# NCERT Solutions for Class 11 Physics Chapter 4 - Motion in a Plane

4.1. State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Ans. (a) Volume : Scalar quantity since it has only magnitude without any direction.

(b) Mass, Speed : Scalar quantity because it is specified only by magnitude.

(c) Acceleration : Vector quantity as it has both magnitude and direction associated.

(d) Density : Scalar quantity as it is specified only by its magnitude.

(e) Number of moles : Scalar quantity as it is specified only by its magnitude.

(f) Velocity : Vector quantity as it has both magnitude and direction.

(g) Angular frequency : Scalar quantity as it is specified only by its magnitude.

(h) Displacement : Vector quantity since it has both magnitude and associated direction.

(i) Angular velocity : Vector quantity as it has both magnitude and direction.

4.2. Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Ans. The two scalar quantities are work and current, as these two don't follow laws of vectors addition and they only have magnitude not direction.

4.3. Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Ans. Impulse is the only vector quantity as it is the product of two vector quantities. Also, it has an associated direction as well as magnitude.

4.4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :

(b) Adding a scalar to a vector of the same dimensions

(c) Multiplying any vector by any scalar

(d) Multiplying any two scalars

(f) Adding a component of a vector to the same vector.

Ans. (a) Adding any two scalars is meaningful if the two have the same unit or both represent the same physical quantity.

(b) Adding a scalar to a vector of the same dimension is meaningless as vector quantity has associated direction.

(c) Multiplication of vector with a scalar is meaningful as it just increases the magnitude of vector quantity and direction remains the same.

(d) Multiplication of scalars is valid and meaningful. This is because, if we have two different physical quantity then their units will also get multiplied.

(e) Adding two vectors is meaningful if they represent the same physical quantity. This is because their magnitude will get added and direction will remain the same.

(f) Adding a component of a vector to the same vector is meaningful as this represents the same case of adding vectors with the same dimensions. In this, the magnitude of the resultant vector will increase and the direction will remain the same.

4.5. Read each statement below carefully and state with reasons, if it is true or false :

(a) The magnitude of a vector is always a scalar,

(b) Each component of a vector is always a scalar,

(c) The total path length is always equal to the magnitude of the displacement vector of a particle.

(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,

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

Ans. (a) True, As the magnitude of a vector will not have any direction (it is a number), so it will be scalar quantity.

(b) False,

The components of a vector will always be a vector as vector also have a direction.

(c) False,

This is true only in case when the particle is moving in a straight line. This is because path length is a scalar quantity whereas displacement is vector quantity.

(d) True,

From the above part (c) it is clear that total path length is either equal or greater than the displacement. As a result, the given statement is true.

(e) True,  Since they don't lie in the same plane so they cannot give null vector after addition.

4.6. Establish the following vector inequalities geometrically or otherwise :

(a) |a + b| < |a| + |b|

(b) |a + b| > ||a| −|b||

(c) |a − b| < |a| + |b|

(d) |a − b| > ||a| − |b||

When does the equality sign above apply?

Ans. (a) Suppose the image given below :

In OCB,

OB < OC + BC

(In , sum of two sides in always greater than third side)

or |a + b| < |a| + |b| ...(i)

But if a and b are in a straight line

then |a + b| = |a| + |b | ...(ii)

From (i) an (ii), we can conclude that,

|a + b| ≤ |a| + |b|

(b) Take the image below :

In OCB, We have

or      OB + BC > OC

(Sum of two sides of a triangle is
greater than the length of another side.)

or      OB > |OC – BC|

or     |a + b| > ||a| – |b|| ...(i)

Also, if a and b are in a straight line but in the opposite direction then

|a + b| = ||a| – |b|| ...(ii)

From (i) and (ii), we get :

|a + b| ≥ ||a| – |b||

(c) Suppose the image as

In OAB, we have

OB <OA + AB

or    |a – b| < |a| + |b| ...(i)

For vector in a straight line,

|a – b| = |a| + |b| ...(ii)

From (i) and (ii) we get :

|a – b| ≤ |a| + |b|

(d) Suppose the image :

In OAB, we have :

OB + AB > OA

or    OB > |OA – AB|

or    |a – b| > ||a| – |b|| ...(i)

Also, if the vectors are in a straight line then :

|a – b| = ||a| – |b|| ...(ii)

From (i) and (ii), we can conclude :

|a – b| ≥ ||a| – |b||

4.7. Given a + b + c + d = 0, which of the following statements are correct :

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of (b + d),

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ?

Ans. (a) Incorrect : In a plane, sum of three vectors can be zero, it is not a necessary condition that all of a, b, c, d should be a null vector.

(b) Correct : We are given that a + b + c + d = 0

So, a + c = – (b + d)

Thus, magnitude of a + c is equal to the b + d.

(c) Correct : We have

a + b + c + d = 0

b + c + d = – a

It is clear that magnitude of a cannot be greater than the sum of the other three vectors.

(d) Correct : Sum of three vectors is zero if they are coplanar.

Thus,   a + b + c + d = 0

or    a + (b + c) + d = 0

Hence, (b + c) must be coplanar with a and d.

4.8. Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?

Ans. The displacement vector is defined as the shortest distance between or body initial and final position.

In this case, the shortest distance between these points is the diameter of the circular ice ground.

Thus, displacement = 400 m.

Girl B had travelled along the diameter so path travelled by her is equal to the displacement.

4.9. A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?

Ans. (a) In this case, the net displacement will be zero because the initial and final position is the same.

∴ Net displacement

= Final position – initial position

(b) Average velocity is defined as the net displacement per unit time. Since, we have the net displacement to be zero. So, the average velocity will also be zero.

(c) For finding average speed we need to calculate the total path travelled.

Total path = OP + arc PQ + OQ

$$= 1 + \frac{1}{4}(2\pi × 1) + 1 \\ \begin{bmatrix}\because \text{Arc PQ = }\frac{1}{4}\text{(Circumference)} \end{bmatrix}$$

= 3.57 km.

Time taken in hour

$$= \frac{10}{60} = \frac{1}{6} \text{hour}$$

So, we can calculate average speed as :

$$= \frac{3.57}{\frac{1}{6}}\\ = \text{ 21.42 km/h.}$$

4.10. 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.

Ans. The track is shown in the figure given below :

Let us assume that the trip starts at point A.

The third turn will be taken at D.

So displacement will be

= 500 + 500 = 1000 m

Total path covered

= AB + BC + CD

= 500 + 500 + 500 = 1500 m

Now, the sixth turn is at A, so, the displacement will be zero. And total path covered will be

= 6 (500) = 3000 m

The eighth turn will be at C.

So, the displacement

= AC

$$= \sqrt{AB^2 + BC^2 + 2(AB)(BC) cos \space60 \degree}\\ = \sqrt{(500)^2+ (500)^2 + 2(500)(500)cos\space 60\degree}$$

= 866.03 m

And the total distance covered

= 3000 + 1000 = 4000 m

= 4 km.

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

Ans. (a) We known that, average speed of taxi is given by :

$$= \frac{\text{Total path travelled}}{\text{Total time taken}}\\ = \frac{23 \text{km}}{\frac{28}{60}\text{h}}\\ \begin{bmatrix}\because \text{1 hr = 60 min, 1 min = } \frac{1}{60}\text{hr} \end{bmatrix}\\ \begin{bmatrix} \text{So, 28 min = } \frac{28}{60} \text{h}\end{bmatrix}$$

= 49.28 km/h

(b) Total displacement

= 10 km    [Actual distance]

Total time taken in hours :

$$= \left( \frac{28}{60} \text{hr}\right) \\ \text{Average velocity : } \\ \left(\frac{10}{\frac{28}{60}}\text{hr} \right)$$

= 21.43 km/h

It can be clearly seen the average speed and average velocity is not the same.

4.12. Rain is falling vertically with a speed of 30 m s–1. A woman rides a bicycle with a speed of 10 m s–1 in the north to south direction. What is the direction in which she should hold her umbrella ?

Ans. The given situation is shown in the figure :

We need to find the relative velocity of rain with respect to woman.

= vrain + (– vwoman)

= 30 + (– 10)

= 20 m/s

And the angle is given by :

$$\text{tan} \theta = \frac{v_{\text{woman}}}{v_{\text{rain}}} \\ tan \theta = \frac{10}{30} \\ q ≈ 18°$$

Hence, woman needs hold an umbrella at 18° from vertical towards the south.

4.13. A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank ?

Ans. The speed of man = 4 km/h.

Time taken to cross the river

$$= \frac{\text{Distance}}{\text{Speed}} \\ = \frac{1}{4}\text{hr} = \text{15 min.}$$

Total distance covered due to the flow of the river

= Speed of river × Time taken

$$= 3 × \frac{1}{4} \\ = \text{0.75 km.}$$

4.14. In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

Ans. According to the question the figure is given below :

The angle between velocity of wind and opposite of velocity is (90° + 45°) = 135°

Using geometry.

$$\text{tan} \space β = \frac{\text{51 sin (90°+45°)}}{\text{72 + 51 cos (90°+45°)}} \\ \text{tan}\space β = \frac{51}{50.8} \\ \text{Thus, } β = \text{tan}^{-1} \frac{51}{50.8} \\ =\text{45.11°}$$

So, the flag will be just 0.11° from the perfect east direction.

4.15. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s–1 can go without hitting the ceiling of the hall ?

Ans. We know that the maximum height reached by a particle in projectile motion is given by :

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

Here,   u = 40 m/s, g = 9.8 m/s2, h = 25 m

Putting the given values in the above questions :

$$\Rightarrow \space \space \text{25 = } \frac{40^2\text{sin}^2 \theta}{2× 9.8}$$

So, we get

sin θ = 0.5534 and θ = 33.60°

Now the horizontal range can be found from :

$$R = \frac{u^2 sin\space 2\theta}{g}\\ = \frac{40^2 \text{sin (2 × 33.6)}}{9.8} \\ \text{= 150.51 m}$$

4.16. 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 ?

Ans. The range of projectile motion, is given by :

$$\text{R} = \frac{u^2 sin\space 2\theta}{g}$$

Substituting the values in the formula we get

$$\Rightarrow \space \space 100 = \frac{u^2 sin \space 90\degree}{g} \\ So, \space \space \frac{u^2}{g} = 100$$

Now since deceleration is also acting on the ball in the downward direction :

v2 – u2 = – 2gh

Since, final velocity is 0, so maximum height is given by :

$$H = \frac{u^2}{2g} \\ = \frac{1}{2}×100 \\ \text{= 50 m}$$

4.17. 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 ?

Ans. Frequency is given by :

$$\text{Frequency} = \frac{\text{No. of revolution}}{\text{Total time taken}} \\ = \frac{14}{25}\text{hz}$$

And, the angular frequency is ω

= 2πf

$$= 2×\frac{22}{7}× \frac{14}{25}\\ = \frac{88}{25}\text{rad/s}$$

Hence, the acceleration is given by :

a = ω2r

$$= \left(\frac{88}{25}\right)^2 × 0.8 \\ = \frac{88 × 88}{25 × 25}× 0.8 \\ \text{= 9.91 m/s}$$

4.18. 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.

Ans. Here, radius of loop, r = 1 km = 1000 m

Speed of aircrafts,

s = 900 km/hr

= 250 m/s

Now, acceleration of aircraft is

$$a = \frac{v^2}{\text{r}} \\ = \frac{(250)^2}{1000} = \frac{250× 250}{1000} \\ \text{= 62.5 m/}s^2 \\ \therefore \space \space \text{Required ratio} \\ \frac{a}{y} = \frac{62.5}{9.8} \\ \text{= 6.38}$$

4.19. Read each statement below carefully and state, with reasons, if it is true or false :

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Ans. (a) False : The net acceleration is not directed only along the radius of the circle. It also has a tangential component.

(b) True : Because particle moves on the circumference of the circle, thus at any point its direction should be tangential to the path in order to move in a circular orbit.

(c) True : In a uniform circular motion, acceleration is radially outward along the circular path. In one complete revolution, all the vectors are cancelled and the null vector is obtained.

4.20. The position of a particle is given by

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

where t is in seconds and the coefficients have the proper units for r to be in metres.

(a) Find the v and a of the particle?

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

$$\textbf{Ans. (a)} \text{ Here, r = } (3.0t\space \hat{i}\space - \space 2.0t^2\space \hat{j}\space +\space 4.0\space \hat{k}\space )\text{m}$$

The velocity vector is given by :

$$v = \frac{dr}{dt} \\ \space v = \frac{d\space (3.0 t \space \hat{i} \space -\space 2.0 t^2\space \hat{j}\space +\space 4.0\space \hat{k} \space)}{dt} \\ v\space = 3\space \hat{i} \space -\space 4t\space \hat{j}$$

Acceleration is given by

$$v = \frac{dv}{dt} \\ = d\space \frac{3\space \hat{i}\space - \space 4t\space \hat{j}}{dt}\space = -4\space \hat{j}$$

(b) Put the value of time t = 2 in the velocity vector as given below :

$$v = 3\space \hat{i}\space -\space 4t\space \hat{j} \\ v = 3\space \hat{i} \space- \space 4(2)\hat{j}\space = 3\space \hat{i}\space -\space 8\space \hat{j}$$

Thus, the magnitude of velocity is :

$$= \sqrt{3^2 + (-8)^2} = 8.54 \text{m/s} \\ \therefore \text{Direction, } \theta = tan^{-1}\frac{8}{3} = -69.45\degree$$

4.21. A particle starts from the origin at t = 0 s with a velocity of 10.0

$$\hat{j}\space \textbf{m/s and moves in the x–y plane with a constant acceleration of } (8.0\space\hat{i}\space + 2.0\space \hat{j}\space) \textbf{ms}^{-2}$$

(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time ?

Ans. (a) We are given the velocity of the particle as

$$10.0\space \hat{j}\space \text{ m/s. }\\ \text{And the acceleration is given by } (8.0\space\hat{i}\space +\space 2.0\space \hat{j}) \text{m/}s^2$$

So, the velocity due to acceleration will be

$$a = \frac{dv}{dt} \\ \text{Then } \space \text{dv} = ( 8.0\space\hat{i}\space + 2.0\space\hat{j})\space dt$$

On integrating both sides,

$$v = 8.0t \space \hat{i}\space + 2.0t\space \hat{j} \space + \space\text{u}$$

Here u is the initial velocity (at t = 0 sec).

Now,

$$v =\frac{dr}{dt} \\ dr = (8.0t\space \hat{i} + 2.0t\space \hat{j}\space + u)dt$$

On integrating both sides, we get

$$r = 8.0 × \frac{1}{2} t^2\space \hat{i} + 2.0 × \frac{1}{2}t^2\space \hat{j} + (10t)\space \hat{j} \\ r = 4t^2\space \hat{i}\space + (t^2 + 10t)\space \hat{j} \\ \Rightarrow \space \space x\space\hat{i} + y\space\hat{j} = 4t^2\space \hat{i} + (t^2 + 10t)\space\hat{j}$$

On comparing coefficients, we get

x = 4t2 and y = 10t + t2

But x = 16

∴ 16 = 4t2

⇒ t2 = 4

⇒ t = 2 sec

∴ y = 22 + 10(2) = 24 m.

(b) Here, acceleration of particle is

$$(8.0\space\hat{i} + 2.0\space\hat{j})\space \text{m/s}$$

Then the velocity of the particle is given by,

$$v = 8.0 t\space \hat{i} + 2.0 t\space \hat{j} + u$$

where, u is the initial velocity.

Put t = 2 sec, then velocity is :

$$v = 8.0(2)\space \hat{i}\space + 2.0(2)\space \hat{j} + 10\space \hat{j} \\ \Rightarrow \space \space v = 16\space\hat{i} + 14 \space\hat{j}$$

Now, the magnitude of velocity gives :

$$\text{| v |} = \sqrt{16^2 + 14^2} \\ = \sqrt{256 + 196} \\ = 21.26 \text{ m/s}$$

$$\textbf{4.22. } \hat{i} \textbf{ and } \hat{j}$$

are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors

$$\hat {i}\space +\space \hat{j}, \textbf{ and } \hat{i}\space -\space \hat{j}\space \textbf{?}$$

What are the components of a vector A =

$$2\space \hat{i}\space + 3\space \hat{j}\space \textbf{along the directions of }\space \hat{i} + \space\hat{j} \textbf{ and }\space \hat{i}\space -\space \hat{j}\space \textbf{[ You may use graphical method]}$$

Ans. Let B be a vector such that :

$$\overrightarrow{B} \space = \space \hat{i} \space + \space \hat{j}$$

Then the magnitude of vector B is given by :

$$\text{| B |} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Now let us assume that the angle made between vector B and x-axis is θ

Then we have :

$$\theta = tan^{-1}\left(\frac{1}{2}\right) = 45\degree$$

Similarly, let C be a vector such that :

$$\overrightarrow{C} = \space \hat{i}\space - \space \hat{j}$$

The magnitude of vector C is

$$\text{| C |} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

Let α be the angle between vector C and x-axis :

$$α = tan^{-1}\left(\frac{-1}{1}\right) = -\space 45\degree\\ \text{Now given } \space \space \overrightarrow{A} = 2\space \hat{i} + 3\space \hat{j}$$

Then the required components of a vector C along the directions of

$$(\space \hat{i} \space+\space \hat{j}) \space \text{is} = \frac{2 + 3}{\sqrt{2}}= \frac{5}{\sqrt{2}}$$

and the required components of a vector C along the directions of

$$(\hat{i}\space - \space \hat{j}) \text{ is : } \\ \frac{2 - 3}{\sqrt{2}} = \frac{-1}{\sqrt{2}}$$

4.23. For any arbitrary motion in space, which of the following relations are true :

(a) vaverage = (1/2) (v(t1) + v(t2))

(b) vaverage = [r(t2) – r(t1)] /(t2 – t1)

(c) v(t) = v(0) + a t

(d) r(t) = r(0) + v(0)t + (1/2) a t2

(e) aaverage = [ v(t2) – v(t1)] /(t2 – t1

(The ‘average’ stands for average of the quantity over the time interval t1 to t2)

Ans. (a) False : Since, it is an arbitrary motion. Hence, the following relation cannot hold all the arbitrary relations.

(b) True : This is true as this relation relates displacement with time correctly.

(c) False : The given equation is valid only in case of uniform acceleration motion.

(d) False : The given equation is valid only in case of uniform acceleration motion.

(e) True : This is the universal relation between acceleration and velocity-time, as the definition of acceleration is given by this.

4.24. Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that

(a) Is conserved in a process

(b) Can never take negative values

(c) Must be dimensionless

(d) Does not vary from one point to another in space

(e) Has the same value for observers with different orientations of axes.

Ans. (a) False : For e.g. energy is a scalar quantity but it is not conserve in inelastic collisions.

(b) False : For example temperature can take negative values in °C.

(c) False : Since, speed is a scalar quantity but has dimensions.

(d) False : Gravitational potential varies in space from point to point.

(e) True : Since it doesn't have direction.

4.25. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ?

Ans. The given situation is shown in the figure :

For finding the speed of aircraft we just need to find the distance AC as we are given t = 10 sec.

Consider DABD,

$$\text{tan } 15\degree = \frac{\text{AB}}{\text{BD}}$$

AB = BD × tan 15°

AC = 2AB = 2BD × tan 15°

= 2 × 3400 × tan 15°

= 1822.4 m

Thus, the speed of aircraft :

$$= \frac{1822.4}{\text{10}}= 182.24 \space \text{m/s}$$

4.26. A vector has magnitude and direction. Does it have a location in space ? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.

Ans. No, a vector doesn't have a definite location as a vector can be shifted in a plane by maintaining or changing its magnitude and direction according to situation.

Vector can change with time. For ex : displacement vector, force vector etc.

No, two equal vectors at a different location may not have identical physical effects. For ex : two equal force vectors at a different location may have different torque but when they are applied together the net torque would be different or may be zero.

4.27. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?

Ans. The main condition for a physical quantity to be a vector is that it should follow the law of vector addition. Also, the vector has both direction and magnitude but this is not the sufficient condition. For e.g. current has both magnitude and direction but it is a scalar quantity as it doesn't follow the law of vector addition.
Rotation is not a vector on a large basis, as it is measured by an angle which follows the law of scalar addition.

4.28. Can you associate vectors with

(a) the length of a wire bent into a loop

(b) a plane area

(c) a sphere ? Explain.

Ans. (a) No, the length of a wire bent into a loop cannot be expressed in vector from as it does not have any direction associated with it.

(b) The plane area can be expressed in vector from as direction can be associated as pointing outward or inward (normal to the plane) of the area.

(c) No, vector cannot be associated with a sphere, as sphere cannot be associated with direction anyhow.

4.29. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance.

Ans. The range of bullet is given to be

R = 3 km = 3000 m

$$R = \frac{u^2\space \text{sin}\space 2\theta}{g} \\ = \frac{u^2\space \text{sin}\space 2 × 30\degree}{g} \\ \Rightarrow \space \space 3 = \frac{u^2\space \text{sin}\space 60\degree}{g} \\ \Rightarrow \space \space 3 = \frac{u^2}{g} × \frac{\sqrt{3}}{2}\\ \Rightarrow \space \space \frac{u^2}{g} = \frac{2 × 3}{\sqrt{3}} = 2\sqrt{3} \space \space ...\text{(i)}$$

Now, we will find the maximum range (maximum range occurs when the angle of projection is 45°)

$$\text{Then, }\space \text{R}_{\text{max}} = \frac{u^2 \text{sin 2}(45\degree)}{g} \\ = \frac{u^2 \text{sin } 90 \degree}{g} = 2\sqrt{3}× 1 = 3.46 \space \text{km}$$

Thus, the bullet cannot travel up to 5 km.

4.30. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s–1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit?

(Take g = 10 m s–2 ).

Ans. According to the question,

Now, the horizontal distance travelled by the shell

= Distance travelled by plane

⇒ u sin θt = vt

$$\Rightarrow \space \space \text{sin } \theta = \frac{v}{u} = \frac{200}{600} = \frac{1}{3} \\ \therefore \space \space \theta = 19.5\degree$$

So, the required height will be :

$$\text{H} = \frac{u^2 \space \text{sin}^2(90 \space - \theta)}{2g}\\ = \frac{600^2 \text{ cos}^2\space \theta}{2g} = \frac{600^2 × \text{cos}^2\space 19.5\degree}{2 × 9.8} \\ \text{= 16006.48 m = 16 Km}$$

4.31. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Ans. Speed of cycle = 27 km/h

$$= \frac{27 × 5}{18} \text{m/s} = 7.5 \text{m/s}$$

The situation is shown in the figure below :

The centripetal acceleration is given by,

$$a_c = \frac{v^2}{r} = \frac{(7.5)^2}{80} = 0.7 \space \text{m/}s^2$$

And the tangential acceleration is given as 0.5 m/s2.

So, the net acceleration becomes :

$$\text{a} = \sqrt{a_c^2 + a_T^2} = \sqrt{0.7^2 + 0.5^2}$$

= 0.86 m/s2

Now for direction,

= 0.86 m/s2

Now for direction,

$$\text{tan }\theta = \frac{a_c}{a_T} = \frac{0.7}{0.5} \\ \text{Thus, }\space \space \theta = 54.46\degree$$

4.32. (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

$$\theta (t) = \text{tan}^{-1} \left(\frac{v_{0y}-gt}{v_{0x}} \right)$$

(b) Shows that the projection angle θ0 for a projectile launched from the origin is given by

$$\theta_0 = \text{tan}^{-1} \left(\frac{4h_{m}}{R} \right)$$

where the symbols have their usual meaning.

Ans. (a) Using the equation of motion in both horizontal and vertical direction.

vy = v0y = gt and vr = v0x

Now,

$$\text{tan θ} = \frac{v_y}{v_x} \\= \frac{v_{0y}-gt}{v_{0x}} \\ \text{Thus }\space \space \theta = tan^{-1}\left(\frac{v_{0y}-gt}{v_{0x}}\right)$$

(b) The maximum height is given by :

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

And the horizontal range is given by :

$$R = \frac{u^2 \text{sin}^22\theta}{g}$$

= u2 × 2 sin θ cos θ

Dividing both, we get :

$$\frac{h}{R} = \frac{\text{\text{tan}}\space \theta} {4} \\ \text{Hence, } \space \space \theta \space =\space \text{tan}^{-1} \left( \frac{4h}{R}\right)$$