CHAPTER > 04

Motion in a Plane
KEY NOTES

Scalars and Vectors Å Multiplying a vector A by a negative number − λ gives another

Å A scalar quantity is a quantity that has magnitude only. It vector whose direction is opposite to the direction of A and
whose magnitude is λ times|A|.
is specified completely by a number, along with the proper
unit. e.g. Distance, mass of an object, etc.
Å A vector quantity is a quantity that has both a magnitude
Addition and Subtraction of Vectors
and a direction. e.g. Displacement, velocity, acceleration,
Å Sum of two vectors A and B can be found, if we place vector B,
etc. so that its tail is at the head of the vector A, then join the tail of
A to the head of B. This line OQ represents a vector R, the sum
Position and Displacement Vectors of vectors A and B as shown in the figure.
Å If P and P′ be the positions of the object at time t and t′, P
A
respectively as shown, then
Y O
P¢ B
R=

P
A+
B

r¢ Q
r

O X Since, in this procedure of vector addition, vectors are

arranged head-to-tail, this graphical method is known as
(i) OP is the position vector of the object at time t.
head-to-tail method. This is also known as triangle
(ii) If the object moves from P to P′, the vector PP′is called method of vector addition.
the displacement vector corresponding to motion
Å Following are important points related to vector addition
from point P to point P′.
(i) Vectors addition is commutative,
Å The magnitude of displacement is either less or equal to
the path length of an object between two points. i.e. A + B = B + A
(ii) Vector addition is associative,
Equality of Vectors and Multiplication i.e. (A + B) + C = A + ( B + C)
of Vectors by Real Numbers Å When two equal and opposite vectors are added (says A and
Å Two vectors A and B are said to be equal, if and only, if −A), as their magnitudes are the same but the directions are
they have the same magnitude and the same direction. opposite. So, the resultant vector will have zero magnitude
and is represented by 0 called a null vector or a zero vector,
Å Multiplying a vector A with a positive number λ gives a
i.e.
vector whose magnitude is changed by the factor λ but the
direction is the same as that of A, i.e.|λ A| = λ |A|, if λ > 0 A − A = 0 and|0|= 0
Å Since, the magnitude of a null vector is zero, its direction Å If a vector A lies in xy-plane such that it is inclined at an
cannot be specified. The main properties of 0 are angle θ with X-axis as shown in the figure, then
(i) A + 0 = A Y
(ii) λ 0 = 0
(iii) 0A = 0 A yj
Å Parallelogram method of vector addition is equivalent to A
the triangle method. In this vectors to be added, says A
and B, have their tail at common origin O, then according θ
X
to this law, the sum of A + B is represented by the diagonal A xi
(OS) of the parallelogram directed from O as shown in the
figure below (i) A = Αxi$ + A y $j
Q S
(ii) Component along X-axis is
B
A+ A x = A cos θ.
B R=
(iii) Component along Y-axis is
A y = A sin θ.
O P
A Ay
(iv) A = A x2 + A y2 and tan θ = .
Å Subtraction of vectors can be defined in terms of addition Ax
of vectors. We define the difference of two vectors A and B Å If a vector A lies in space, such that α, β & γ are the angles
as the sum of two vectors A and −B, i.e.
between A and X, Y & Z-axes respectively, as shown in the
A − B = A + ( − B) figure, then we have

Resolution of Vectors Y

Å Vector A can be resolved along two coplanar vectors a and

b as shown in figure below as A = λ a + µb, where λ and µ
are real numbers. Ay
Ay A
P
β
γ α Ax
X
Az
A mb Az
Ax
Z
O
la Q (i) A x = A cosα; A y = A cosβ; A z = A cosγ
Å A unit vector is a vector of unit magnitude and points in a (ii) A = A i$ + A $j + A k$
x y z
particular direction. It has no dimension and unit. It is
used to specify a direction only. (iii) |A|= A = A x2 + A y 2 + A z2
A
A unit vector associated with a vector A is n$ = , Å Vectors can be added by analytical method, i.e. adding the
| A| vectors by combining their respective components along
where n$ is along A. the coordinate axis.
Å Unit vectors along the X, Y and Z-axes of a rectangular If two vectors are A = Ax i$ + Ay $j + Az k$
coordinate systems are denoted by $i, $j and k,
$ respectively.
and B = Bx i$ + By $j + Bz k$
Since, these are unit vectors, we have
|$i|=|$j|=|k$ |= 1 Then, resultant, R = A + B = R x $i + R y $j + Rz k$

These unit vectors are perpendicular to each other. where, R x = Ax + Bx , R y = Ay + By

Å If we multiply a unit vector, say n$ by a scalar λ, the result and Rz = Az + Bz .
is a vector λ = λn.
$

KEY NOTES
Law of Cosines and Sines Å v in component form can be expressed as
dx $ dy $
Å If two vectors A and Binclined at an angleθ, as shown below v= i+ j = vx $i + vy $j ⇒ |v| = vx2 + vy2
S
dt dt
Q
and direction of v is given by
R β
M vy  vy 
tan θ = or θ = tan −1  
B vx  vx 
θ θ Å Average acceleration a of an object for a time-interval ∆t
α
O A P N
moving in xy-plane is the change in velocity divided by
the time interval, i.e.
Then, from parallelogram method of vector addition,
∆v ∆v x $ ∆v y $
(i) Resultant vector, R = A + B. a= = i+ j = ax$i + ay $j
∆t ∆t ∆t
(ii) R = A 2 + B 2 + 2AB cosθ is known as law of cosines. Å The acceleration (instantaneous acceleration) is the
R A B limiting value of the average acceleration as the
(iii) = = is known as law of sines.
sin θ sin β sin α time-interval approaches zero i.e.,
B sin θ ∆ v dv
(iv) Angle of R from A, tan α = .
A + B cosθ a = lim =
∆ t→ 0 ∆t dt
In one-dimension, the velocity and acceleration of an
Motion in a Plane Å

object are always along the same straight line (either in the
Å Position vector r of a particle P located in a plane (two same direction or in the opposite direction). However, for
dimensions) with reference to the origin of an xy-reference motion in two or three dimensions, velocity and
frame is given by r = xi$ + y$j, where x and y are components acceleration vectors may have any angle 0° and 180°
of r along X and Y-axes. between them.
If the positions of a particle are P and P′ at time t and t′,
Å
Motion in a Plane with Constant
respectively as shown below, then its displacement,
∆r = r ′ − r = ( x ′ $i + y ′ $j) − ( xi$ + y$j) = $i∆x + $j∆y
Acceleration
Å Suppose an object is moving in xy-plane with constant
where, ∆x = x ′ − x , ∆y = y ′ − y. acceleration a. Let the position and velocity of the object be
Y r0 and v 0 at time t = 0 and r and v at any time t. Then,
v +v
Direction of v (i) v = v 0 + at (ii) v = 0
P¢ 2
∆y P 1 2
∆r (iii) r = r0 + v 0t + at .
2
r′
1
r In component form, x = x 0 + v 0xt + axt 2
2
1 2
y = y 0 + v 0yt + ayt
O ∆x X 2
Å The average velocity v of an object is the ratio of the Å The motion in a plane (two-dimensions) can also be
displacement ∆r and the corresponding time-interval ∆t. treated as two separate simultaneous one dimensional
Mathematically, motions with constant acceleration along two
∆r $ ∆x $ ∆y perpendicular directions.
v= =i +j or v = vx $i + vy $j
∆t ∆t ∆t Relative Velocity in Two-dimensions
The direction of the average velocity is same as that of ∆r. Å Suppose two objects A and B are moving with velocities
Å The velocity (instantaneous velocity) is given by the v A and v B respectively (each with respect to some
limiting value of the average velocity as the time-interval common frame of reference, say ground), then velocity of
approaches zero, i.e. object A relative to that of B is v AB = v A − v B.
∆r dr Similarly, the velocity of object B relative to that of A is
v = lim v = lim =
∆t→ 0 ∆t→ 0 ∆t dt v BA = v B − v A
Å The direction of velocity at any point on the path of an Therefore, v AB = − v BA
object is tangential to the path at that point and is in the
direction of motion. ⇒ |v AB| =|v BA|

KEY NOTES
 Note Tf = 2t m , which is expected because of the symmetry of
Projectile Motion the parabolic path.
Å An object that is in flight after being thrown or projected is Å Maximum height of a projectile is the maximum height
called a projectile. Such a projectile might be a football, a
hm reached by the projectile. It is given as
cricket ball, a baseball, etc.
(v0 sin θ 0 ) 2
Å If a projectile (particle or body) moves in a horizontal as hm =
well as vertical direction simultaneously, the motion of 2g
particle is known as projectile motion. Å The horizontal distance travelled by a projectile from its
Å For an object, after being projected with initial velocity v 0 initial position to the position, where it passes y = 0 during
that makes an angle θ 0 with X-axis as shown below its fall is called horizontal range R of projectile. It is the
distance travelled during the time of flight Tf . It is given
y
v 2 sin 2θ 0
as R= 0 .
g

v0 a = - gj R is maximum when 2θ is maximum, i.e. when θ 0 = 45 °,

v0 sin θ0

v2
then its maximum value is R max = 0 .
g
θ0
O v0 cos θ0 x Uniform Circular Motion
Å When an object follows a circular path at a constant speed,
(i) Acceleration acting on it is that due to gravity which the motion of the object is called uniform circular motion.
is directed vertically downward, i.e.
Å The acceleration of an object moving with speed v in a
a = −g$j or a = 0 and a = − g
x y v2
circle of radius R has a magnitude and is always
(ii) Components of its velocity at time t can be given by R
v x = v 0 cosθ 0 directed towards the centre. This acceleration is called
v y = v 0 sinθ 0 − gt centripetal acceleration.
(iii) One of the component of velocity, i.e. x-component Å Since, v and R are constants, the magnitude of centripetal
remains constant throughout the motion and only the acceleration is also constant. However, the direction
y-component changes. changes. Therefore, a centripetal acceleration is not a
vy constant vector.
(iv) At maximum height, v y = 0 and therefore tan −1 = 0. Å The resultant acceleration of an object in circular motion is
vx
towards the centre only, if the speed is constant.
Å The equation of path of a projectile given by Å Angular speed is defined as the time rate of change of
gx 2 angular displacement. It is given as
y = x tan θ 0 −
(v0 cos θ 0 ) 2 ∆θ
ω=
∆t
This is the equation of a parabola, i.e. the path of the
projectile is a parabola.
Å Relation between linear speed and angular speed is
given as
Å The shape of trajectory of the motion of an object is not
determined by the acceleration alone but also depends on v = Rω
the initial conditions of motion (initial and final velocity). So, centripetal acceleration, ac = ω 2 R .
For example, the trajectory of an object moving under Å The time taken by an object to make one revolution is
the same acceleration due to gravity can be straight line known as its time period T and the number of revolution
or a parabola depending on the initial conditions. made in one second is called its frequency ν( = 1 / T).
Å Time of maximum height is the time taken by the Å In term of frequency ν , we have
projectile to reach the maximum height. It is given as
v sinθ 0 ω = 2πν
tm = 0 ac = 4π 2 ν 2 R
g
Note The kinematic equations for uniform acceleration do not
Å Time of flight is the total time Tf during which the apply to the case of uniform circular motion. Since, in this
2v sinθ 0 case, the magnitude of acceleration is constant but the
projectile is in flight. It is given as Tf = 0 .
g direction is changing.

KEY NOTES
 Mastering NCERT
MULTIPLE CHOICE QUESTIONS

TOPIC 1 ~ Scalars and Vectors

1 Amongst the following quantities, which is not a 7 Choose the correct option regarding the given figure.
vector quantity?
(a) Force (b) Acceleration A
(c) Temperature (d) Velocity

B
=
A
2 In order to describe the motion in two or three

.5
–1
dimensions, we use
(a) positive sign (a) B = A (b) B = − A
(b) vectors (c) | B | = | A | (d) | B | ≠ | A |
(c) negative sign 8 A and B are two inclined vectors and R is their sum.
(d) Both (b) and (c) Choose the correct figure for the given description.
3 If length and breadth of a rectangle are 1.0 m and 0.5 m P P
A A
respectively, then its perimeter will be a
(a) O B (b) O B
(a) free vector (b) scalar quantity
R= R=
(c) localised vector (d) Neither (a) nor (b) A+ A+
B B
4 Set of vectors A and B, P and Q are as shown below Q Q
P P
X X¢ A A
A
Y
B P (c) O B (d) O B
O Y¢
Q R= R=
O A+ A+
O B B
O Q Q

Length of A and B is equal, similarly length of P and 9 Among the following properties regarding null vector
Q is equal. Then, the vectors which are equal, are which is incorrect?
(a) A and P (b) P and Q (a) A + 0 = A (b) λ 0 = λ
(c) A and B (d) B and Q (c) 0 A = 0 (d) A − A = 0
5 | λ A| = λ | A| , if 10 Suppose an object is at point P at time t moves to P ′
(a) λ > 0 (b) λ < 0 and then comes back to P. Then, displacement is a
(c) λ = 0 (d) λ ≠ 0 (a) unit vector (b) null vector
(c) scalar (d) None of these
6 If a vector is multiplied by a negative number, we get a
vector whose 11 Find the correct option about vector subtraction.
(a) magnitude and direction both are changed (a) A−B=A+B
(b) only direction is changed (b) A+B= B− A
(c) only magnitude is changed (c) A − B = A + (− B )
(d) only direction is reversed (d) None of the above
TOPIC 2 ~ Resolution of Vectors
12 If A is a vector with magnitude A, then the unit vector their is a change in which of the following with
a$ in the direction of vector A is regard to R?
A
(a) AA (b) A ⋅ A (c) A × A (d) (a) Magnitude
|A | (b) Direction
13 Unit vector of 4$i − 3$j + k$ is (c) Both magnitude and direction
(a) $i − $j + k$ (b) 26$i − 26$j + 26k$ (d) None of the above
4 i$ − 3$j + k$ 19 Consider vectors a, b and c as
(c) (d) 5$i − 4 $j + 5k$
26 a = a x $i + a y $j + a z k$
14 Consider a vector A that lies in xy-plane. If Ax and A y
b = b $i + b $j + b k$
x y z
are the magnitudes of its x and y -components
respectively, then the correct representation of A can c = cx $i + c y $j + c z k$
be given by
Then, for a vector T = a + b − c has its y-component
Y Y
in the form
(a) a y + b y + c y (b) − a y + b y − c y
A A
(a) A sin θ j (b) A sin θ j (c) a y + b y − c y (d) a y − b y + c y
θ
θ
O X O X 20 Unit vector in the direction of the resultant of vectors
A cos θ î A cos θ î
A = − 3$i − 2$j − 3k$ and B = 2$i + 4$j + 6k$ is
Y
− 3$i + 2$j − 3k$
(a) (b) − $i + 2$j + 3k$
A 14
(c) A cos θ j (d) None of these − $i + 2$j + 3k$
θ (c) (d) − 2$i − 4 $j + 8k$
O X 14
A sin θ î
21 Two forces P and Q of magnitude 2F and 3F ,
15 Magnitude of a vector Q is 5 and magnitude of its
respectively,are at an angle θ with each other. If the
y-component is 4. So, the magnitude of the
force Q is doubled, then their resultant also gets
x-component of this vector is
doubled. Then, the angle θ is JEE Main 2019
(a) 8 (b) 3 (c) 6 (d) 8
(a) 60° (b) 120°
16 A vector is inclined at an angle 60° to the horizontal. (c) 30° (d) 90°
If its rectangular component in the horizontal direction 22 It is found that A + B = A . This necessarily implies
is 50 N, then its magnitude in the vertical direction is
(a) |B| = 0 (b) A , B are parallel
(a) 25 N (b) 75 N (c) 87 N (d) 100 N
(c) A , B are perpendicular (d) A; B ≤ 0
17 Three vectors are given as P = 3$i − 4$j, Q = 6$i − 8$j and
23 Find the value of difference of unit vectors A and B
R = (3/ 4) $i − $j, then which of the following is correct? whose angle of intersection is θ.
(a) P, Q and R are equal vectors (a) 2 sin(θ / 2 ) (b) 2cos(θ / 2 )
(b) P and Q are parallel but R is not parallel (c) sin(θ / 2 ) (d) cos(θ / 2 )
(c) P, Q and R are parallel 24 Given, | A + B | = P , | A − B | = Q. The value of
(d) None of the above
P 2 + Q 2 is
18 Two vectors P and Q are inclined at an angle θ and R
(a) 2( A 2 + B 2 ) (b) A 2 − B 2
is their resultant as shown in the figure.
(c) A 2 + B 2 (d) 2( A 2 − B 2 )
Q
25 For two vectors A and B, | A + B | = | A − B | is
R
always true, when
q
a (a) | A | = | B | ≠ 0
O P (b) | A | = | B | ≠ 0 and A and B are parallel or anti-parallel
Keeping the magnitude and the angle of the vectors (c) when either | A | or | B | is zero
same, if the direction of P and Q is interchanged, then (d) None of the above
26 Two vectors A and B have equal magnitudes. The 27 Rain is falling vertically with a speed of 35 ms −1 .
magnitude of ( A + B) is ‘n’ times the magnitude of Winds starts blowing after sometime with a speed of
( A − B). The angle between A and B is 12 ms −1 in east to west direction. In which direction
 n 2 − 1  n − 1 from vertical should boy waiting at a bus stop hold
(a) sin −1  2  (b) sin −1  
 n + 1  n + 1 his umbrella?
 n 2 − 1  n − 1 (a) tan −1 (0.45) , west (b) tan −1 (0.343) , west
(c) cos −1  2  (d) cos −1  
 n + 1  n + 1 (c) tan −1 (0.343) , east (d) tan −1 (0.24) , east

TOPIC 3 ~ Motion in a Plane

28 Position vector r of a particle P located in a plane 31 The direction of instantaneous velocity is shown by
with reference to the origin of an x y-plane as shown v
n of fv
in the figure below is given by Y ctio
Y tion o
Dire Direc
P2
y ∆r3
P P P3
(a) ∆r2 (b)
r r2 r r1
P
O X O X
4 units r Y of v
ct ion
Dire
(c) P (d) None of these
x
O 2 units
r
(a) 2$i + 4 $j (b) 4 $i + 2$j (c) 6k$ (d) $i + $j + 4 k$ O X
29 Suppose a particle moves along a curve shown by the
thick line and the positions of particle are represented 32 The position of a particle is given by
by P at t and P ′ at t′. r = 3t$i + 2t 2 $j + 5k$ , then the direction of
y v (t) at t =1 s in
(a) 45° with X -axis (b) 63° with Y -axis
P′ (c) 30° with Y -axis (d) 53° with X -axis
∆y P
∆r 33 The x and y-coordinates of the particle at any time are
r′ x = 5 t − 2 t 2 and y =10 t respectively, where x and y
r are in metres and t in seconds. The acceleration of the
particle at t = 2 s is NEET 2017
(a) 0 (b) 5 ms −2 (c) −4 ms −2 (d) −8 ms −2
O ∆x
34 The position vector of particle changes with time
where, coordinates of P is (2, 3) and P ′ is (5, 6). Net according to the relation r( t ) = 15t 2 $i + ( 4 − 20t 2 ) $j.
displacement of the particle will be What is the magnitude of the acceleration (in ms −2 )
(a) zero (b) 7$i + 9$j at t =1? JEE Main 2019
(c) 10i$ + 18$j (d) 3i$ + 3$j (a) 50 (b) 100 (c) 25 (d) 40
30 A particle is moving such that its position coordinates 35 In three dimensional system, the position coordinates
(x, y) are (2m, 3m) at t = 0 s, (6m, 7m) at time 2s and of a particle (in motion) are given below
(13m, 14m) at time t = 5s. Average velocity vector x = a cos ωt JEE Main 2019
(v av ) from t = 0s to t = 5 s is y = a sin ωt
1 $ 11 $ $
(a) (13i + 14 $j ) (b) (i + j) z = aωt
5 5
7 The velocity of particle will be
(c) 2( i$ + $j ) (d) ( i$ + $j ) (a) 2 aω (b) 2 aω (c) aω (d) 3 aω
3
36 A particle starts from origin at t = 0 with a velocity 37 When an object is shot from the bottom of a long
5.0 $i ms −1 and moves in xy-plane under action of smooth inclined plane kept at an angle 60º with
horizontal, it can travel a distance x1 along the plane.
force which produces a constant acceleration of
But when the inclination is decreased to 30º and the
(3 .0 $i + 2 .0 $j) ms −2 . What is the y-coordinate of the same object is shot with the same velocity, it can travel
particle at the instant, if x-coordinate is 84 m? x 2 distance. Then x1 : x 2 will be NEET 2019
(a) 36 m (b) 24 m (c) 39 m (d) 18 m (a) 2:1 (b) 1: 3 (c) 1: 2 3 (d) 1: 2

TOPIC 4 ~ Relative Velocity in Two-dimensions

38 If two objects P and Q move along parallel straight 42 The stream of a river is flowing with a speed of 2 km/h.
lines in opposite direction with velocities v P and v Q A swimmer can swim at a speed of 4 km/h. What
respectively, then relative velocity of P w.r.t Q, should be the direction of the swimmer with respect to
(a) vPQ = vP = vQ (b) vP − vQ the flow of the river to cross the river straight ?
JEE Main 2019
(c) vP + vQ (d) vQ − vP
(a) 60° (b) 120° (c) 90° (d) 150°
39 Buses A and B are moving in the same direction with 43 A man standing on a road has to hold his umbrella at
velocities 20$i ms − 1 and 15$i ms − 1 , respectively. 30° with the vertical to keep the rain away. He throws
Then, relative velocity of A w.r.t. B is the umbrella and starts running at10 kmh − 1 . He finds
(a) 35 $i ms− 1 (b) 5 $i ms− 1 (c) 5 $j ms− 1 (d) 35 $j ms− 1 that raindrops are hitting his head vertically. The
actual speed of raindrops is
40 Rain is falling vertically with a speed of 35 ms −1 . A
(a) 20 kmh −1 (b) 10 3 kmh −1
woman rides a bicycle with a speed of 12 ms −1 in east
(c) 20 3 kmh −1 (d) 10 kmh −1
to west direction. The direction in which she should
hold her umbrella is 44 A girl can swim with speed 5kmh −1 in still water. She
(a) at cos −1 (0.343) with vertical towards east crosses a river 2 km wide, where the river flows
(b) at tan −1 (0.343) with vertical towards west steadily at 2 kmh −1 and she makes strokes normal to
the river current. Find how far down the river she go
(c) at cos −1 (0.343) with vertical towards west
when she reaches the other bank.
(d) at tan −1 (0.343) with vertical towards east (a) 1 km (b) 2 km (c) 800 m (d) 750 m
41 A car driver is moving towards a fired rocket with a 45 A girl riding a bicycle with a speed of 5 ms −1 towards
velocity of 8$i ms −1 . He observed the rocket to be east direction sees raindrops falling vertically
moving with a speed of 10 ms −1 . A stationary observer downwards. On increasing the speed to 15 ms −1 , rain
will see the rocket to be moving with a speed of appears to fall making an angle of 45° of the vertical.
(a) 5 ms −1 (b) 6 ms −1 Find the magnitude of velocity of rain.
(c) 7 ms −1 (d) 8 ms −1 (a) 5 ms –1 (b) 5 5 ms –1 (c) 25 ms –1 (d) 10 ms –1

TOPIC 5 ~ Projectile Motion

46 The motion of an object that is in flight after being 48 A ball is projected with velocity10 ms − 1 in a direction
projected is a result of two simultaneously occurring making an angle 30° with the horizontal, what is the
components of motion, which are the components in position coordinate (in metres) of the ball after 1s?
(a) horizontal direction with constant acceleration (a) (8.66, 0.1) (b) (9.8, 1.0)
(b) vertical direction with constant acceleration (c) (4.26, 5.29) (d) (0.4, 8.66)
(c) horizontal direction without acceleration
49 When a ball is thrown obliquely from the ground
(d) Both (b) and (c)
level, then the x-component of the velocity
47 At the top most point of the trajectory for an object (a) decreases with time
that has been projected at an angle θ with the (b) increases with time
horizontal, has acceleration (c) remains constant
(a) > g (b) < g (c) zero (d) = g (d) zero
50 At which point of a projectile motion, acceleration and 58 The ceiling of a hall is 30 m high. A ball is thrown
velocity are perpendicular to each other? with 60 ms −1 at an angle θ, so that it could reach the
(a) At the point of projection ceiling of the hall and come back to the ground. The
(b) At the point of drop angle of projection θ that the ball was projected is
(c) At the top most point given by
1 1
(d) Anywhere in between the point of projection and top (a) sin θ = (b) sin θ =
most points 8 6
1
51 The equations of motion of a projectile are given by (c) sin θ = (d) None of these
x =18t and 2 y = 54t − 98
. t 2 . The angle θ of projection is 3
2 2 59 Two projectiles having different masses m1 and m2
(a) tan −1 ( 3 ) (b) tan −1 (15
. ) (c) sin −1 (d) cos −1 are projected at an angle α and (90° − α ) with the
3 3
same speed from some point. The ratio of their
52 A projectile is fired from the surface of the earth with a maximum heights is
velocity of 5 ms −1 at an angle θ with the horizontal. (a) cot α : sin α (b) 1 : 1
Another projectile fired from another planet with a (c) tan 2 α : 1 (d) 1: tan α
velocity of 3 ms −1 at the same angle, follows a
60 Two projectiles A and B thrown with speeds in the
trajectory of the projectile fired from the earth. The
value of the acceleration due to gravity on the planet is ratio 1 : 2 acquired the same height. If A is thrown
(in ms −2 ) is (given, g = 98
. ms −2 ) CBSE AIPMT 2014
at an angle of 45° with the horizontal, then angle of
projection of B will be
(a) 3.5 (b) 5.9 (c) 16.3 (d) 110.8
(a) 0° (b) 60° (c) 30° (d) 45°
53 A projectile is given an initial velocity of ( $i + 2$j) ms −1 , 61 What is the range of a projectile thrown with
where $i is along the ground and $j is along vertical. If velocity 98 ms −1 with angle 30° from horizontal?
g =10 ms −2 , the equation of its trajectory is JIPMER 2018

(a) y = x − 5x 2
(b) y = 2x − 5x 2 (a) 490 3 m (b) 245 3 m (c) 980 3 m (d) 100 m

(c) 4 y = 2x − 5x 2
(d) 4 y = 2x − 25x2 62 Given below figure show three paths of a rock with
different initial velocities. The correct increasing
54 Amongst the following graphs, which graph represents order for the respective initial horizontal velocity
the correct relation between the height of projectile ( h) component (ignoring the effect of air resistance) is
and time ( t ), when a particle (projectile) is thrown from JEE Main 2013
the ground obeliquely?
Y
h h

(a) (b)

O t O t
h h
1 2 3
X
(c) (d) O

O t O t (a) 1 < 2 < 3 (b) 3 < 2 < 1 (c) 2 < 1 < 3 (d) 3 < 1 < 2
55 Two stones were projected simultaneously in the same 63 A man can throw a stone to a maximum distance
vertical plane from same point obliquely, with different of 80 m. The maximum height to which it will rise, is
speeds and angles with the horizontal. The trajectory of (a) 30 m (b) 20 m (c) 10 m (d) 40 m
path followed by one, as seen by the other, is 64 If a person can throw a stone to maximum height of
(a) parabola (b) straight line (c) circle (d) hyperbola h metre vertically, then the maximum distance
56 Time taken by a stone to reach the maximum height is through which it can be thrown horizontally at an
5.8 s, then total time taken by the stone during which it angle θ by the same person is
was in flight is h
(a) (b) h (c) 2h (d) 3h
(a) 5.8 s (b) 11.6 s (c) 2.9 s (d) 4.2 s 2
57 An aircraft flying horizontally with the speed 480 kmh − 1 65 Find angle of projection with the horizontal in terms
of maximum height attained and horizontal range.
releases a parachute at a height of 980 m from the ground. 2H 4R 4H H
It will strike the ground at (use, g = 10 ms − 2 ) (a) tan −1 (b) tan −1 (c) tan −1 (d) tan −1
R H R R
(a) 1 km (b) 2 km (c) 2.8 km (d) 1.867 km
66 The speed of a projectile at the maximum height is 67 A body is projected at t = 0 with a velocity 10 ms −1 at
(1/2) of its initial speed. Find the ratio of range of an angle of 60° with the horizontal. The radius of
projectile to the maximum height attained. curvature of its trajectory at t =1 s is R. Neglecting air
4 resistance and taking acceleration due to gravity
(a) 4 3 (b)
3 g =10 ms −2 , the value of R is JEE Main 2019
3 (a) 10.3 m (b) 2.8 m (c) 5.1 m (d) 2.5 m
(c) (d) 6
4

TOPIC 6 ~ Uniform Circular Motion

68 In a uniform circular motion, velocity and 73 Two particles A and B are moving in uniform circular
acceleration vectors are motion in concentric circles of radii rA and rB with
(a) perpendicular to each other speed v A and v B , respectively. Their time period of
(b) in same direction rotation is the same. The ratio of angular speed of A
(c) in opposite direction to that of B will be NEET 2019
(d) in arbitrary direction (a) v A : vB (b) rB : rA (c) 1 : 1 (d) rA : rB
69 If a car is executing a uniform circular motion, then 74 Two cars A and B move along a concentric circular
its centripetal acceleration represents path of radius rA and rB with velocities v A and v B
(a) a scalar quantity v
(b) constant vector maintaining constant distance, then A is equal to
vB
(c) not a constant vector rB rA r 2A r 2B
(a) (b) (c) (d)
(d) constant direction and changing magnitude rA rB r 2B r A2
70 The displacement of a particle moving on a circular
75 If the length of the second’s hand in a stop clock is
path when it makes 60° at the centre is 3 cm, then linear velocity of its tip is
(a) 2r (b) r
(a) 4.28 ms −1 (b) 01047
. ms −1
(c) 2r (d) None of these
(c) 0.00314 ms −1 (d) 1424
. ms −1
71 Two particles A and B are moving on two concentric
76 What is the centripetal acceleration of a point mass
circles of radii R1 and R 2 with equal angular speed ω.
which is moving on a circular path of radius 5m with
At t = 0, their positions and direction of motion are
shown in the figure speed 23 ms −1 ?
Y (a) 106 ms −2 (b) 90 ms −2
(c) 60 ms −2 (d) None of these
77 Two bodies of masses m1 and m2 , respectively are
A
moving in circles of radii r1 and r2 , respectively. Their
R1
X speeds are such that they make complete circles in the
R2 same time t. The ratio of their centripetal
B
accelerations is
(a) m1 r1 : m2 r2 (b) m1 : m2 (c) r1 : r2 (d) 1 : 1
78 A car revolves uniformly in a circle of diameter 080
. m
π and completes 100 rev min −1 . Its angular velocity is
The relative velocity v A − v B at t = is given by
2ω (a) 10.467 rads −1 (b) 0.6 rads −1
JEE Main 2019 (c) 46.26 rads −1 (d) 8 rads −1
(a) ω ( R1 + R2 )i$ (b) − ω ( R1 + R2 )$i 79 A car is revolving at the rate of 72 revolutions per
(c) ω ( R1 − R2 )i$ (d) ω ( R2 − R1 )$i minute in a uniformly circular path. Find the velocity
of the car which is 25 cm away from its centre.
72 A particle is moving with 10 ms −1 in a circle of (a) 1 ms –1 (b) 1.5 ms –1 (c) 2 ms –1 (d) 2.5 ms –1
radius 5m, find out the magnitude of average velocity
80 A particle is revolving at 1200 rpm in a circle of
if particle is moved by 60° in 1 s. JIPMER 2019
radius 30 cm. Then, its acceleration is
(a) 5 ms −1 (b) 10 ms −1 (a) 1600 ms −2 (b) 4740 ms −2
(c) 5 3 ms −1 (d) 20 ms −1 (c) 2370 ms −2 (d) 5055 ms −2
 SPECIAL TYPES QUESTIONS

I. Assertion and Reason 88 Unit vector

■ Direction (Q. Nos. 81-86) In the following questions, I. has dimensions and a unit.
a statement of Assertion is followed by a corresponding II. has unit magnitude.
statement of Reason. Of the following statements, choose III. when multiplied by a scalar results in a scalar.
the correct one.
Which of the following statement(s) is/are incorrect?
(a) Both Assertion and Reason are correct and Reason
(a) Both I and II (b) Only II
is the correct explanation of Assertion.
(c) Both I and III (d) Both II and III
(b) Both Assertion and Reason are correct but Reason
is not the correct explanation of Assertion. 89 A motorboat is racing towards north at 25 kmh −1 and
(c) Assertion is correct but Reason is incorrect. the water current in that region is 10 kmh −1 in the
(d) Assertion is incorrect but Reason is correct. direction of 60° east of south. Then, study the
81 Assertion Magnitude of resultant of two vectors may statement regarding the resultant velocity of boat.
be less than the magnitude of either vector. I. Its magnitude is 22 kmh −1 .
Reason Vector addition is commutative. II. It makes an angle 23.2° from velocity of boat.
82 Assertion Vector addition of two vectors is always III. Its magnitude is 36 kmh −1 .
greater than their vector subtraction. IV. It makes an angle 25° from velocity of boat.
Reason At θ = 90°, addition and subtraction of Which of the following statement(s) is/are correct?
vectors are equal. (a) Only I (b) Both I and II
(c) Only III (d) Only IV
83 Assertion The maximum height of projectile is
always 25% of the maximum range. 90 I. The motion of a projectile may be thought of as the result of
two separate, simultaneously occurring components of
Reason For maximum range, projectile should be motions.
projected at 90°. AIIMS 2018
II. At the maximum height for a projectile, the
84 Assertion The range of a projectile is maximum at x-component of the velocity is zero.
45°. III. In projectile motion, one component is along a
Reason At θ = 45°, the value of sin θ is maximum. horizontal direction without any acceleration and the
85 Assertion Sum of maximum height for angles α and other along the vertical direction with constant
90°−α is independent of the angle of projection. acceleration due to the force of gravity.
Which of the following statement(s) is/are correct?
Reason For angles α and 90°−α, the horizontal
(a) Both I and II (b) Both I and III
range R is same. (c) Only I (d) Only III
86 Assertion Uniform circular motion is uniformly 91 I. When an object follows a circular path at a constant
accelerated motion. speed, the motion of the object is called uniform circular
Reason Kinematics equations for uniform motion.
acceleration motion cannot be applied in the case of II. The acceleration of an object in uniform circular motion
uniform circular motion. is always directed away the centre of the circle.
Which of the following statement(s) is/are correct?
II. Statement Based Questions (a) Only I (b) Only II
87 I. Magnitude of a vector is always a scalar. (c) Both I and II (d) Neither I nor II
II. Each component of a vector is always a scalar. 92 Which of the following statement(s) is/are correct?
III. Vectors obey triangle law of addition. (a) Displacing a vector parallel to itself leaves the vector
unchanged.
Which of the following statement(s) is/are correct?
(b) Three equal vectors cannot add upto zero.
(a) Both I and II (b) Both II and III
(c) If | A + B | ≠ | A − B |, they are perpendicular vectors.
(c) Both III and I (d) I, II and III
(d) Both (b) and (c)
93 For two vectors A and B which lie in a plane. Which 99 When the projectile is projected obliquely from the
of the following statement is correct? ground level with initial velocity v 0 and if the air
(a) If magnitude of A and B vector is 3 and 4 and they add resistance is not neglected, then choose the correct
upto give vector having magnitude of 5, then they must statement.
be perpendicular vector. (a) The projectile traverses a perfect parabolic path.
(b) If they add up to give more than 5, then they must be (b) It will hit the ground with the same speed with which it
inclined at obtuse angle. was projected from it.
(c) If they add upto give less than 5, then they must be v2
inclined at acute angle. (c) Range of the projectile < 0 .
(d) None of the above g
 v sin θ 0 
94 Given, a + b + c + d = 0, which of the following (d) Maximum height of the projectile >  0 .
 2g 
statements are incorrect?
(a) a , b, c and d must each be a null vector. 100 For a particle performing uniform circular motion,
(b) The magnitude of ( a + c ) equals the magnitude of choose the incorrect statement(s) from the following.
( b + d ).
(a) Magnitude of particle velocity (speed) remains constant.
(c) The magnitude of a can never be greater than the sum
(b) Particle velocity remains directed parallel to radius
of the magnitudes of b, c and d.
vector.
(d) b + c must lie in the plane of a and d, if a and d are not (c) Direction of acceleration keeps changing as particle moves.
collinear and in the line of a and d, if they are collinear. (d) None of the above
95 A particle moves in a plane such that its coordinate
changes with time as x = at 2 + bt and y = ct, III. Matching Type
101 Match the Column I (example of motion) with
where a, b and c are constants.
Column II (type of motion) and select the correct
Then, which of the following statment is correct? answer from the codes given below.
(a) Acceleration of particle is constant.
(b) Velocity of particle is constant. Column I Column II
(c) Acceleration depends only on its x-component. A. Free fall 1. One-dimensional motion
(d) Acceleration depends only on its y-component.
B. Projectile motion 2. Two-dimensional motion
96 Three particles A, B and C are projected from the C. Circular motion 3. Three-dimensional motion
same point with the same initial speeds making angle
D. Motion along a straight road
30°, 45° and 60°, respectively with the horizontal.
Which of the following statement is correct? A B C D
(a) A , B and C have unequal ranges. (a) 2 1 3 1
(b) Ranges of A and C are less than that of B . (b) 1 2 3 2
(c) 1 2 2 1
(c) Ranges of A and C are equal and greater than that of B. (d) 1 3 2 1
(d) A , B and C have equal ranges.
102 Match the Column I (magnitude of vectors A and B)
97 A biker stands on the edge of a cliff 490 m above the with Column II (relationship between A and B) and
ground and throws a stone horizontally with an initial select the correct answer from the codes given below.
speed of 15 ms −1 . Which one of the following
Column I Column II
statement is correct?
(a) The time taken by the stone to reach the ground is 30 s. A. |A| = l B| = 2
l→ 1. A= −B
  → |
(b) The time taken by the stone to reach the ground is 20 s.
|B|= l A= B
(c) The speed with which it hits the ground is 99 ms −1 . B.
|A| = l
  → ← 2.
(d) None of the above
C. |A | = 2 l B=l 3. 2A = B
98 An insect trapped in a circular groove of radius 12 cm  → ←
moves along the groove steadily and completes D. |A| = l | B| = l
 →   → 4. A=−2B
7 revolutions in 100 s. Which one of the following
statement is correct? A B C D
(a) The angular speed of insect is 1.68 rad/s. (a) 2 4 1 2
(b) The linear speed of the motion is 5.3 cm/s. (b) 3 1 4 2
(c) The magnitude of acceleration is 6.78 cm/s 2 . (c) 3 1 2 4
(d) The acceleration vector is a constant vector. (d) 2 3 4 1
103 Match the Column I (the relations between vectors a, 104 A vector is given by A = 4$i + 3$j + 5k$ (where α, β and
b and c) with Column II (the orientations of a, b and c γ are the angles made by A with coordinate axes, then
in the xy- plane) and select the correct answer from match the Column I with Column II (respective
the codes given below. values) and select the correct answer from the codes
Column I Column II given below.
A. a + b= c 1. Y
Column I Column II
c A. α 1. cos−1 (1 / 2 )
a b B. β 2. cos−1 (4 / 5 2 )
X C. γ 3. cos−1 (3 / 5 2 )
B. a −c= b 2. Y
A B C A B C
(a) 1 2 3 (b) 2 3 1
c (c) 3 2 1 (d) 1 1 2
b
105 Match the Column I (respective change in values of
a X ux , horizontal component of initial velocity of
C. b− a = c 3. Y
projectile and u y vertical component of initial
velocity) with Column II (respective change in
b a horizontal range R and maximum height H ) and select
c
the correct answer from the codes given below.
X
Column I Column II
D. a + b+ c= 0 4.
A. Only ux is doubled 1. R will be two times
a X B. ux is doubled, u y is halved 2. H will be one-fourth
b c C. ux is halved, u y is doubled 3. H will be four times
4. R will be halved

A B C D A B C D A B C A B C
(a) 3 2 1 4 (b) 4 3 1 2 (a) 2 4 1 (b) 1 2 3
(c) 1 4 3 2 (d) 3 2 1 4 (c) 3 1 2 (d) 2 4 1

NCERT & NCERT Exemplar

MULTIPLE CHOICE QUESTIONS

NCERT 108 A cricketer can throw a ball to a maximum horizontal

106 Which of the following algebraic operations is not distance of 100 m. How much high above the ground can
meaningful? the cricketer throw the same ball?
(a) 100 m (b) 50 m (c) 25 m (d) 200 m
(a) Multiplying any vector by any scalar
(b) Multiplying any two scalars 109 An aircraft executes a horizontal loop of radius 1 km with a
(c) Adding any two vectors speed of 900 kmh −1 . What is the ratio of its centripetal
(d) None of the above acceleration with the acceleration due to gravity?
107 The ceiling of a long hall is 25 m high. What is the (a) 6.38 (b) 3.19 (c) 12.76 (d) 5.38
maximum horizontal distance that a ball thrown 110 In a harbour, wind is blowing at the speed of 72 kmh −1 and
with a speed of 40 ms −1 can go without hitting the the flag on the mast of a boat anchored in the harbour
ceiling of the hall? flutters along north-east direction. If the boat starts
(a) 150.5 m (b) 250.5 m moving at the speed of 51 kmh −1 of the north, what is the
(c) 130.2 m (d) 100.5 m direction of the flag on the mast of the boat?
(a) East (b) North (c) South (d) West
 111 A bullet fired at an angle of 30° with the horizontal 115 Figure shows the orientation of two vectors u and v
hits the ground 3 km away. By adjusting its angle of in the xy-plane. If u = a$i + b$j and v = p$i + q$j, then
projection, it can (assume, the muzzle speed to be which of the following option is correct?
fixed and neglect air resistance.) Y
(a) hit a target at 5 km
(b) hit a target at 6 km v
(c) cannot hit a target at 5 km u
(d) None of the above
112 An aircraft is flying at a height of 3400 m above the O X
ground. If the angle subtended at a ground
observation point by the aircraft positions 10 s apart (a) a and p are positive, while b and q are negative
is 30°, then what is the speed of the aircraft? (b) a, p and b are positive, while q is negative
(a) 140 m/s (b) 196.30 m/s (c) a, q and b are positive, while p is negative
(c) 9.8 m/s (d) 120 m/s (d) a, b, p and q are all positive
116 The component of a vector r along X-axis will have
113 A stone tied to the end of a string 80 cm long is
maximum value, if
whirled in a horizontal circle with a constant speed.
(a) r is along positive Y-axis
If the stone makes 14 revolutions in 25 s, what is the
(b) r is along positive X-axis
magnitude of its centripetal acceleration?
(c) r makes an angle of 45° with the X-axis
(a) 8.2 m/s 2 (b) 2 m/s 2 (d) r is along negative Y-axis
(c) 8.8 m/s 2 (d) 9.91 m/s 2
117 The horizontal range of a projectile fired at an angle
of 15° is 50 m. If it is fired with the same speed at an
NCERT Exemplar angle of 45°, its range will be
114 Which one of the following statement is correct? (a) 60 m (b) 71 m
(a) A scalar quantity is the one that is conserved in a (c) 100 m (d) 141 m
process. 118 In a two-dimensional motion, instantaneous speed v 0
(b) A scalar quantity is the one that can never take negative is a positive constant, then which of the following are
values. necessarily correct?
(c) A scalar quantity is the one that does not vary from one (a) The average velocity is not zero at any time
point to another in space.
(b) Average acceleration must always vanish
(d) A scalar quantity has the same value for observers with
different orientation of the axes. (c) Displacements in equal time intervals are equal
(d) Equal path lengths are traversed in equal intervals

Answers
> Mastering NCERT with MCQs
1 (c) 2 (b) 3 (b) 4 (c) 5 (a) 6 (a) 7 (d) 8 (d) 9 (b) 10 (b)
11 (c) 12 (d) 13 (c) 14 (a) 15 (b) 16 (c) 17 (c) 18 (b) 19 (c) 20 (c)
21 (b) 22 (a) 23 (a) 24 (a) 25 (c) 26 (c) 27 (c) 28 (a) 29 (d) 30 (b)
31 (c) 32 (d) 33 (c) 34 (a) 35 (a) 36 (a) 37 (b) 38 (c) 39 (b) 40 (b)
41 (b) 42 (b) 43 (a) 44 (c) 45 (b) 46 (d) 47 (d) 48 (a) 49 (c) 50 (c)
51 (b) 52 (a) 53 (b) 54 (c) 55 (b) 56 (b) 57 (d) 58 (b) 59 (c) 60 (c)
61 (a) 62 (a) 63 (b) 64 (c) 65 (c) 66 (b) 67 (b) 68 (a) 69 (c) 70 (b)
71 (d) 72 (a) 73 (c) 74 (b) 75 (c) 76 (a) 77 (c) 78 (a) 79 (c) 80 (b)

> Special Types Questions

81 (b) 82 (d) 83 (c) 84 (c) 85 (b) 86 (d) 87 (c) 88 (c) 89 (b) 90 (b)
91 (a) 92 (a) 93 (a) 94 (a) 95 (a) 96 (b) 97 (c) 98 (b) 99 (c) 100 (b)
101 (c) 102 (b) 103 (b) 104 (b) 105 (b)

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106 (c) 107 (a) 108 (b) 109 (a) 110 (a) 111 (c) 112 (b) 113 (d) 114 (d) 115 (b)
116 (b) 117 (c) 118 (d)
 Hints & Explanations
1 (c) Temperature is not a vector quantity because it has 11 (c) To subtract B from A, we can add – B and A.
magnitude only. So, A + ( − B ) = A − B = R 2 . This is as shown below
However, force, acceleration and velocity have both a
magnitude and a direction. So, these are vectors in
nature. A –B
B =A –B
R2
2 (b) In order to describe two-dimensional or three- A
–B
dimensional motions, we use vectors.

R1
However, direction of the motion of an object along a B

=
A+
straight line is shown by positive and negative signs.

B
(a) (b)
3 (b) The perimeter of the rectangle would be the sum of
. m + 0.5 m + 10
the lengths of the four sides, i.e. 10 . m Hence, option (c) is the correct about vector subtraction.
+ 0.5 m = 3.0 m. 12 (d) In general, a vector A can be written as
Since, length of each side is a scalar, thus the perimeter A = | A | n$ … (i)
is also a scalar.
where, n$ is a unit vector along A.
4 (c) Two vectors are said to be equal, if and only if they
If a$ is a unit vector along A, then from Eq. (i) we can
have the same magnitude and direction. A
Among the given vectors A and B are equal vectors as write, a$ =
|A |
they have same magnitude (length) and direction.
However, P and Q are not equal even though they are of 13 (c) Given, A = 4 $i − 3$j + k$
same magnitude because their directions are different. |A |= Ax2 + A y2 + A z2
5 (a) | λ A | = λ | A |, if λ > 0, as multiplication of vector
= ( 4 )2 + ( −3 )2 + (1)2 = 26
A with a positive number λ gives a vector whose
magnitude is changed by the factor λ but the direction $ = A = 4i − 3j + k
$ $ $
is same as that of A. ∴ Unit vector, A
|A| 26
6 (a) Multiplying a vector A with a negative number λ
14 (a) Vector along X-axis (x-component)
gives a vector whose magnitude is changed by the
factor λ but direction is reversed. = Ax i$ = | A |cos θi$ = A cos θi$
7 (d) | B | = − 1.5 | A |. So when A is multiplied by − 1.5, Vector along Y- axis (y-component)
then its direction gets reversed and magnitude would be = A y $j = | A |sin θ$j = Asin θj$
1.5 times | A |. This can be shown as
Thus, | B | ≠ | A |.
Y
8 (d) Vectors by definition obey the triangle law of
addition. According to which, if vector B is placed with
its tail at the head of vector A. Then, when we join the
N P
tail of A to the head of B. The line OQ represents a
vector R, i.e. the sum of the vectors A and B. Thus, A
A sin θ j
figure given in option (d) is correct.
P θ
A O X
O A cos θ î M
Hints & Explanations

B
R=
A+ 15 (b) Given, | Q | = 5
B
Q Qy = 4
9 (b) Null vector 0 is a vector whose magnitude is zero
Qx = ?
and its direction cannot be specified. So, it means, |0 | = 0 As, |Q | = Qx2 + Q y2
Thus, λ 0 = 0.
⇒ | Q |2 = Qx2 + Q 2y
Hence, property given in option (b) is incorrect.
Substituting the given values, we get
10 (b) Since in the given case, the initial and final positions
coincides, so the displacement will be zero. Thus, it is a ( 5 )2 = Qx2 + 4 2
null vector. ⇒ Qx = 9= 3
 16 (c) Given, vector can be shown below as In first case F1 = 2F and F2 = 3 F
Y ⇒ Fr2 = 4 F 2 + 9F 2 + 2 × 2 × 3F 2 cos θ
⇒ Fr2 = 13F 2 + 12F 2 cos θ … (ii)
In second case F1 = 2F and F2 = 6 F
Ay A
(Q Force Q gets doubled)
θ and Fr′ = 2Fr (Given)
Ax X
0
By putting these values in Eq. (i), we get
where, θ = 60°
Ay ( 2Fr )2 = ( 2F )2 + ( 6F )2 + 2 × 2 × 6 F 2 cos θ
Then, tan θ = or A y = Ax tan θ
Ax ⇒ 4 Fr2 = 40F 2 + 24 F 2 cos θ … (iii)
⇒ A y = 50 tan 60° = 50 × 3 (Q 3 = 1732
. ) From Eq. (ii) and Eq. (iii), we get;
= 86.6 −
~ 87 N 52F 2 + 48F 2 cos θ = 40F 2 + 24 F 2 cos θ
17 (c) Given, P = 3$i − 4 $j ⇒ 12 + 24 cos θ = 0 or cos θ = − 1 / 2
or θ = 120º (Qcos 120º = − 1 / 2 )
and Q = 6i$ − 8$j = 2( 3$i − 4 $j ) = 2P
2 2
3 1 P 22 (a) Given that A + B = A or A + B = A
Also, R = $i − $j = ( 3i$ − 4 $j ) =
4 4 4 ⇒ A
2
+ B
2
+ 2 A B cos θ = A
2

So, P, Q and R are parallel with unequal magnitude.

Thus, they are not equals vectors. where, θ is angle between A and B.
⇒ B ( B + 2 A cos θ ) = 0
18 (b) Since, the magnitude and angle between the vectors
⇒ B = 0 or B + 2 A cos θ = 0
is unchanged, so the magnitude of the resultant R will
B
be same. However, the direction of R will get changed. ⇒ cos θ = − ...(i)
2A
19 (c) T = a + b − c
= ( ax i$ + a y $j + a z k$ ) + ( bx i$ + b y $j + b z k$ ) If A and B are anti-parallel, then θ = 180°
− ( cx $i + c y $j + c z k$ ) Hence, from Eq. (i),
B
= ( ax + bx − cx )$i + ( a y + b y − c y )$j + ( a z + b z − c z ) k$ cos 180° = − 1 = − ⇒ B = 2A
…(i) 2A
As, T = Tx $i + Ty $i + Tz k$ …(ii) Hence, the given condition can only be implied of either
On comparing Eqs. (i) and (ii), we get B = 0 or A and B are anti-parallel provided B = 2 A .
y-component of T, Ty = a y + b y − c y 23 (a) Difference of vectors A and B can be given as
20 (c) Resultant vector of A and B is |( A − B )|2 = A 2 + B 2 − 2 AB cos θ
R = A + B = ( − 3$i − 2$j − 3k$ ) + ( 2$i + 4 $j + 6k$ )
= 1 + 1 − 2cosθ
= − $i + 2$j + 3k$ [Q A and B are unit vectors]
| R | = ( − 1)2 + ( 2 )2 + ( 3 )2 = 2 − 2cos θ = 2(1 − cos θ )
 θ
= 1 + 4 + 9 = 14 = 2 × 2sin 2 (θ / 2 ) = 4 sin 2  
 2
Unit vector in the direction of R is
|A − B| = 2sin (θ / 2 )
R − $i + 2$j + 3k$
R$ = = 24 (a) According to the question, representation of vectors
|R | 14
Hints & Explanations

A and B can be shown as follows

21 (b) Resultant force Fr of any two forces F1
(i.e. P) and F2 (i.e. Q) with an angle θ between them can B
be given by vector addition as
θ
F1=P
–B (180º–θ) A

F2=Q Fr
F2=Q
θ
α
Given, |A + B| = P
F1=P
⇒ | A + B |2 = P 2
Fr2 = F12 + F22 + 2F1 F2 cos θ … (i) P 2 = A 2 + B 2 + 2 AB cosθ …(i)
 Also, | A − B | = Q ⇒ | A − B |2 = Q 2 Using the rule of vector addition, we see that the
resultant of vr and vw is R as shown in the figure.
⇒ A + B + 2 AB cos(180°−θ ) = Q
2 2 2
The magnitude of R is
⇒ A 2 + B 2 − 2 AB cosθ = Q 2 …(ii)
|R | = v2r + v2w
Adding Eqs. (i) and (ii), we get
P 2 + Q 2 = 2( A 2 + B 2 ) = 352 + 122 = 37 ms −1
The direction θ that R makes with the vertical is given
25 (c) Given, A + B = A − B
by
⇒
2
+ B
2
+ 2 A B cos θ v 12
A tan θ = w = = 0.343 or θ = tan −1 ( 0.343 )
vr 35
2 2
= A + B − 2 A B cos θ Therefore, the boy should hold his umbrella in the
2 2 vertical plane at an angle of about tan −1 ( 0.343 ) with the
⇒ A + B + 2 A B cos θ
vertical towards the east.
2 2
= A + B − 2 A B cos θ
28 (a) Position vector r of an object in x y-plane at point P
⇒ 4 A B cos θ = 0 ⇒ A B cos θ = 0 with its components along X and Y-axes as x and y,
⇒ A = 0 or B = 0 or cos θ = 0 ⇒ θ = 90° respectively is given as r = x$i + y$j.
Thus, | A + B | = | A − B | is always true, when either | A | Given, x = 2 units and y = 4 units
or | B | is zero or A and B are perpendicular to each other.
So, position vector at P will be given as r = 2$i + 4 $j.
26 (c) Given, |A |= |B |
29 (d) Position vector of the particle at P, r = 2$i + 3$j
or A=B …(i)
Let magnitude of ( A + B ) is R and for ( A − B ) is R′. Position vector of the particle at P ′, r ′ = 5$i + 6$j
Now, R=A+B ∴ Displacement of the particle is ∆r = r ′− r
and R 2 = A 2 + B 2 + 2 AB cos θ ⇒ ∆r = ( 5$i + 6$j ) − ( 2$i + 3$j )
R 2 = 2 A 2 + 2 A 2 cos θ …(ii) = ( 5 − 2 )i$ + ( 6 − 3 )$j = 3$i + 3$j
[Q using Eq. (i)] 30 (b) Position vector of the particle at
Again, R′ = A − B t = 0 s, r0 s = 2i$ + 3$j
⇒ R ′ 2 = A 2 + B 2 − 2 AB cosθ
t = 2 s, r = 6i$ + 7$j
2s
R ′ 2 = 2 A 2 − 2 A 2 cosθ …(iii)
and t = 5 s, r5 s = 13$i + 14 $j
[Q using Eq. (i)]
2 Displacement in t = 0 s to t = 5 s,
R ∆r = r5 s − r0 s
Given, R = nR ′ or   =n
2
 R ′
= (13 − 2 )i$ + (14 − 3 )$j = 11i$ + 11$j
Dividing Eq. (ii) with Eq. (iii), we get
∆r 11$i + 11$j 11 $ $
n 2 1 + cos θ Average velocity, v = = = (i + j)
= ∆t 5 5
1 1 − cos θ
31 (c) The direction of instantaneous velocity at any point
n2 − 1 (1 + cos θ ) − (1 − cos θ ) on the path of an object is tangential to the path at that
or =
n2 + 1 (1 + cos θ ) + (1 − cos θ ) point and is in the direction of motion. Also, direction
of average velocity is same as that of ∆r.
n2 − 1 2cos θ
⇒ = = cos θ So, amongst the given figures we can say that, options
n +12
2
(a) and (b) are depicting the direction of averge velocity
Hints & Explanations

 n 2 − 1 but option (c) is correctly depicting the direction of

or θ = cos − 1  2  instantaneous velocity.
 n + 1
32 (d) Given, r = 3ti$ + 2t 2 $j + 5k$
27 (c) The velocity of the rain and wind are represented by
dr d
vectors vr and vw in the figure below. ∴ v( t ) = = ( 3ti$ + 2t 2 $j + 5k$ ) = 3i$ + 4 t$j
vw
dt dt
N At t = 1s, v = 3$i + 4 $j
θ  vy   4
Thus, its direction is θ = tan −1   = tan −1  
R vr W E  vx   3
≅ 53° with X -axis
S
 33 (c) Given, x = 5t − 2t 2 36 (a) Given, initial velocity of the particle at t = 0 s,
Velocity of the particle, v0 = 5.0 i$ ms −1 , acceleration, a = ( 3.0$i + 2.0$j ) ms −2
dx d The position of the particle is given by
vx = = ( 5t − 2t 2 ) = 5 − 4t
dt dt 1
r ( t ) = v0 t + at 2 = 5.0 $i t + (1/ 2 )( 3.0$i + 2.0$j )t 2
dv
Acceleration, ax = x = − 4 ms −2 2
dt = ( 5.0t + 1.5t 2 )i$ + 1.0t 2 $j …(i)
Also, y = 10t ,
As, r ( t ) = x( t )i$ + y( t )$j …(ii)
dy d
Velocity, v y = = (10t ) = 10 Comparing Eqs. (i) and (ii), we get
dt dt
dv y x( t ) = 5.0t + 15
. t 2 and y ( t ) = + 1.0t 2
∴ Acceleration, a y = =0 Given, x( t ) = 84m
dt
∴ Net acceleration of the particle, ⇒ 5.0t + 1.50 t 2 = 84
a net = ax $i + a y $j = ( −4 ms −2 ) i$ + 0 or . t 2 + 5.0t − 84 = 0
150
or a net = − 4 $i ms −2 or a net = − 4 ms −2 Solving the above quadratic equation, the value of t is
given as,
34 (a) Position vector of particle is given as
− b ± b 2 − 4 ac −5 ± 52 − 4 (150
. ) ( −84 )
r = 15t 2 i$ + ( 4 − 20t 2 )$j t= =
2a 2(150
. )
Velocity of particle is
dr d −5 ± 25 + 504 −5 ± 529 −5 ± 23
v= = [15t 2 $i + ( 4 − 20t 2 )$j] = = =
dt dt 3 3 3
= 30t i$ − 40 t $j = 6 or − 9.33
(Neglecting the negative values as t can never be
Acceleration of particle is
negative)
d d
a = ( v ) = ( 30ti$ − 40t$j ) ⇒ t = 6s
dt dt
At t = 6s, y = 1.0( 6 )2 = 36 m
= 30$i − 40$j
So, magnitude of acceleration at t = 1s is 37 (b) The motion of object shot in two cases can be
depicted as below
| a |t = 1 s = ax2 + a 2y = 302 + 402
= 50 ms− 2 Case I

35 (a) Given that the position coordinates of a particle

x = a cos ωt  u
 60°
y = a sin ωt  …(i)
z = aωt  g sin 60°
60° g cos 60°
g
So, the position vector of the particle is
r$ = xi$ + y$j + zk$
Case II
u
⇒ r$ = a cos ωt $i + a sin ωt $j + a ωt k$ 30°
r$ = a[cos ωt $i + sin ωt $j + ωt k$ ] 30° g cos 30°
g sin 30°
therefore, the velocity of the particle is g

dr d [ a ] [cos ωt $i + sin ωt$j + ωt k$ ]

Hints & Explanations

Q v$ = = Using third equation of motion, v2 = u 2 − 2gh … (i)

dt dt As the object stops finally, so
⇒ v$ = − aω sin ωt i + aω co sωt $j + aωk$ )
$ v=0
The magnitude of velocity is For inclined motion, g = g sin θ and h = x
| v | = vx2 + v2y + v2z Substituting these values in Eq. (i), we get
u2
or | v | = ( − aω sin ωt )2 + ( aω cos ωt )2 + ( aω )2 ⇒ u 2 = 2g sin θ x ⇒ x =
2g sin θ
= ω a ( − sin ω t )2 + (cosω t )2 + (1)2 u2
For Case I, x1 =
= 2ωa 2g sin 60°
 For Case II, x2 =
u2 Velocity of rocket, v y $j = (6$j ) ms −1
2g sin 30° ∴ Relative speed of rocket w.r.t. a stationary observer
x1 u2 2g sin 30° = 6 – 0 = 6 ms –1
⇒ = ×
x2 2g sin 60° u2 42 (b) Let the velocity of the swimmer is
1 2 1 vs = 4 km/h
= × = or 1 : 3
2 3 3 and velocity of river is vr = 2 km/h
Also, angle of swimmer with the flow of the river (down
38 (c) Relative velocity of P w.r.t. Q is given by stream) is α as shown in the figure below
vPQ = vP − ( − vQ ) = vP + vQ
39 (b) Given, v A = 20i$ ms− 1 α=90°+θ vr
vr
vB = 15$i ms− 1
vsr θ vs
Relative velocity of A w.r.t. B,
v AB = v A − vB
= 20i$ − 15i$ = 5$i ms− 1 From diagram, angle θ is
v 2 km / h 1
40 (b) In figure below vr represents the velocity of rain and sin θ = r = =
vsr 4 km / h 2
vb is the velocity of the bicycle, the woman is riding.
Both these velocities are with respect to the ground. ⇒ θ = 30°
vb
Clearly, α = 90° + 30° = 120°
43 (a) When the man is at rest with respect to the ground,
N the rain comes to him at an angle 30° with the vertical.
q
This is the direction of the velocity of raindrops with
vrb
respect to the ground.
W E
vr This is as shown below
vmg
–vb S

Since, the woman is riding a bicycle, the velocity of 30°

rain as experienced by her is the velocity of rain relative
to the velocity of the bicycle, she is riding, i.e.
vrb = vr − vb vrm vrg
The angle θ made by the relative velocity vrb with the
vertical given by Here, vr g = velocity of the rain with respect to the
v 12 ground,
tan θ = b = = 0.343
vr 35 vmg = velocity of the man with respect to the ground
or θ = tan −1 ( 0.343 ) and vrm = velocity of the rain with respect to the man.
Here, when the man throws the umbrella and starts
Therefore, the woman should hold her umbrella at an
running, then | vmg | = 10 kmh −1
angle of about tan −1 ( 0.343 ) with the vertical towards
the west. From the above figure, we can write
vrg sin 30° = vmg
Note The difference between this question and the Q.27 is 10
that, in Q.40, the boy experiences the resultant (vector or vrg = = 20 kmh −1
sum) of two velocities while in this question, the sin 30°
woman experiences the velocity of rain relative to the 44 (c) Given, speed of girl, vg = 5 km h −1
Hints & Explanations

bicycle (the vector difference of the two velocities).

Speed of river, vr = 2 km h −1
41 (b) The velocity of car driver = 8 $i ms −1
Width of river, d = 2 km
Velocity of rocket = v y $j ms −1
The given condition is as shown in the figure below
Relative velocity of rocket w.r.t. car = 8$i − v y $j
B vr C
Since, the speed of the rocket observed by the car driver
is 10 ms −1 .
d vg
∴ ( v y )2 + ( 8 )2 = (10 )2 α

v2y = 100 − 64 = 36 A
⇒ v y = 6 ms −1
 Since, the girl dive the river normal to the flow of the 1 1
− × 9.8 = 5 − 4.9 = 01
= 10 × . m
river, time taken by the girl to cross the river, so 2 2
d 2 km 2
t= = = h So, the position coordinate of the ball after 1s is
vg 5 kmh −1 5 (8.66 m, 0.1 m).
In this time, the girl will go down the river by the 49 (c) After the object has been projected, the x-component
distance AC due to river current. of the velocity remains constant throughout the motion
∴ Distance travelled along the river and only the y-component changes, like an object in
2 free-fall in vertical direction. This is shown graphically
= vr × t = 2 × at few instants below
5
4 4000 Y
= km = m = 800 m vy=0
5 5 v v=v0x^i
^ q=0
45 (b) Given, velocity of girl, v = 5i$ ms −1
g
vy j
v0x^i
Let velocity of rain, vr = vx $i + v y $j ms −1
v0x^i
Relative velocity of rain = vr − vg = ( vx − 5 )$i + v y $j v0 ^ v
v0y^
j vy j
vx − 5
Now, it is vertical, so tan θ = =0 q0
vy v0x^i
X
v0x ^
i q=- q0
⇒ vx − 5 = 0 ⇒ vx = 5 …(i) ^ ^
vy j = - v0y j
v
On increasing the speed of the girl, relative velocity
becomes ( vx − 15 )i$ + v y $j 50 (c) At the top most point of the projectile, there is only
v − 15 horizontal component of velocity and acceleration due
tan θ = tan 45° = x =1 to the force of gravity in vertically downward direction.
vy
So, velocity and acceleration are perpendicular to each
⇒ vx − 15 = v y ⇒ v y = −10 [using Eq. (i)] other at the top most point.
∴ Velocity of rain = ( 5$i − 10$j) ms −1 51 (b) Given, equations of motion are
⇒ Magnitude of velocity of rain x = 18t , 2 y = 54 t − 9.8t 2
= ( 5 ) + (10 )
2 2
General equations of projectile are
= 125 = 5 5 ms −1 1
x = u cosθ ⋅ t and y = u sin θ ⋅ t − gt 2
2
46 (d) An object that is in flight after being thrown or where, θ is the angle of projection.
projected is called a projectile. The motion of projectile
Comparing it with given equation, we have
may be thought of as the result of two separate,
54
simultaneously occurring components of motions. One u cos θ = 18 and u sin θ =
component along a horizontal direction without any 2
acceleration u sin θ 54 / 2
⇒ =
and the other along the vertical direction with constant u cos θ 18
54
acceleration due to the force of gravity. ∴ tan θ = = 1.5 ⇒ θ = tan −1 (1.5)
47 (d) When an object projected at an angle θ with the 2 × 18
horizontal, then the acceleration acting on it is that due 52 (a) The trajectory of a projectile projected at an angle θ
to gravity which is directed vertically downward and with the horizontal direction from ground is given by
remains constant throughout. gx2
y = x tan θ − 2
a = − g$j
Hints & Explanations

i.e., 2u cos 2 θ
Thus, at the top most point value of a = g . For same trajectories with equal angle of projection,
48 (a) Given, u = 10 ms− 1 , θ = 30°, t = 1s g
= constant
Horizontal distance, x = u cosθt = 10cos 30° × 1 u2
10 × 3 g earth g ′planet
= = 5 3 m = 8.66 m ⇒ 2
= 2
2 uearth u planet
Similarly, vertical distance,
1 Given, g earth = g = 9.8 ms −2 , uearth = 5 ms −1 and
y = u sin θt − gt 2
2 u planet = 3 ms −1
1 Let, g ′planet = g ′
= 10sin 30° × 1 − × 9.8 × 12
2
 So, substituting these values in Eq. (i), we get Distance at which the parachute strikes the ground
9.8 g ′ = Horizontal velocity × t
=
52 32 = 480 × 14 ×
1
=
6720
= 1867
. km
9.8 × 9 3600 3600
⇒ g′ = = 3.5 ms−2
25 58 (b) Given, u = 60 ms −1
53 (b) Given, initial velocity, u = ( $i + 2$j ) ms −1 Maximum height H that the ball will achieve
Magnitude of velocity, = Height of ceiling of the hall = 30 m
u = (1)2 + ( 2 )2 = 5 ms −1 u 2 sin 2 θ
As, maximum height, H =
2g
Equation of trajectory of projectile,
( 60 ) sin θ
2 2
gx2 ⇒ 30 =
y = x tan θ − 2 2g
2u cos 2 θ
30 × 2g 10
gx2 sec2 θ ⇒ sin 2 θ = = [Q g = 10 ms −2 ]
= x tan θ − 60 × 60 60
2u 2 1
gx2 ⇒ sin θ =
= x tan θ − 2 (1 + tan 2 θ ) [Qsec2 θ = 1 + tan 2 θ] 6
2u
u 2 sin 2 α
Substituting the given values, we get 59 (c) Maximum height, H =
2g
10( x )2  uy 2 
∴y= x× 2− 2
[1 + ( 2 )2 ] Q tan θ = = = 2 For same speed of projection,
2( 5 )  ux 1 
H ∝ sin 2 α
10( x2 )
= 2x − (1 + 4 ) = 2x − 5x2 H1 sin 2 α
2× 5 ∴ =
H 2 sin ( 90° − α )
2

54 (c) When a particle is thrown from the ground

obeliquely, i.e. at an angle θ with the X -axis, then sin 2 α
= = tan 2 α
firstly height attained by it would increase gradually cos 2 α
with time. At a certain instant, it would reach the 60 (c) Given condition, h1 = h2
maximum height. After that, with the increase in the
time, the particle’s height decreases gradually and then u 2 sin 2 θ
For projectile maximum height attained, h =
finally comes to zero. This has been correctly depicted 2g
in the graph shown in option (c). ⇒ u12 sin 2 45° = u22 sin 2 θ
55 (b) Velocities of the stones at some instant t can be
u12 1 1 1
given as ⇒ sin 2 θ = sin 2 45° = ⋅ =
v1 = u1 cos θ1 i$ + ( u1 sin θ1 − gt ) $j u22 2 2 4
 u1 1
v2 = u2 cos θ 2 i$ + ( u2 sin θ 2 − gt )$j
Qgiven, u =
and

⇒ Relative velocity, v1 − v2 = ( u1 cos θ1 − u2 cos θ 2 )i$ 1  2 2
⇒ sin θ =
+ ( u sin θ − u cos θ ) $j 2
1 1 2 2
⇒ θ = 30°
= constant
So, angle of projection of B will be 30°.
Since, their relative velocity is constant.
So, the trajectory of path followed by one as seen by 61 (a) Given, u = 98 ms −1 and θ = 30°
other will be straight line, making a constant angle with Q Range of a projectile,
horizontal. u 2 sin 2θ 98 × 98 × sin 60°
R= = = 490 3 m
Hints & Explanations

56 (b) Due to symmetry of the parabolic path transversed g 9.8

by a stone during its flight, 2t m = T f
u 2 sin 2θ 2u 2 sin θ cos θ
where, t m is the time-taken by the stone to reach the 62 (a) Range of a projectile, R = =
maximum height and T f is the total flight of the stone. g g
Given, t m = 5.8s 2( u sin θ ) ( u cos θ ) 2ux u y
= =
⇒ T f = 2 × 5.8 = 116
. s g g

57 (d) Time taken by the parachute to fall through a ⇒ R ∝ horizontal initial velocity component ( ux )
height h of 980 m ∴ From the given plot, we can see that for path 3, range
is maximum. This implies that the rock has the
2h 2 × 980
t= = = 14 s maximum horizontal velocity component in this path.
g 10 Thus, the correct order will be 1 < 2 < 3.
 63 (b) Given ,maximum horizontal range, R max = 80 m Now, angle made by the velocity vector at time of t = 1 s
vy |10 − 5 3 |
u 2 sin 2θ | tan α | = =
As, range of a projectile, R = vx 5
g
and it is maximum θ = 45° ⇒ tan α = |2 − 3 |
u2 or α = 15º
∴ = 80 m
g ∴ Radius of curvature of the trajectory of the projected
u 2 sin 2 θ body R = v2 / g cosα
Maximum height, h =
2g ( 5 )2 + (10 − 5 3 )2
=
80 1 10 × 0.97
= (sin 2 45° ) = 40 × = 20 m [Q v2 = vx2 + v2y and cos 15º = 0.97]
2 2
64 (c) When, stone is thrown vertically upward, then ⇒ R = 2.77 m ≈ 2.8 m
u2 68 (a) In a uniform circular motion, velocity at each point
Maximum height, h = ⇒ u 2 = 2gh …(i)
2g is along the tangent at that point in the direction of
motion. However, acceleration is directed towards the
Maximum horizontal distance covered by the stone
centre at each point of the circular path.
when it is thrown horizontally at an angle θ is
∴ Velocity and acceleration vectors are perpendicular to
u2
R max = (when θ = 45°), each other.
g
69 (c) For a uniform circular motion,
⇒ R max = 2h [from Eq. (i)]
v2
u sin θ
2 2 centripetal acceleration, ac =
65 (c) Maximum height, H = … (i) R
2g
Since, v and R are constants, the magnitude of the
u 2 sin 2θ centripetal acceleration of the car is also constant.
Horizontal range, R = … (ii)
g However, the direction changes pointing towards the
centre. Therefore, a centripetal acceleration is not a
Dividing Eq. (i) by Eq. (ii), we get
constant vector.
H tan θ
=
R 4 70 (b) In the figure, AB is the required displacement of the
4H particle.
⇒ θ = tan −1
R In triangle OAB, OA = OB and ∠ AOB = 60°
A
66 (b) Let u be the initial speed, so speed at hightest point
u
= u cosθ = ⇒ θ = 60° r Displacement
2 60°
B
u 2 sin 2θ O r
Q Horizontal range, R =
g
u 2 sin 2 θ
and maximum height, H = Therefore, ∆AOB is an equilateral triangle, so
2g
OA = OB = r = AB.
R 4 4 4
⇒ = = =
H tan θ tan 60° 3 71 (d) Angle covered by each particle in time duration 0 to
π
67 (b) Components of velocity at an instant of time t of a
is
2ω
body projected at an angle θ is π π
θ=ω × t=ω × = rad
Hints & Explanations

vx = u cosθ + g x t and v y = u sin θ + g y t 2ω 2

Here, components of velocity at t = 1 s, is π
So, positions of particles at t = is as shown below
vx = u cos 60º + 0 [as g x = 0] 2ω
1
= 10 × = 5 m / s
2
A
and v y = u sin 60º + ( −10 ) × (1)
R1 t=0
3
= 10 × + ( − 10 ) × (1)
2 R2
= 5 3 − 10
⇒ | v y | = |10 − 5 3 | m/s B
 π 76 (a) Given, speed, v = 23 ms −1
Velocities of particles at t = are
2ω
and radius, r = 5 m
vA = − ωR1 i$ and v B = − ωR2 $i
v2 23 × 23
The relative velocity of particles is Centripetal acceleration, ac = =
r 5
v A − v B = −ω R1 $i − ( −ω R2 $i ) = 105.8 ≈ 106 ms −2
= − ω ( R1 − R2 )$i = ω ( R2 − R1 )$i v2
−1
77 (c) As, centripetal acceleration is given as ac =
72 (a) Given, velocity of particle, v = 10 ms r
Radius, r = 5m v12
For the first body of mass m1 , ac1 =
Angular displacement, θ = 60° r1
π v22
∴ θ = 60° × rad For the second body of mass m2 , ac2 =
180° r2
π
⇒ θ= rad Also time taken to complete one revolution by both the
3 bodies is same.
When particle moved by an angle of 60°, i.e. from A to 2πr1 2πr2 v r
B on the circular path, then according to figure, Hence, T1 = T2 ⇒ = ⇒ 1 = 1 …(i)
v1 v2 v2 r2
displacement AB is equal to radius r because it forms an
equilateral triangle. v12 r
i.e. ac 1 : a c 2 = × 22
So, displacement ( AB ) = 5 m r1 v2
Given, t = 1s r12 r2 r1
A = × = = r1 : r2 [from Eq. (i)]
r22 r1 r2
60º B Diameter 0.80 m
60º 78 (a) Radius, r = = = 0.40 m
2 2
O 100
Frequency, ν = 100 revmin −1 = revs −1
60
1 60
Time period, T = = = 0.60
ν 100
Displacement ( AB )
∴ Average velocity = 2π 2 × 314
.
Time ∴ Angular velocity, ω = = = 10.467 rads −1
T 0.60
r
= = 5 ms−1 72
1 79 (c) Given, frequency, ν = 72 revmin −1 = revs −1 ,
60
2π
73 (c) Angular speed, ω = , where T is the time period radius r = 25 cm = 25 × 10−2 m,
of rotation. T
Linear velocity, v = ωr = ( 2πν )r
2π
For particle A , ω A = 22 72 25
TA = 2× × ×
7 60 100
2π
For particle B, ω B = 44 6 25 66 −1
TB = × × = ms
7 5 100 35
ω A 2π TB TB 1
∴ = × = = or 1 : 1 = 2 ms −1 (approx)
ω B TA 2π TA 1
1200
[QTA = TB (given)] 80 (b) Given, ν = 1200 rpm = rps,
60
Hints & Explanations

74 (b) We know that, linear speed v = ωr, where angular 30

velocity ω is constant. r = 30 cm = m
100
vA r
∴ v ∝ r or = A Acceleration of the particle
vB rB
= Centripetal acceleration = ω 2 r = ( 2πν )2 r
75 (c) Given, r = 3 cm = 3 × 10−2 m 2
 22 1200 30
Time period, T = 60 s = 2 × ×  × ≈ 4740 ms −2
2π 2π  7 60  100
Angular velocity, ω = = = 01047
. rads −1
T 60 81 (b) Resultant of two vectors A and B is given as
Linear velocity, v = ωr = 01047
. × 3 × 10−2
−1
R= A 2 + B 2 + 2 AB cos θ
= 0.00314 ms
 ∴ We can say that u 2 sin 2θ
As, horizontal range, R =
(i) If θ is an obtuse angle, then magnitude of R will be g
less than magnitude of the either vectors A or B.
So, for same value of initial velocity, horizontal range
e.g. if | A | = 4 , | B | = 3 and θ = 120°, then of projectile is same for complementary angles.
| R | = 4 2 + 32 + 2 × 4 × 3cos (120° ) Therefore, Assertion and Reason are correct but Reason
 1 is not the correct explanation of Assertion.
= 25 − 12 = 13 Qcos 120° = − 
 2 86 (d) In uniform circular motion the velocity of the object
is changing continuously in direction, the object
∴ |R | < |A |
undergoes uniform acceleration which is not a constant
(ii) If the vectors are in opposite direction and are equal vector. However, for a uniformly accelerated motions,
in magnitude, then also the magnitude of R will be
the acceleration of the object should be constant. Hence,
less than the magnitude of either vectors A or B.
it is not an example of uniformly accelerated motion.
e.g., if | A | = | B | = a (say) and θ = 180°
Kinematic equations for constant acceleration is not
then, | R | = a 2 + a 2 − 2a 2 cos (180° ) applicable for uniform circular motion. Since, in this
case the magnitude of acceleration is constant but its
= 2a 2 − 2a 2 [Qcos 180° = − 1]
direction is changing.
∴ | R | < | A | or |B| Therefore, Assertion is incorrect and Reason is correct.
Also, vector addition is commutative in nature.
87 (c) Statements I and III are correct but II is incorrect
A+B=B+A
and it can be corrected as
Therefore, Assertion and Reason are correct but Reason
Each component of a vector is always a vector.
is not the correct explanation of Assertion.
88 (c) A unit vector is a vector of unit magnitude and
82 (d) | A + B | = A 2 + B 2 + 2 AB cos θ points in a particular direction.
|A − B |= A + B − 2 AB cos θ
2 2
It has no unit and dimensions. It is just used to specify a
direction only.
| A + B | = | A − B |= A2 + B2 [At θ = 90°]
If we multiply a unit vector, say n$ by a scalar λ, then
Therefore, Assertion is incorrect but Reason is correct. the result is a vector = λn.
$
83 (c) To obtain maximum range, angle of projection must So, statements I and III are incorrect but II is correct.
be 45°, i.e., θ = 45°. 89 (b) The vector vb representing N
u 2 sin 2 × 45° u 2 sin 90° u 2 the velocity of the motorboat
So, R max = = = …(i)
g g g and the vector vc representing R
2 the water current are shown in vb
u 2 sin 2 45° u 2  1  u 2 R max
∴ H max = =   = = figure in direction specified φ
2g 2g  2  4g 4 by the problem. Using the θ
[from Eq. (i)] parallelogram method of W vc E
So, H max is 25% of R max . addition, the resultant 60°
Therefore, Assertion is correct but Reason is incorrect. velocity of boat R is S
u 2 sin 2θ obtained in the direction shown in the figure.
84 (c) Horizontal range, R =
g where, θ = 90° + 30° = 120°
We can obtain the magnitude of R using the law of cosine
At, θ = 45°, sin 2θ = 1
u2 R = vb2 + vc2 + 2vb vc cos 120°
∴ R max = = maximum range
g = 252 + 102 + 2 × 25 × 10 ( − 1/ 2 )
sin θ = 1(maximum), at θ = 90°
Hints & Explanations

Q ~ 22 kmh −1
=
Therefore, Assertion is correct but Reason is incorrect. To obtain the direction, we apply the law of sines
u 2 sin 2 α R v
85 (b) Maximum height, H1 = = c
2g sin θ sin φ
u 2 sin 2 ( 90°−α ) u 2 cos 2 α v 10 × sin 120°
and H2 = = or sin φ = c sin θ =
2g 2g R 22
u2 u2 10 3
⇒ H1 + H 2 = (sin 2 α + cos 2 α ) = = = 0.394 ⇒ φ = 23.2°
2g 2g 2 × 22
So, statements I and II are correct but III and IV are
Thus, the sum of height for angle of projections α and incorrect.
90°−α is independent of the angle of projection.
90 (b) Statements I and III are correct but II is incorrect Therefore, a also can never be greater than the sum
and it can be corrected as of the magnitudes of b , c and d.
For a projectile motion, one of the component of the (d) Since a + b + c + d = 0
velocity, i.e. the x-component remains constant or a + (b + c ) + d = 0
throughout the motion and only the y-component The resultant of three vectors a, ( b + c ) and d can
changes, like an object in free fall in vertical direction. be zero only when they lie in a plane and can be
This means, at the maximum height, v y = 0. represented by the three sides of triangle taken in
one order.
91 (a) Statement I is correct but II is incorrect and it can be
If a and d are collinear, then ( b + c ) must be in the
corrected as
line of a and d, only then the vector sum of all the
The acceleration of an object in uniform circular motion vectors will be zero.
is always directed towards the centre of the circle. Thus, the statement given in option (a) is incorrect,
92 (a) Three equal vectors can add upto zero when each of rest are correct.
them is inclined at 120° with the other. 95 (a) Position vector of the particle,
If | A + B | = | A − B |, they are inclined at 90° as shown r = x$i + y$j = ( at 2 + bt )$i + ( ct )$j
below.
N dr
Velocity, v = = ( 2at + b )$i + c$j
dt
B A +B It means, velocity is not constant as it varies with time.
dv
Acceleration, a = = ( 2a )i$
P M dt
A
This means, acceleration of the particle is constant
–B
A–B throughout the motion.
Thus, the statement given in option (a) is correct, rest
O are incorrect.
In this figure, length of MN is equal to length of MO.
96 (b) When a particle is projected at an angle θ with the
So, | A + B | = | A − B |. horizontal with initial velocity u, then the horizontal
Thus, the statement given in option (a) is correct, rest u 2 sin 2θ
are incorrect. range R of projectile is .
g
93 (a) Since, ( 5 )2 = ( 3 )2 + ( 4 )2
Clearly, for maximum horizontal range, sin 2θ = 1
which is in accordance to or 2θ = 90° or θ = 45°. Hence, in order to achieve
pythagoras theorem. So, the A+B (5) maximum range, the particle should be projected at 45°.
vectors can be shown in the (4) B
u2
figure as In this case, R max =
g
∴ A and B are perpendicular.
A (3)
However, if the length of A + B Hence, ranges of A and C are less than that of B.
vector is more than or less than 5, then they should be Thus, the statement given in option (b) is correct, rest
inclined at acute and obtuse angle, respectively. are incorrect.
Thus, the statement given in option (a) is correct, rest 97 (c) The equation of motion
are incorrect. x ( t ) = x0 + v0 x t
94 (a)  1
y ( t ) = y0 + v0 y t +   a y t 2
(a) a + b + c + d can be zero in many ways other than  2
a , b , c and d must each be a null vector, e.g. if the
vectors are in different directions, then their Here, x0 = y0 = 0, v0 y = 0
a y = − g = − 9.8 ms −2
Hints & Explanations

resultant will be zero.

(b) Since a + b + c + d = 0 v0 x = 15 ms −1
∴ a + c = − (b + d )
The stone hits the ground when y ( t ) = − 490 m
or |a + c | = | b + d |
Substituting the given values in Eq. (i), we get
(c) Since a + b + c + d = 0
 1
∴ a = − (b + c + d ) − 490 = −   ( 9.8 ) t 2
 2
or |a | = | b + c + d |
Therefore, the magnitude of a is equal to the t = 10 s
magnitude of ( b + c + d ). As magnitude of ∴ The time taken by the stone to reach the ground is 10 s.
( b + c + d ) can be equal or less than the sum of The velocity component for the stone are vx = v0 x and
magnitudes of b , c and d but can never be greater. v y = v0 y − gt .
 So, when the stone hits the ground B. As, | A | = | B | and but both are in opposite direction,
vx = v0 x = 15 ms −1 so A = − B
v y = 0 − 9.8 × 10 = − 98 ms −1 C. As, | A | = 2 | B | but both are in opposite direction,
Therefore, the speed of the stone, so A = − 2B.
~ 99 ms −1
v = vx2 + v2y = (15 )2 + ( − 98 )2 = D. | A | = | B | and both are in same direction,
so A = B.
Thus, the speed with which it hits the ground is 99 ms −1 . Hence, A → 3, B→ 1, C→ 4 and D→ 2.
Thus, the statement given in option (c) is correct, rest
are incorrect. 104 (b) Magnitude of | A | = Ax2 + A y2 + A z2
98 (b) Given, R = 12 cm and T = 100 s = ( 4 )2 + ( 3 )2 + ( 5 )2
2 π 2π × 7
Angular speed, ω = = = 0.44 rad/s = 16 + 9 + 25 = 5 2
T 100
Linear speed, v = ωR = 0.44 × 12 = 5.3 cm/s Angles made by A with coordinate axes,
The direction of velocity v is along the tangent to the A 4  4 
circle at every point. The acceleration is directed cos α = x = or α = cos −1  
|A | 5 2  5 2
towards the centre of the circle. Since, this direction
changes continuously, hence acceleration here is not Ay 3  3 
cos β = = or β = cos −1  
constant vector. |A | 5 2  5 2
The magnitude of acceleration,
Az 5 1
a = ω 2 R = ( 0.44 )2 × 12 = 2.3 cm/s 2 cos γ = = =
|A | 5 2 2
Thus, the statement given in option (b) is correct, rest
 1
are incorrect. or γ = cos −1  
 2
99 (c) Air resistance is a form of dissipative force. In the
presence of such an opposing force, any object will lose Hence, Α → 2, B → 3 and C → 1.
some part of its initial energy and consequently, 2ux u y u 2y
momentum too. Thus, a projectile that traverses a 105 (b) As, Ri = and H i =
parabolic path would certainly show deviation from its g g
idealised trajectory. A. When only ux is doubled, then
In the presence of air resistance, it will not hit the 2( 2ux )u y 4u y ux
R= = = 2Ri
ground with the same speed with which it was projected g g
from it.
and H = H i
In the absence of air resistance, the x-component of the So, only R will be changed. It will be two times.
velocity remains constant and it is only the
B. When ux is doubled, u y is halved, then
y-component that undergoes a continuous change.
However, in the presence of air resistance, both of these  uy 
2( 2ux )  
would get affected. That would mean that the range  2
R= = Ri
v2 g
would be less than 0
g 2
 uy  u 2y 1
and maximum height attained would be less than H =  /g = = Hi
 2 4g 4
( v0 sin θ 0 )2
. So, only H will change to one-fourth of the original
2g value.
Thus, the statement given in option (c) is correct, rest C. When ux is halved, u y is doubled, then
Hints & Explanations

are incorrect. u 
2  x  ( 2u y )
100 (b) Statement given in option (b) is incorrect and it can  2
be corrected as R= = Ri
g
For a particle performing uniform circular motion,
velocity will be tangential in the direction of motion at a ( 2u y )4
H= = 4Hi
particular point. g
Rest statements are correct. So, only H will change to four times the original
102 (b) value.
A. As, | B | = 2 | A | and they both are in the same Hence, A → 1, B → 2 and C → 3.
direction, so 2A = B.
106 (c) Adding any two vectors is not meaningful because Acceleration due to gravity, g = 9.8 ms−2
only vectors of same dimension, i.e. having same unit Centripetal acceleration ( a ) 62.5
can be added. ∴ = = 6.38
Acceleration due to gravity ( g ) 9.8
107 (a) Given, initial velocity, u = 40 ms −1
110 (a) When the boat is anchored in the harbour, the flag
Height of the hall, H = 25 m flutters along the north-east direction, therefore the
Let the angle of projection of the ball be θ, when velocity of the wind is along north-east direction.
maximum height attained by it be 25 m. Velocity of the wind, vw = 72 kmh −1 (north-east)
u 2 sin 2 θ
Maximum height attained by the ball, H = Velocity of the boat, vb = 51 kmh −1
2g
When the boat starts moving along north direction, the
( 40 ) sin θ
2 2
⇒ 25 = flag will flutter along the direction of the relative
2 × 9.8 velocity of wind w.r.t. boat ( vwb ). Angle between
25 × 2 × 9.8 vw and − vb = 45° + 90° = 135°. Actually when the
or sin 2 θ = = 0.3063 boat moves in northern direction, flag flutters in
1600
southern direction.
or sin θ = 0.5534 = sin 33.6° or θ = 33.6°
If relative velocity of wind w.r.t. the boat ( vwb ) makes
u 2 sin 2θ angle β with the direction of wind, i.e. north-east, then
∴ Horizontal range, R =
g vb sin 135°
tan β =
( 40 )2 sin ( 2 × 33.6° ) vw + vb cos 135°
=
9.8 Qsin 135° = sin (180° − 45° ) = sin 45° 
1600 × sin 67.2°  and cos 135° = cos (180° − 45° ) = − cos 45°
=  
9.8
1600 × 0.9219 1
= = 150.5 m 51 ×
51 sin 45° 2
9.8 = =
72 + 51( − cos 45° )  1
108 (b) Horizontal range of a projectile is given by 72 + 51  − 
 2
u 2 sin 2θ
R= =
51
= 1.0034
g 72 2 − 51
If θ = 45°, then R is maximum and is equal to
u2 = tan (45.1° ) or β = 45.1°
R max = N
g
Given, R max = 100 m vb vw
u2
∴ 100 = …(i) 45°
β
g O
W E
When cricketer throws the ball vertically upward, then –vb 135°
it goes upto height H. vwb

Using equation of motion, v2 = u 2 + 2as

S
( 0 )2 = u 2 + 2( − g )H
Angle with respect to east direction
u2 1  u2  1
or H= =   = × 100 = 45.1° − 45° = 0.1°
2g 2  g  2
[using Eq. (i)] Therefore, the flag will flutter almost due to east.
= 50 m 111 (c) Angle of projection, θ = 30°
So, the cricketer can throw the ball 50 m above the Horizontal range, R = 3 km = 3000 m
Hints & Explanations

ground.
As, we know that
109 (a) Radius of horizontal loop, r = 1km = 1000 m u 2 sin 2θ
R=
5
Speed of aircraft, v = 900 kmh −1 = 900 × ms −1 g
18 u2 R
 −1 5 −1  or =
Q 1 kmh = ms  g sin 2θ
 18 
= 250 ms −1 u2 3000 3000
or = =
g sin 60° 3/ 2
Centripetal acceleration of the aircraft,
v2 ( 250 )2 62500 u 2 6000
a= = = = 62.5 ms −2 = …(i)
r 1000 1000 g 3
 When bullet is fired at an angle of projection 45°, then 115 (b) Clearly from the diagram, u = a$i + b$j
horizontal range is maximum.
As u is in the first quadrant, hence both components a
u 2 sin( 2 × 45° ) u 2
∴ R max = = and b will be positive.
g g For v = p i$ + q$j, as it is in positive x-direction and
6000 located downward hence x-component p will be positive
= = 2000 3 = 3464 m
3 but y-component q will be negative.
Therefore, bullet cannot be fired up to 5000 m or 116 (b) Let r makes an angle θ with positive X -axis.
6000 m with the same muzzle speed because it can Component of r along X-axis is
cover maximum horizontal distance upto 3464 m.
rx = r cos θ
112 (b) Let A be the position of the aircraft at time t = 0 and ( rx ) max = r (cos θ ) max
at t = 10 s it is at position B as shown below. = r cos 0° = r
Given, ∠ AOB = 30° and height of the aircraft above
the ground, h = 3400 m (Q cos θ is maximum at θ = 0°)
AB As θ = 0°, r is along positive X-axis.
In right angle triangle OAB, tan 30° =
OA 117 (c) Given, θ = 15 ° and R = 50 m
A B u 2 sin 2 θ
We know that, range, R = … (i)
g
3400 m

where, θ is angle of projection.

30°
Putting all the given values in Eq. (i), we get
u 2 sin ( 2 × 15° )
⇒ R = 50 m =
O g
or AB = OA tan 30° ⇒ 50 × g = u sin 30° = u ×
2 2 1
1 3400 2
= 3400 × = m
3 1732
. ⇒ 50 × g × 2 = u 2

∴ Speed of the aircraft =

Distance travelled ⇒ u 2 = 50 × 9.8 × 2
Time taken = 100 × 9.8 = 980
3400 1 ⇒ u = 980 = 49 × 20
= × = 196.30 m/s
1732
. 10
= 7 × 2 × 5 ms −1
113 (d) Radius of the horizontal circle r
= 14 5 ms −1
= Length of the string = 80 cm = 0.80 m
14 −1 For θ = 45°,
Frequency of revolution, ν = s
25 u 2 sin ( 2 × 45° ) u 2
R= = (Q sin 90° = 1)
Angular speed of the revolution of the stone, g g
22 14 88
ω = 2πν = 2 × × = rad/s (14 5 )2 14 × 14 × 5
7 25 25 ⇒ R= = = 100 m
g 9.8
Centripetal acceleration of the stone, a = rω 2
total distance travelled
2 118 (d) We know that, speed, v0 =
 88 88 88
= 0.80 ×   = 0.80 × × = 9.91 m/s 2 time taken
 25 25 25 Hence, total distance travelled = path length = speed ×
time taken
114 (d) A scalar quantity is independent of direction and
Hints & Explanations

hence has the same value for observers with different If v0 is positive, so equal path lengths are traversed in
orientations of the axes. equal intervals.
Also, it is completely specified by a number only, so it As it is independent of displacement, so it will not give
can take negative and positive values both. any information regarding average velocity and
acceleration.