).

Can the magnitude of the resultant vector of two given vectors be less than the magnitude of any of the given vectors ?

Answer

Yes, this is possible when the angle between the two given vectors is more than 90°. For example, if two vectors of magnitudes 3 units and 4 units are oppositely directed (θ = 180°), the magnitude of the resultant is

$\text R = 4 - 3 = 1\ \text{unit}$

which is less than the magnitude of either of the given vectors.

If a vector $\vec{A} \space$is multiplied by a scalar m, what will be the resultant vector ?

Answer

The resultant will be a vector $\text m\vec{\text A} \space$whose magnitude is m times the magnitude of $\vec{\text A}$. Its direction is the same as that of $\vec{\text A} \space$if m is positive, and opposite to that of $\vec{\text A} \space$if m is negative. If m is a pure number, the resultant represents the same physical quantity as $\vec{\text A}$; if m carries a unit, the resultant represents a new physical quantity.

Force $\vec{F} \space$and displacement $\vec{s} \space$both are vector quantities. Which type is the quantity work $W = \vec{F} \cdot \vec{s}$?

Answer

Work is a scalar quantity. It is the scalar (dot) product of the two vectors $\vec{\text F} \space$and $\vec{\text s}$, and the scalar product of two vectors is always a scalar,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

Can the scalar product of two vectors be negative ?

Answer

Yes. The scalar product is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$. When the angle between the two vectors is obtuse, that is, when 90° < θ ≤ 180°, cos θ is negative and hence the scalar product is negative. For example, the work done by the force of friction is negative because the friction acts opposite to the displacement.

Under what condition the scalar product of two vectors is maximum?

Answer

The scalar product $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$ is maximum when cos θ is maximum, that is, when θ = 0° and the two vectors are parallel (in the same direction). Its maximum value is then

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 0^\circ = \text{AB}$

In which condition, is the scalar product of two vectors numerically minimum?

Answer

The scalar product is numerically minimum, that is, zero, when the two vectors are mutually perpendicular (θ = 90°), because

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 90^\circ = 0$

Thus the scalar product vanishes even though neither of the two vectors is a null vector.

What is the angle between the directions of velocity and acceleration at the highest point of a projectile path?

Answer

90°. At the highest point of the trajectory the vertical component of the velocity is zero and only the horizontal component u cos θ₀ remains, so the velocity is directed horizontally. The acceleration of the projectile is the acceleration due to gravity, which acts vertically downwards. Hence the angle between the velocity and the acceleration there is 90°.

At what point of the projectile-path the speed is minimum? At which point maximum?

Answer

The speed is minimum at the highest point of the path, since there the vertical component of velocity is zero and only the horizontal component u cos θ₀ is left.

The speed is maximum at the point of projection (and again at the point where the projectile strikes the ground, since it returns to the ground with the same speed with which it was projected).

Is it important in the long jump that how much height you take for jumping?

Answer

Yes, it is important. The horizontal range of the jumper is

$\text R = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$

which is maximum for θ₀ = 45°. To be projected at 45° the jumper must gain a certain height, since the height attained is $\text h = \dfrac{\text u^2 \sin^2 \theta_0}{2\text g}$. If he takes too little height the angle of projection is too small, and if he takes too much height the angle is too large; in both cases the range becomes less than the maximum.

Two bombs of 5 and 10 kg are thrown from a cannon with the same velocity in the same direction. Which bomb will reach the ground first? If the bombs are thrown in the same direction with different velocities, then?

Answer

Both bombs will reach the ground at the same time. The time of flight is

$\text T = \dfrac{2\text u \sin \theta_0}{\text g}$

which does not contain the mass of the projectile. Since both bombs are thrown with the same velocity in the same direction, their time of flight is the same, whatever their masses.

If the bombs are thrown in the same direction but with different velocities, their vertical components of velocity are different. The bomb having the greater vertical component u sin θ₀ will have the longer time of flight, so the bomb thrown with the smaller velocity will reach the ground first.

There are two displacement vectors, one of magnitude 3 m and the other of 4 m. How should the two vectors be added so that the magnitude of the resultant vector be :

(a) 7 m

(b) 1 m and

(c) 5 m ?

Answer

The magnitude of the resultant of two vectors is $\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$, with A = 3 m and B = 4 m.

(a) For R = 7 m, the resultant is maximum, so the two vectors must be added in the same direction, that is, θ = 0°.

(b) For R = 1 m, the resultant is minimum, so the two vectors must be added in opposite directions, that is, θ = 180°.

(c) For R = 5 m,

$5 = \sqrt{(3)^2 + (4)^2 + 2 \times 3 \times 4 \times \cos \theta} \\[1em] 25 = 25 + 24 \cos \theta \quad \Rightarrow \quad \cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ$

So the two vectors must be added at right angles to each other.

Can two vector quantities of different magnitudes be added in such a way that their resultant be zero ?

Answer

No. The minimum magnitude of the sum of two vectors is (A − B), which is obtained when they are oppositely directed. This is zero only if A = B. Hence two vectors of different magnitudes cannot be added to get a zero resultant.

Can three vector quantities be added in such a way that their resultant be zero ?

Answer

Yes. If three vectors can be represented in magnitude and direction by the three sides of a triangle taken in the same (cyclic) order, they themselves complete a closed figure and their sum-vector cannot be drawn. This means that the sum of these three vectors is zero.

If the magnitudes of two vectors are kept unchanged and angle between them is changed then what will be the effect on their resultant vector ?

Answer

Both the magnitude and the direction of the resultant will change. Since

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta} \quad \text{and} \quad \tan \alpha = \dfrac{\text B \sin \theta}{\text A + \text B \cos \theta}$

the magnitude of the resultant decreases from its maximum value (A + B) at θ = 0° to its minimum value (A − B) at θ = 180°, and for other angles it lies between these two values. The angle α made by the resultant with $\vec{\text A} \space$also changes.

Which statement is true?

(a) $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \space$or

(b) $\vec{A} \cdot \vec{B} = -\vec{B} \cdot \vec{A}$?

Answer

Statement (a) is true. The scalar product of two vectors is commutative, since

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = \text{BA} \cos \theta = \vec{\text B} \cdot \vec{\text A}$

(It is the vector product that is not commutative, for which $\vec{\text A} \times \vec{\text B} = -\vec{\text B} \times \vec{\text A}$.)

Two non-zero vectors $\vec{A} \space$and $\vec{B} \space$are such that (a) $\vec{A} \cdot \vec{B} = 0$, (b) $\vec{A} \cdot \vec{B} = AB$. What information do we get about $\vec{A} \space$and $\vec{B} \space$in each case?

Answer

(a) Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = 0$ and neither vector is a null vector, cos θ = 0, so θ = 90°. Hence the two vectors are mutually perpendicular.

(b) Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = \text{AB}$, we get cos θ = 1, so θ = 0°. Hence the two vectors are parallel, that is, they are in the same direction.

If $\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C}$, then are A and B always equal ?

Answer

No. From the given relation,

$\vec{\text A} \cdot \vec{\text C} - \vec{\text B} \cdot \vec{\text C} = 0 \quad \Rightarrow \quad (\vec{\text A} - \vec{\text B}) \cdot \vec{\text C} = 0$

This is satisfied either when $\vec{\text A} = \vec{\text B}$, or when $(\vec{\text A} - \vec{\text B}) \space$is perpendicular to $\vec{\text C}$. Hence $\vec{\text A} \space$and $\vec{\text B} \space$need not always be equal.

Under what condition the sum and difference of two vectors will be equal in magnitude ?

Answer

Equating the magnitudes,

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

which gives 4AB cos θ = 0, so cos θ = 0 and θ = 90°.

Hence the sum and the difference of two vectors are equal in magnitude when the two vectors are mutually perpendicular.

Under what condition the directions of sum and difference of two vectors will be the same ?

Answer

The sum $\vec{\text A} + \vec{\text B} \space$and the difference $\vec{\text A} - \vec{\text B} \space$can point in the same direction only when $\vec{\text B} \space$lies along the same line as $\vec{\text A}$, that is, when the two vectors are collinear (θ = 0° or θ = 180°).

If θ = 0°, the sum has magnitude A + B and the difference has magnitude A − B, both directed along $\vec{\text A} \space$provided A > B.

If θ = 180°, the sum has magnitude A − B and the difference has magnitude A + B, and again both are directed along $\vec{\text A} \space$provided A > B.

In either case, if A < B the difference reverses relative to the sum.

Hence the directions of the sum and the difference are the same when the two vectors are collinear, that is θ = 0° or θ = 180°, and the magnitude of the vector being subtracted is the smaller of the two, that is A > B.

Give two examples of scalar product of two vectors.

Answer

(i) Work is the scalar product of force and displacement, $\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$.

(ii) Power is the scalar product of force and velocity, $\text P = \vec{\text F} \cdot \vec{\text v} = \text{Fv} \cos \theta$.

Give two examples of the vector product of two vectors.

Answer

(i) Torque is the vector product of the position vector and the force, $\vec{\tau} = \vec{\text r} \times \vec{\text F}$, and its magnitude is τ = rF sin θ.

(ii) Angular momentum is the vector product of the position vector and the linear momentum, $\vec{\text L} = \vec{\text r} \times \vec{\text p}$, and its magnitude is L = rp sin θ.

A force $\vec{F} \space$displaces a stationary body by a distance $\vec{d}$. Show this happening with vector notation.

Answer

The work done by the force is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text d} = \text{Fd} \cos \theta$

where θ is the angle between the directions of $\vec{\text F} \space$and $\vec{\text d}$.

A body is so projected in air that the horizontal range covered by the body is equal to the vertical height attained by the body. Find the angle of projection. (tan 76° = 4)

Answer

Given,

• Horizontal range = maximum height, that is, R = h

Equating the two,

$\dfrac{\text u^2 \sin 2\theta_0}{\text g} = \dfrac{\text u^2 \sin^2 \theta_0}{2\text g}$

so that

$2 \sin 2\theta_0 = \sin^2 \theta_0 \\[1em] 2 \times 2 \sin \theta_0 \cos \theta_0 = \sin^2 \theta_0 \\[1em] \tan \theta_0 = 4$

Since tan 76° = 4,

$\theta_0 = 76^\circ$

Hence, the angle of projection is 76°.

Though force $\vec{F} \space$and displacement $\vec{s} \space$are vector quantities, yet work W is a scalar quantity, why ? If $\vec{F} \space$and $\vec{s} \space$are not zero, yet it is possible that the value of W is zero, why ?

Answer

Work is the scalar product of force and displacement,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

The scalar product of two vectors is always a scalar, because the product Fs cos θ has magnitude only and no direction is associated with it. This is why work is a scalar even though both force and displacement are vectors.

If $\vec{\text F} \space$and $\vec{\text s} \space$are both non-zero, the work can still be zero when θ = 90°, since cos 90° = 0. Thus when the force acting on a body is perpendicular to its displacement, the work done by that force is zero. For example, the centripetal force does no work on a body in circular motion, since it is always perpendicular to the displacement.

Can the vector product of two vectors be negative?

Answer

No. The vector product of two vectors is itself a vector, and a vector cannot be called positive or negative. Its magnitude AB sin θ is non-negative, since sin θ is never negative for any angle from 0° to 180°. Only the direction of the vector product is reversed when the order of the vectors is reversed, because

$\vec{\text A} \times \vec{\text B} = -\vec{\text B} \times \vec{\text A}$

The horizontal range (R) of a projectile is given by $R = \left(\dfrac{u^2}{g}\right) \times \sin 2θ$, where u is the velocity of projection and θ is the angle of projection with respect to the horizontal. Find out with the help of this formula, the angle of projection for the maximum range of the projectile.

Answer

For a given velocity of projection u, the range

$\text R = \left(\dfrac{\text u^2}{\text g}\right) \sin 2\theta$

is maximum when sin 2θ has its greatest value, that is, when

$\sin 2\theta = 1 \quad \Rightarrow \quad 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ$

Hence, the range is maximum for an angle of projection of 45°, and then $\text R_{max} = \dfrac{\text u^2}{\text g}$.

At what angle with the horizontal should a player throw a ball so that it may go to a maximum distance?

Answer

The player should throw the ball at an angle of 45° with the horizontal. For a given speed of projection the horizontal range $\text R = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$ is maximum when sin 2θ₀ = 1, that is, when θ₀ = 45°. This is also the reason why a long-jumper takes his jump at an angle of about 45°.

What is the difference between the following two data?

(i) 5 (6 km/h, west).

(ii) (5 h) (6 km/h, west).

Answer

(i) Here the vector (6 km/h, west) is multiplied by the pure number 5. Since a pure number has no unit, the product represents the same physical quantity, that is, a velocity,

$5 \times (6\ \text{km h}^{-1},\ \text{west}) = 30\ \text{km h}^{-1},\ \text{west}$

(ii) Here the vector is multiplied by the scalar quantity 5 h, which carries a unit. The product therefore represents a new physical quantity, that is, a displacement,

$(5\ \text h) \times (6\ \text{km h}^{-1},\ \text{west}) = 30\ \text{km},\ \text{west}$

Hence, the first is a velocity of 30 km h^-1 towards the west, while the second is a displacement of 30 km towards the west.

Can the magnitude of the resultant vector of two given vectors be less than the magnitude of any of the given vectors? Explain with examples.

Answer

Yes, this happens when the angle between the two given vectors is more than 90°, because then the term 2AB cos θ in

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

is negative and it reduces the magnitude of the resultant.

For example, let A = 3 units and B = 4 units.

If θ = 180°, the magnitude of the resultant is

$\text R = \text B - \text A = 4 - 3 = 1\ \text{unit}$

which is less than the magnitude of either of the given vectors.

Again, if θ = 120°,

$\text R = \sqrt{(3)^2 + (4)^2 + 2 \times 3 \times 4 \times \left(-\dfrac{1}{2}\right)} \\[1em] = \sqrt{9 + 16 - 12} = \sqrt{13} \\[1em] = 3.6\ \text{units}$

which is less than 4 units, the magnitude of the larger vector.

Two forces of equal magnitude of 3 N act on a body making angle θ. If the resultant force of these two forces is also 3 N, then what will be the value of θ ?

Answer

Given,

• Magnitude of each force, A = B = 3 N
• Magnitude of the resultant, R = 3 N

By the parallelogram law of vector addition,

$\text R^2 = \text A^2 + \text B^2 + 2\text{AB}\cos \theta$

Substituting the values,

$(3)^2 = (3)^2 + (3)^2 + 2 \times 3 \times 3 \times \cos \theta \\[1em] 9 = 9 + 9 + 18 \cos \theta \\[1em] 18 \cos \theta = -9 \\[1em] \cos \theta = -\dfrac{1}{2}$

Therefore,

$\theta = 120^\circ$

Hence, the angle between the two forces is 120°.

A ball is thrown horizontally and at the same time another ball is dropped down from the top of a tower.

(a) Will both the balls reach the ground at the same time?

(b) Will both strike the ground with the same velocity?

Answer

(a) Yes. The horizontal and vertical motions of a projectile are independent of each other. The initial vertical velocity of both the balls is zero, and both fall through the same height under the same vertical acceleration g. Since the time of fall is $\text T = \sqrt{\dfrac{2\text h}{\text g}}$, which does not depend on the horizontal velocity, both balls reach the ground at the same time.

(b) No. On striking the ground the vertical components of the velocities of the two balls are equal, but the ball thrown horizontally also has a horizontal component of velocity which remains unchanged throughout. Hence its resultant velocity $\text v = \sqrt{\text u^2 + \text g^2\text T^2}$ is greater in magnitude and is inclined to the vertical, whereas the dropped ball strikes the ground vertically.

"The path of a projectile projected from earth is a parabola. The velocity of the projectile will be minimum at the highest point." Explain whether this statement is true or false?

Answer

The statement is true.

The equation of the trajectory of a projectile is

$\text y = \text x \tan \theta_0 - \dfrac{\text{gx}^2}{2\text u^2 \cos^2 \theta_0}$

which is quadratic in x and linear in y, and therefore represents a parabola. Hence the path of a projectile is parabolic.

The horizontal component of the velocity u cos θ₀ remains constant throughout the flight, while the vertical component goes on decreasing as the projectile rises and becomes zero at the highest point. Hence at the highest point only the horizontal component is left, and the speed there,

$\text v = \text u \cos \theta_0$

is the least value of the speed anywhere on the path.

Answer the following questions with reasons :

(i) The velocity of a body is continuously changing. Can its speed remain constant ? If speed is changing, can the velocity remain constant ?

(ii) Is it possible for a body to have a constant speed in an accelerated motion ?

(iii) A particle is moving with a uniform speed on a circular path. State, with reason, whether it has an acceleration or not.

(iv) Can a body move on a curved path without having acceleration?

(v) A particle is moving in a straight line. What can you say about its acceleration?

(vi) A body is moving on a curved path with a constant speed. What is the nature of its acceleration?

(vii) The speed of a body is constant. Can it have a path other than a circular or straight-line path?

(viii) If both the speed of a body and the radius of its circular path are doubled, what will happen to the centripetal force?

(ix) A body is moving with a constant speed in a horizontal circle. Which quantity among its velocity, acceleration and kinetic energy remains unchanged?

Answer

(i) Yes, the speed can remain constant while the velocity changes, since the velocity may change on account of a change of direction alone, as in uniform circular motion. No, if the speed is changing the velocity cannot remain constant, because the speed is the magnitude of the velocity.

(ii) Yes. It is possible if the body is moving on a curved path. Then the direction of the velocity is changing and the motion is accelerated, although the magnitude of the velocity remains constant, as in uniform circular motion.

(iii) Yes, it has an acceleration. Although its speed is constant, the direction of its velocity changes continuously, so its velocity is changing. This gives rise to the centripetal acceleration $\text a = \dfrac{\text v^2}{\text r}$, directed along the radius towards the centre.

(iv) No. A curved path means that the direction of the velocity is continuously changing, and a change of velocity necessarily means acceleration.

(v) The acceleration is either zero, when the body moves with a uniform velocity, or its direction is always along the line of motion, when the speed is changing. There can be no component of acceleration perpendicular to the path.

(vi) The acceleration is perpendicular to the direction of motion, that is, it is directed along the radius towards the centre of curvature. It changes only the direction of the velocity and not its magnitude, and is therefore called centripetal acceleration.

(vii) Yes, all curved paths are possible. If the acceleration is zero the body moves with constant speed along a straight line; if the magnitude of the acceleration is constant and its direction remains perpendicular to the motion, the body moves with constant speed on a circle; and if the magnitude of the perpendicular acceleration changes, the body moves with constant speed along some other curved path.

(viii) The centripetal force is $\text F = \dfrac{\text{mv}^2}{\text r}$. On doubling both v and r,

$\text F' = \dfrac{\text m(2\text v)^2}{2\text r} = \dfrac{4\text{mv}^2}{2\text r} = 2\text F$

so the centripetal force becomes twice its original value.

(ix) The kinetic energy remains unchanged, since the speed is constant. The velocity and the acceleration both change continuously, as their directions keep changing.

Write the formula of the scalar product of two vectors, explaining the symbols used.

Answer

The scalar product (or dot product) of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$

where,

• A is the magnitude of the vector $\vec{\text A}$
• B is the magnitude of the vector $\vec{\text B}$
• θ is the angle between the directions of $\vec{\text A} \space$and $\vec{\text B}$

The product AB cos θ is a scalar, so the scalar product of two vectors is always a scalar quantity.

In terms of the rectangular components of the two vectors, the same product is written as

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Write the formula of the vector product of two vectors with an example.

Answer

The vector product (or cross product) of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\vec{\text A} \times \vec{\text B} = \text{AB} \sin \theta\ \hat{n}$

where,

• A and B are the magnitudes of $\vec{\text A} \space$and $\vec{\text B}$
• θ is the angle between the two vectors
• $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{\text A} \space$and $\vec{\text B}$, its sense being decided by the right-hand screw rule

Example : The torque acting on a particle about the origin O is the vector product of the position vector $\vec{\text r} \space$and the force $\vec{\text F}$,

$\vec{\tau} = \vec{\text r} \times \vec{\text F}$

whose magnitude is τ = rF sin θ and whose direction is perpendicular to the plane containing $\vec{\text r} \space$and $\vec{\text F}$.

Write down the units and dimensional formula of angular velocity.

Answer

The angular velocity is the time rate of change of angular displacement,

$\omega = \dfrac{\text d\theta}{\text{dt}}$

S.I. unit : radian per second (rad s^-1)

Dimensional formula : Since the angular displacement is a dimensionless quantity,

$[\omega] = \dfrac{[\text M^0\text L^0\text T^0]}{[\text T]} = [\text M^0\text L^0\text T^{-1}]$

Write down the relationship between linear velocity and angular velocity of a particle moving on a circular path.

Answer

If a particle moves on a circular path of radius r with angular velocity ω, its linear velocity is

$\text v = \text r\omega$

In vector form the relation is written as

$\vec{\text v} = \vec{\omega} \times \vec{\text r}$

Thus, for a given angular velocity the linear velocity of the particle is directly proportional to its distance from the centre; the greater the distance of the particle from the centre, the greater is its linear velocity.

Write the formula of centripetal acceleration of a particle moving on a circular path in terms of the angular velocity.

Answer

The centripetal acceleration of a particle moving on a circular path of radius r is

$\text a = \dfrac{\text v^2}{\text r}$

Substituting v = rω,

$\text a = \dfrac{(\text r\omega)^2}{\text r} \\[1em] = \omega^2 \text r$

Hence, in terms of the angular velocity the centripetal acceleration is a = ω²r, directed along the radius towards the centre of the circular path.

A body is moving with a uniform speed on a circular path of radius r. If n is the frequency of circulation, then write down the centripetal acceleration in terms of n and r.

Answer

The angular velocity is related to the frequency n by

$\omega = 2\pi \text n$

The centripetal acceleration is

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

Substituting the value of ω,

$\text a = (2\pi \text n)^2 \text r \\[1em] = 4\pi^2 \text n^2 \text r$

Hence, the centripetal acceleration in terms of n and r is a = 4π²n²r.

The vertical component of a force inclined at 30° from the horizontal is 200 N. Find the magnitude of the force.

Answer

Given,

• Angle of the force with the horizontal, θ = 30°
• Vertical component of the force, F_y = 200 N

If a force F makes an angle θ with the horizontal, its vertical component is

$\text F_y = \text F \sin \theta$

Substituting the values,

$200 = \text F \sin 30^\circ \\[1em] 200 = \text F \times \dfrac{1}{2} \\[1em] \text F = 400\ \text N$

Hence, the magnitude of the force is 400 N.

The cross product of two vectors $\vec{A} \space$and $\vec{B} \space$is written as follows :

$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$

Expand the right hand side of the equation.

Answer

Expanding the determinant along the first row,

$\vec{\text A} \times \vec{\text B} = \hat{i}(\text A_y\text B_z - \text A_z\text B_y) - \hat{j}(\text A_x\text B_z - \text A_z\text B_x) + \hat{k}(\text A_x\text B_y - \text A_y\text B_x)$

which may also be written as

$\vec{\text A} \times \vec{\text B} = (\text A_y\text B_z - \text A_z\text B_y)\hat{i} + (\text A_z\text B_x - \text A_x\text B_z)\hat{j} + (\text A_x\text B_y - \text A_y\text B_x)\hat{k}$

Explain when the resultant of two equal vectors (i) will be zero, (ii) will be equal to each ?

Answer

The magnitude of the resultant of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

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

Here the two vectors are equal in magnitude, so B = A and

$\text R = \sqrt{2\text A^2(1 + \cos \theta)}$

(i) For R = 0 :

$2\text A^2(1 + \cos \theta) = 0 \\[1em] 1 + \cos \theta = 0 \quad \Rightarrow \quad \cos \theta = -1 \quad \Rightarrow \quad \theta = 180^\circ$

Hence the resultant is zero when the two vectors are oppositely directed.

(ii) For R = A :

$\text A^2 = 2\text A^2(1 + \cos \theta) \\[1em] \dfrac{1}{2} = 1 + \cos \theta \quad \Rightarrow \quad \cos \theta = -\dfrac{1}{2} \quad \Rightarrow \quad \theta = 120^\circ$

Hence the resultant is equal in magnitude to each of them when the angle between the two vectors is 120°.

Find the condition under which $(\vec{a} \cdot \vec{b})^2 = a^2 b^2$.

Answer

The scalar product of the two vectors is $\vec{\text a} \cdot \vec{\text b} = \text{ab} \cos \theta$. Therefore,

$(\vec{\text a} \cdot \vec{\text b})^2 = \text a^2\text b^2 \cos^2 \theta$

Putting this equal to a²b²,

$\text a^2\text b^2 \cos^2 \theta = \text a^2\text b^2 \\[1em] \cos^2 \theta = 1 \quad \Rightarrow \quad \cos \theta = \pm 1 \quad \Rightarrow \quad \theta = 0^\circ \text{ or } 180^\circ$

Hence, the given condition holds when the two vectors are parallel or antiparallel, that is, when they are collinear.

If $\vec{A} \times \vec{B} = 0$ and $\vec{B} \times \vec{C} = 0$, prove that $\vec{A} \times \vec{C} = 0$.

Answer

The vector product of two parallel vectors is a null vector, since

$\vec{\text A} \times \vec{\text B} = \text{AB} \sin \theta\ \hat{n} = 0 \quad \text{when } \theta = 0^\circ \text{ or } 180^\circ$

Given that $\vec{\text A} \times \vec{\text B} = 0$ and neither is a null vector, $\vec{\text A} \space$is parallel to $\vec{\text B}$.

Given that $\vec{\text B} \times \vec{\text C} = 0$, similarly $\vec{\text B} \space$is parallel to $\vec{\text C}$.

Since $\vec{\text A} \space$is parallel to $\vec{\text B} \space$and $\vec{\text B} \space$is parallel to $\vec{\text C}$, the vectors $\vec{\text A} \space$and $\vec{\text C} \space$must also be parallel to each other. Hence the angle between them is 0° (or 180°), and

$\vec{\text A} \times \vec{\text C} = \text{AC} \sin 0^\circ\ \hat{n} = 0$

Hence proved.

Prove that : $|\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = (AB)^2$

Answer

Let θ be the angle between the two vectors. Then

$|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta \quad \text{and} \quad |\vec{\text A} \cdot \vec{\text B}| = \text{AB} \cos \theta$

Squaring and adding,

$|\vec{\text A} \times \vec{\text B}|^2 + |\vec{\text A} \cdot \vec{\text B}|^2 = \text A^2\text B^2 \sin^2 \theta + \text A^2\text B^2 \cos^2 \theta \\[1em] = \text A^2\text B^2(\sin^2 \theta + \cos^2 \theta) \\[1em] = \text A^2\text B^2 = (\text{AB})^2$

Hence proved.

A body is projected such that the vertical height attained by it and the horizontal range are equal. Find the angle of projection with the horizontal. (tan 76° = 4).

Answer

Given,

• Maximum height = horizontal range, that is, h = R

Equating the two,

$\dfrac{\text u^2 \sin^2 \theta_0}{2\text g} = \dfrac{\text u^2 \sin 2\theta_0}{\text g}$

so that

$\sin^2 \theta_0 = 2 \sin 2\theta_0 \\[1em] \sin^2 \theta_0 = 2 \times 2 \sin \theta_0 \cos \theta_0 \\[1em] \dfrac{\sin \theta_0}{\cos \theta_0} = 4 \quad \Rightarrow \quad \tan \theta_0 = 4$

Since tan 76° = 4,

$\theta_0 = 76^\circ$

Hence, the angle of projection with the horizontal is 76°.

Compute the angular velocity of a geostationary satellite.

Answer

A geostationary satellite always remains above the same point of the earth's surface, so its period of revolution around the earth is equal to the period of the axial rotation of the earth.

Given,

• Time period, T = 24 h = 24 × 3600 = 86400 s

The angular velocity is

$\omega = \dfrac{2\pi}{\text T}$

Substituting the value,

$\omega = \dfrac{2 \times 3.14}{86400} \\[1em] = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$

Hence, the angular velocity of a geostationary satellite is 7.27 × 10^-5 rad s^-1, which is the same as $\dfrac{\pi}{12}$ rad h^-1.

Teacher explains the topic 'Motion in a Plane' and to evaluate the understanding of the students, he asks some questions. State whether the responses of the students, in each case are correct or incorrect. Give the reason for your answers.

(i) TEACHER : Can three vector quantities be added in such a way that their resultant be zero?

STUDENT : Yes, if these can be represented in magnitude and direction by the sides of a triangle taken in cyclic order.

(ii) TEACHER : Can you associate vectors with the volume of a sphere?

STUDENT : No, a vector cannot be associated with a sphere, because the direction of the normal drawn outwards on the surface of the sphere is changing at every point.

(iii) TEACHER : What is a scalar product?

STUDENT : The scalar product of two vectors is defined as a scalar quantity equal to the product of their magnitudes and sine of the angle between the given vectors.

Answer

(i) Correct. If three vectors are represented in magnitude and direction by the three sides of a triangle taken in the same (cyclic) order, they complete a closed figure. Their sum-vector cannot then be drawn, which means that their resultant is zero.

(ii) Correct. The volume of a sphere is a scalar quantity and no unique direction can be associated with it, because the outward normal to the surface of the sphere points in a different direction at every point of the surface.

(iii) Incorrect. The scalar product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them, that is,

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$

The sine of the angle occurs in the magnitude of the vector product, $|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta$.

Vectors have both magnitude and direction, they cannot be added by ordinary algebra. A vector can be represented by an arrow. The length of the arrow represents the magnitude and the arrow-head represents the direction. If two vectors are represented in magnitude and direction by two sides of a triangle taken in the same order, then their resultant is represented by the third side of the triangle taken in the opposite order.

(i) The magnitude of resultant of two vectors is equal to each of them. Find the angle between the vectors.

(ii) If two forces 6 N and 8 N act at a point at an angle of 90° with each other. Determine the magnitude of the resultant force.

(iii) Give a few examples of vector quantities.

Answer

(i) The magnitude of the resultant of two vectors is

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

Here R = A = B, so

$\text A^2 = \text A^2 + \text A^2 + 2\text A^2 \cos \theta \\[1em] \text A^2 = 2\text A^2(1 + \cos \theta) \\[1em] \dfrac{1}{2} = 1 + \cos \theta \quad \Rightarrow \quad \cos \theta = -\dfrac{1}{2}$