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

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 4
Chapter Name : Motion in a plane

Q4.1 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.

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

Page : 85 , Block Name : Exercise

Q4.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear
momentum, electric eld, average velocity, magnetic moment, relative velocity

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

Page : 85 , Block Name : Exercise

Q4.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total
path length, energy, gravitational potential, coef cient of friction, charge.

Answer. Impulse
Impulse is given by the product Of force and time. Since force is a vector quantity, its product with time (a
scalar quantity) gives a vector quantity.

Page : 85 , Block Name : Exercise

Q4.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities
are meaningful :
(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector
by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector
to the same vector.

Answer. (a) Meaningful
(b) Not Meaningful
(c) Meaning -ll
(d) Meaningful
(e) Meaningful
(f) Meaningful
Explanation:

(a)The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.
(b)The addition of a vector quantity with a scalar quantity is not meaningful.
(c) A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.
(d) A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having
the same or different dimensions.
(e) The addition of two vector quantities is meaningful only if they both represent the same physical quantity.
(f) A component of a vector can be added to the same vector as they both have the same dimensions.

Page : 85 , Block Name : Exercise

Q4.5 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 (de ned 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

Answer. (a) True
(b) False
(c) False
(d) True
(e) True

Explanation:
(a) The magnitude of a vector is a number. Hence, it is a scalar.
(b) Each component of a vector is also a vector.
(c) Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path
length is always greater than the magnitude of displacement. It becomes equal to the magnitude of
displacement only when a particle is moving in a straight line.
(d) It is because of the fact that the total path length is always greater than or equal to the magnitude Of
displacement Of a particle.
(e) Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the
same order.

Page : 85 , Block Name : Exercise

Q4.6 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 | |

→
Answer. (a) Let two vectors →
a and b be represented by the adjacent sides of a parallelogram OMNP, as shown in
the given gure.

Here, we can write:
→
| OM | = | →
a | . . . (i)
→ →
→
| MN | = | OP | = | b | . . . (ii)
→
→
| ON | = | →
a + b | . . . (iii)
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
→ →
|→a + b | < |→
a | + | b | . . . (iv)
→
If the two vectors → a and b act along a straight line in the same direction, then we can write:
→ →
|→
a + b | = |→
a | + | b | . . . (v)
Combining equations (iv) and (v), we get:
→ →
|→
a + b | ≤ |→
a| + |b|
→
(b) Let two vectors →
a and b be represented by the adjacent sides of a parallelogram OMNP, as shown in the
given gure.

In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
ON + MN > OM
ON + OM > MN
→ → →
| ON | > | OM − OP | (OP = MN)
→ →
|→
a + b | > | |→
a | − | b | | … (iv)

→
If the two vectors →
a and b act along a straight line in the opposite direction, then we can write:
→ →
|→
a + b | = | |→
a | − | b | | … (v)
Combining equations (iv) and (v), we get:
→ →
|→
a + b | ≥ | |→
a| − |b| |
→
(c) Let two vectors →
a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given
gure.

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:
OS < OP + PS
→ →
|→
a − b | < |→
a| + | − b|
→ →
|→
a − b | < |→
a | + | b | . . . (iii)
If the two vectors act in a straight line but in opposite directions, then we can write:
→ →
|→
a − b | = |→
a | + | b | … (iv)
combining equations (iii) and (iv), we get:
→ →
|→
a − b | ≤ |→
a| + |b|
→
(d) Let two vectors →
a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the
given gure.

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both
quantities positive, we take modulus on both sides as:
→ →
‖→
a − b‖ > ‖→
a| − |b| |
→ →
|→
a − b | > | |→
a | − | b | | . . . (iv)
If the two vectors act in a straight line but in the same directions, then we can write:
→ →
|→
a − b | = ‖→
a | − | b | | (v)
Combining equations (iv) and (v), we get:
→ →
|→
a − b | ≥ | |→
a| − |b| |

Page : 85 , Block Name : Exercise

Q4.7 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 ?

Answer. (a) Incorrect
In order to make a + b + c + d = O, it is not necessary to have all the four given vectors to be null vectors. There
are many other ccrnbinations which can give the sunn zero.
(b) Correct
a+b+c+d=0
a + c = − (b + d)
Taking modulus on both the sides, we get:
| a + c | = | − (b + d) | = | b + d |
| a + c | = | − (b + d) | = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d)
(c) Correct
a+b+c+d=0
a = (b + c + d)
Taking modulus both sides, we get:
|a| = |b + c + d|
| a | ≤ | a | + | b | + | c | …(i)
Equation (i) shows that the magnitude of a is equal to or less than the sum of the
magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the
magnitudes of b, c, and d.
(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the free vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a md
d, assuming that these three vectors are represented by the three sides of a triangle. If a and d are collinear,
then it implies that the vector (b + c) is in the line of a md d. This implication holds only then the vector sum
of all the vectors will be zgo.

Page : 86 , Block Name : Exercise

Q4.8 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 Figure What is the

magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?

Answer. Displacement is given by the minimum distance between the initial and nal positions of a particle.
In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will
be equal to the diameter of the ground.
Radius of the ground = 200m
Diameter of the ground = 2 × 200 = 400m
Hence, the magnitude of the displacement for each girl is 400 m, This is equal to the
actual length of the path skated by girl B.

Page : 86 , Block Name : Exercise

Q4.9 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 ?

Answer. Displacement is given by the minimum distance between the initial and nal positions of a body. In
the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement
is zero.
(b) Average velocity is given by the relation:
Net displacement
= Total time
Since the net displacement of the cyclist is zero, his average velocity will also be zero.
(c) Average speed of the cyclist is given by the relation:
Total path length
Average speed = Total time
Total path length = OP + PQ + QO
N
1
= 1 + 4 (2π × 1) + 1
1
= 2 + 2 π = 3.570km

10 1
Time taken = 10min = 60 = 6 h
3.570
= 1 = 21.42km / h
6
1
Average speed 6

Page : 86 , Block Name : Exercise

Q4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m.
Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn.
Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Answer. The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given gure

Let the motorist start from point P. The motorist takes the third turn at S.
EMagnitude of displacement = PS = PV + VS = 500 + 500 = 1000m
Total path length = PQ + QR + RS = 500 + 500 + 500 = 1500m
The motorist takes the sixth turn at point P, which is the starting point,
*magnitude of displacement = O
Total path length = PQ + QR + RS + ST + TU + UP
= 500 + 500 + 500 + 500 + 500 + 500 = 3000m
The motorist takes the eight turn at point R
EMagnitude of displacement = PR
= √PQ 2 + QR 2 + 2(PQ) ⋅ (QR)cos60 ∘
= √500 2 + 500 2 + (2 × 500 × 500 × cos60 ∘ )

= √250000 + 250000 + 500000 ×
( 1
2 )
= 866.03m

β = tan − 1
( 500sin60 ∘
500 + 500cos60 ∘ ) = 30 ∘

Therefore, the magnitude of displacement is 866.03 m at an angle of 300 with PR.
Total path length = Circumference Of the hexagon + PQ + QR
6 x 500 + 500 + 500 - 4000 m
The magnitude of displacement and the total path length coresponding to the required turns is shown in the
given table

Page : 86 , Block Name : Exercise

Q4.11 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 ?

Answer.
(a) Total distance travelled = 23 km
28
Total time taken = 28 min = h
60
Total distance travelled
=
Total time taken
23
= = 49.29km / h

( )
28
60

(b) Distance between the hotel and the station = 10 km = Displacement of the car
10
= 28 = 21.43km / h
10
EAverage velocity 60

Therefore, the two physical quantities (averge speed and average velocity) are not
equal.

Page : 86 , Block Name : Exercise

Q4.12 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 ?

Answer. The described situation is shown in the given gure.

Here,
vc = Velocity of the cyclist
vr = Velocity Of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the
direction of the relative velocity (v) of the rain with respect to the woman.
v = vr + ( − vc )
= 30 + ( − 10) = 20m / s
vc 10
tanθ = =
vr 30

θ = tan − 1
()
1
3

θ = tan − 1 ()
1
3

θ = tan − 1 ()
1
3

= tan − 1
()
1
3
≈ 18 ∘

Hence, the woman must hold the umbrella toward the south, at an angle Of nearly
180 with the vertical.

Page : 86 , Block Name : Exercise

Q4.13 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 ows 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 ?

Answer. Speed of the man, vm = 4 km/h
Width of the river = 1 km
Width of the river
Time taken to cross the river = Speed of the river
1 1
= 4 h = 4 × 60 = 15min
Speed of the river, vr = 3 km/h
Distance covered with ow of the river = vr x t
1 3
= 3 × 4 = 4 km
3
= 4 × 1000 = 750m

Page : 86 , Block Name : Exercise

Q4.14 In a harbour, wind is blowing at the speed of 72 km/h and the ag on the mast of a boat anchored in the
harbour utters 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 ag on the mast of the boat ?

Answer. Velocity of the boat, vb = 51 km/h
Velocity of the wind, = 72 km/h

The ag is uttering in tie north-east direction. It shows Ünat the wind is blowing toward the north-east
direction. When the ship begins sailing toward the north, the ag will move along the direction of the relative
velocity (vwb) of the wind with respect to the boat.

The angle between vw and ( − vb) = 90 ∘ + 45 ∘
51sin(90 + 45)
tanβ =
72 + 51cos(90 + 45)
1
51 ×
51sin45 √2
= = 1
72 + 51( − cos45)
72 − 51 ×
√2
51 51 51
= = =
72√2 − 51 72 × 1.414 − 51 50.800

∴ β = tan − 1(1.0038) = 45.11 ∘
Angle with respect to the east direction = 45.110 — 450 = 0.110
Hence, the ag will utter almost due east.

Page : 86 , Block Name : Exercise

Q4.15 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 ?

Answer. Speed of the ball, u = 40 m/s
Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle
9, is given by the relation:
u 2sin 2 θ
h= 2g
( 40 ) 2sin 2 θ
25 = 2 × 9.8
sin2θ = 0.30625
sinθ = 0.5534
Dθ = sin − 1(0.5534) = 33.60 ∘
u 2sin 2θ
Horizontal range, R = g

(40) 2 × sin2 × 33.60
=
9.8
1600 × sin67.2
=
9.8
1600 × 0.922
= = 150.53m
9.8

Page : 87 , Block Name : Exercise

Q4.16 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 ?

Answer. Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance
when the angle of projection is 450, i.e., 9 = 450.
The horizontal range for a projection velocity v, is given by the relation:
u 2sin 2θ
R= g
u2
100 = g sin90 ∘
u2
g
= 100
The ball will achieve the maximum height when it is thrown vertically upward. For
such motion, the nal velocity v is zero at the maximum height H.
Acceleration, a —g
v 2 − u 2 = − 2gH
1 u2 1
H = 2 × g = 2 × 100 = 50m

Page : 87 , Block Name : Exercise

Q4.17 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 ?

Answer. Length of the string, I — 80 cm = 0.8 m
Number of revolutions= 14
Time taken =25 s
Number of revolutions Number of revolutions 14
v= Frequency,
v= Time taken
= 25 Hz

Frequency. Angular frequency, ω = 2πv
22 14 88
= 2 × 7 × 25 = 25 rads − 1
Centripetal acceleration, a c = ω 2r

=
()
88 2
25 × 0.8

= 9.91m / s 2
The direction Of centripetal acceleration is always directed along the string, toward
the centre, at all points.

Page : 87 , Block Name : Exercise

Q4.18 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

Answer. Radius of the loop, r =1 km = 1000 m
5
Speed of the aircraft, v=900 km/h = 900 × 18 = 250m / s
v2
Centripetal acceleration, a c = r
( 250 ) 2
= 1000 = 62.5m / s 2
Acceleration due to gravity, g = 9.8m / s2
ac 62.5
g
= 9.8 = 6.38
a c = 6.38g

Page : 87 , Block Name : Exercise

Q4.19 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

Answer. (a) False
The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward
the centre. It happens only in the case of uniform circular motion.
(b) True
At a point on a circular path, a particle appears to move tangentially to the circular path, Hence, the velocity
vector of the is always along the tangent at a point.
(c) True
In uniform circular motion (LJCM), the direction of the acceleration vector points toward the centre of the
circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Page : 87 , Block Name : Exercise

Q4.20 The position of a particle is given by
r = 3.0ti − 2.0t 2j + 4.0km
where t is in seconds and the coef cients 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 ?

Answer. Velocity of the particle, →
v = 10.0ĵm / s
Acceleration of the particle ,→
a = (8.0î + 2.0ĵ)
Also,
d →v
But, →
a = dt = 8.0î + 2.0ĵ
d→v = (8.0î + 2.0ĵ)dt
Integrating both sides:
→
v (t) = 8.0tî + 2.0t

Where,
→
u = velocity vector of the particle at t = 0
→
v = velocity vector of the particle at time t
d →r
→
But, v = dt

d→r = →
v dt = (8.0tî + 2.0tĵ + →
u)dt
Integrating the equations with the conditions: and t = 0; r = 0 andat t = t; r = r
1 1
→
r =→ut + 8.0t 2î + × 2.0t 2ĵ
2 2
→
= ut + 4.0t 2î + t 2ĵ
= (10.0ĵ)t + 4.0t 2î + t 2ĵ

(
xî + yĵ = 4.0t 2î + 10t + t 2 ĵ )
Since the motion of the particle is con ned to the x-y plane, on equating the
coef cients of i and j , We get:
x = 4t 2
1

l=
()x
4
2

And y = 10t + t 2
(a) When x = 16 m:
1

t=
()
16
4
2
= 2s

Gy = 10 × 2 + (2)2 = 24m
(b) Velocity of the particle is given by:
→
v(t) = 8.0tî + 2.0tĵ + →
u
at t = 2s
→
v(t) = 8.0 × 2î + 2.0 × 2ĵ + 10ĵ
= 16î + 14ĵ
∴ √(16) 2 + (14) 2
= √(16) 2 + (14) 2
= 21.26m / s

Page : 87 , Block Name : Exercise

Q4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a
constant acceleration of (8.0 2.0 ) i j + m s − 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 ?

Answer. Velocity of the particle, →
v = 10.0ĵm / s
Acceleration of the particle ,→
a = (8.0î + 2.0j)
Also, But ,
¯ d →v
a = dt = 8.0î + 2.0ĵ
→
dv = (8.0î + 2.0ĵ)dt
Integrating both sides:
→
v (t) = 8.0tî + 2.0t

→
u = velocity vector of the particle at t = 0
→
v = Velocity vector of the particle at time t
d →r
→
But, v = dt

d→r = →
v dt = (8.0tî + 2.0tĵ + →
u)dt
Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r
1 1
→
r =→ut + 8.0t 2î + × 2.0t 2ĵ
2 2
→
= ut + 4.0t 2î + t 2ĵ
= (10.0ĵ)t + 4.0t 2î + t 2ĵ

(
xî + yĵ = 4.0t 2î + 10t + t 2 ĵ )
Since the motion of the particle is con ned to the x-y plane, on equating the coef cients of i and j , we get:
x = 4t 2
1

l= ()x
4
2

And y = 10t + t 2
(a) When x = 16m:
1

t= ()16
4
2
= 2s

By = 10 × 2 + (2)2 = 24m
(b) Velocity of the particle is given by:
→
v(t) = 8.0tî + 2.0tĵ + →
u
at t = 2s
→
v(t) = 8.0 × 2î + 2.0 × 2ĵ + 10ĵ
= 16î + 14ĵ
∴ speed of particle
|→
v| = √(16) 2 + (14) 2
= √256 + 196 = √452
= 21.26m / s

Page : 87 , Block Name : Exercise

Q4.22 i and j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the
vectors i j + , and i j − ? What are the components of a vector A= 2 i j + 3 along the directions of i j + and i j − ?
[You may use graphical method]

Answer.
¯
Consider a vector P, given as:
¯
P = î + ĵ
P xî + P yĵ = î + ĵ
On comparing the components on both sides, we get:

Px = Py = 1
→
2 2
1P1 =
√P + P = √1 + 1 = √2
x y
2 2

Hence, the magnitude of the vector î + ĵ is 2√2
√
→
Let θ be the angle made by the vector P, with the x -axis, as shown in the following
figure.

∴ tanθ =
() Qy
Qx

θ = − tan − 1 − 1 ( ) 1
= − 45 ∘

Hence, the vector î − ĵ makes an angle of − 45 ∘ with the x -axis.
It is given that:
→
A = 2î + 3ĵ
A xî + A yĵ = 2î + 3ĵ
On comparing the coefficients of i and ĵ, we have:
A x = 2 and A y = 3
→
|A| = √2 2 + 3 2 = √13

∴ tanθ =
() Ay
Ax

θ = tan − 1 2() 3

= tan − 1(1.5) = 56.31 ∘
Angle between the vectors (2î + 3ĵ) and (î + ĵ), θ = 56.31 − 45 = 11.31 ∘

→ →
Component of vector A , along the direction of P, making an angle θ

( )
( î + ĵ )
= Acosθ ′ P̂ = (Acos11.31)
√2
0.9806
= √13 × (î + ĵ)
√2
= 2.5(i + ĵ)
25
10 √
= × 2

5
=
√2
Let θ 'be the angle between the vectors (2î + 3ĵ) and (î − ĵ)
θ ′′ = 45 + 56.31 = 101.31 ∘
¯
→
A , along the direction of Q, making an angle θ ′′

( ) ( )√
→ î − ĵ
= Acosθ ∗ Q = Acosθ ∗
2

( )√
( î − ĵ )
= √13cos 901.31 ∘
2

13
= −
√ 2
sin11.30 ∘ (î − ĵ)

= − 2.550 × 0.1961(î − ĵ)
= − 0.5(î − ĵ)
5
10 √
= − × 2

1
= −
√2

Page : 87 , Block Name : Exercise

Q4.23 For any arbitrary motion in space, which of the following relations are true :
(a) average = (1/2) (v (t 1 ) + v (t 2 ))
(b) v average = [r(t 2 ) - r(t 1 ) ] /(t 2 – t 1 )
(c) v (t) = v (0) + a t
(d) r (t) = r (0) + v (0) t + (1/2) a t 2
(e) a average =[ v (t 2 ) - v (t 1 )] /( t 2 – t 1 )
(The ‘average’ stands for average of the quantity over the time interval t 1 to t 2 )

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

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Q4.24 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

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

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Q4.25 An aircraft is ying 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 ?

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

Height of the aircraft from ground, OR — 3400 m Angle subtended between the positions, OOQ 300 Time = 10s

In ΔPRO :
PR
tan15 ∘ = OR

PR = ORtan15 ∘
= 3400 × tan15 ∘
△PRO is similar to ΔRQO .
EPR = RQ
PQ = PR + RQ
= 2PR = 2 × 3400tan15 ∘
= 6800 × 0.268 = 1822.4m
1822.4
Speed of the aircraft = 10
= 182.24m / s

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Q4.26 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.

Answer. Generally speaking, a vector has no de nite locations in space. This is because a vector remains
invariant what-I displaced in such a way that its magnitude md
direction remain the same. However, a position vector has a de nite location in space. A vector can vary with
time. For example, the displacement vector of a particle moving with a certain velocity varies with time. Two
equal vectors located at different locations in space need not produce the same physical effect. For example,
two equal forces acting on an Object at different points can cause the body to rotate, but their combination
cannot produce an equal turning

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Q4.27 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 speci ed by the direction of the axis of
rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?

Answer. A physical quantity having both magnitude and direction need not be considered a vector. For
example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a
physical quantity to be considered a vector is that it should follow the law of vector addition.
Generally spending, the rotation of a body about an axis is not a vector quantity as it does not follow the law
of vector addition. However, a rotation by a certain angle follows the law of vector addition and is therefore
considered a vector.

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Q4.28 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ?
Explain.

Answer. No; Yes; No
(a) One cannot associate a vector with the length of a wire bent into a loop.
(b) One can associate an area vector with a plane area. The direction of this vector is normal, inward or to the
plane area.
(c) One cannot associate a vector with the volume Of a sphere. However, an area vector can be associated with
the area of a sphere.

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Q4.29 A bullet red 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 xed, and neglect air
resistance.

Answer.
Range, R = 3km
Angle of projection, θ = 30 ∘
Acceleration due to gravity, g = 9.8m / s2
Horizontal range for the projection velocity u0, is given by the relation:

2
u 0 sin 2θ
R= g
2
u0
3 = g sin60 ∘
2
u0
g = 2√3
The maximum range (Rmax) is achieved by the bullet when it is red at an angle of
450 with the horizontal, that is,
2
u0
R max = g
On comparing equations
R mx = 3√3 = 2 × 1.732 = 3.46km
Hence, the bullet will not hit a target 5 km away.

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Q4.30 A ghter plane ying 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 red for the shell with muzzle speed
600 m s-1 to hit the plane ? At what minimum altitude should the pilot y the plane to avoid being hit ? (Take
g = 10 m s − 2 ).

Answer. Height of the ghter plane 1.5 km 1500 m
Speed of the ghter plane, v = 720 km/h = 200 m/s
Let 9 be the angle with the vertical so that the shell hits the plane, The situation is
shown in the given gure.

Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = uxt
Distance travelled by the plane = Vt
The shell hits the plane. Hence, these two distances must be equal.
uxt = Vt
usinθ = v
v
sinθ =
u
200 1
= = = θ.33
600 3
θ = sin − 1(0.33)
= 19.5 ∘
In order to avoid being hit by the shell, the pilot must y the plane at an altitude (H)
higher than the maximum height achieved by the shell.

u 2sin 2(90 − θ)
∴H=
2g
(600) 2cos 2θ
=
2g
360000 × cos 2 19.5
= 2 × 10

= 18000 × (0.943) 2
= 16006.482m
≈ 16km

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Q4.31 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 ?

Answer.
0.86m / s2; 54.46 ∘ with the direction of velocity
Speed of the cyclist, v = 27kmh = 7.5m / s
Radius of the circular turn, r = 80m
Centripetal acceleration is given as:
v2
ae =
r
(7.5) 2
= = 0.7m / s 2
80

Suppose the cyclist begins cycling from point P and moves toward point Q. At point
Q, he applies the brakes and decelerates the speed of the bicycle by 0.5 m/s 2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist. Since the
angle between
a c and a Tis 900, the resultant acceleration a is given by:
2 2
a=
√a + a
c T

= √(0.7) 2 + (0.5) 2
= √0.74 = 0.86m / s 2
ac
tanθ = a
T

Where θ is the angle of the resultant with the direction of velocity
0.7
tanθ = 0.5 = 1.4

θ = tan − 1(1, 4)
= 54.46 ∘

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Q4.32 2 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

θ(t) = tan − 1
( )
t 0y − gt
b sax

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

θ 0 = tan − 1
( )
4h m
R

where the symbols have their usual meaning.

Answer.

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