CLS Aipmt 17 18 XI Phy Study Package 1 SET 2 Chapter 4

Chapter 4
Motion in a Plane
Solutions
SECTION - A
Objective Type Questions
1. Which of the following is a vector?
(1) Current (2) Time (3) Acceleration (4) Volume
Sol. Answer (3)
Acceleration is a vector quantity.
2. The change in a vector may occur due to
(1) Rotation of frame of reference (2) Translation of frame of reference
(3) Rotation of vector (4) Both (1) & (3)
Sol. Answer (3)
Change in a vector may occur due to rotation of vector and not due to rotation of frame of reference.
3. Which one of the following pair cannot be the rectangular components of force vector of 10 N?
(1) 6 N & 8 N (2) 7 N & 51 N (3) 6 2 N & 2 7 N (4) 9 N & 1 N
Sol. Answer (4)
The vector magnitude = Ax 2  Ay 2
Vector magnitude = 10
But (4) option gives the magnitude
⇒ 92  12  82  10 [by trial method check options]
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164 Motion in a Plane Solution of Assignment (Set-2)
4. The resultant of two vectors at an angle 150° is 10 units and is perpendicular to one vector. The magnitude
of the smaller vector is
(1) 10 units (2) 10 3 units (3) 10 2 units (4) 5 3 units
Sol. Answer (2)
⇒ R 2  A2  B 2 .....(1)
R = 10
Perpendicular
Also tan 30º =
Base
1 R B
 R
3 A
30° 150°
From equation (1) A  10 3 A
102  10 3   B 2
2
B = 20
5. Two vectors, each of magnitude A have a resultant of same magnitude A. The angle between the two vectors
is
(1) 30° (2) 60° (3) 120° (4) 150°
Sol. Answer (3)

| A || B || R |
R = A2  B 2  2 AB cos 
A2 = A2 + A2 + 2A2cos
–A2 = 2A2cos
1
cos  =  ⇒   120º
2
 
6. Let  be the angle between vectors A and B . Which of the following figures correctly represents the angle
?
A B B
(1) (2) (3)  B (4)
B   A 
A A
Sol. Answer (3)

To find angle between vectors, they will be joined either head to head or tail to tail.
Solution of Assignment (Set-2) Motion in a Plane 165
7. A is a vector of magnitude 2.7 units due east. What is the magnitude and direction of vector 4 A ?
(1) 4 units due east (2) 4 units due west (3) 2.7 units due east (4) 10.8 units due east
Sol. Answer (4)

A  2.7 iˆ

Vector 4A
⇒ 4(2.7iˆ)  10.8iˆ or 10.8 units due east.
8. Two forces of magnitude 8 N and 15 N respectively act at a point. If the resultant force is 17 N, the angle
between the forces has to be
(1) 60° (2) 45° (3) 90° (4) 30°
Sol. Answer (3)
R= A2  B 2  2 AB cos 
A = 8, B = 15, R = 17
172 = 82 + 152 + 2 × 8 × 15 × cos 
289 = 64 + 225 + 240 cos 
⇒ 289 = 289 + 24cos 
24cos = 0
cos = 0 ⇒ = 90º
9. A particle is moving in a circle of radius r having centre at O, with a constant speed v. The magnitude of change
in velocity in moving from A to B is
v
B
v
60°
O A
(1) 2v (2) 0 (3) 3v (4) v
Sol. Answer (4)
  2  V  sin ⎛ 60º ⎞ 1
 V   2V sin ⎜ ⎟  2  V  ⇒ V | V |
2 ⎝ 2 ⎠ 2
10. Two forces of 10 N and 6 N act upon a body. The direction of the forces are unknown. The resultant force on
the body may be
(1) 15 N (2) 3 N (3) 17 N (4) 2 N
Sol. Answer (1)
The resultant of two vectors always lie between (A + B) & (A – B).
So the resultant of 10 N & 6 N should lie between 16 N & 4 N.
So answer is 15 N.
11. The vector OA where O is origin is given by OA  2iˆ  2 ˆj . Now it is rotated by 45° anticlockwise about O. What
will be the new vector?
(1) 2 2 ĵ (2) 2 ĵ (3) 2 iˆ (4) 2 2 iˆ
Sol. Answer (1)

OA  2iˆ  2 jˆ
P(2, 2)

OA  4  4 ⇒ 2 2
On rotating by an angle of 45º anticlockwise it will lie along y-axis.
45°
So A  2 2 jˆ
12. A car moves towards north at a speed of 54 km/h for 1 h. Then it moves eastward with same speed for same
duration. The average speed and velocity of car for complete journey is
15 54
(1) 54 km/h, 0 (2) 15 m/s, m/s (3) 0, 0 (4) 0, km/h
2 2
Sol. Answer (2)
54 Km
B N
–1
54 Kmh
d = 54 Km
t = 1h W E
S
A
54 2
Displacement 
Km
Distance = 2 × 54 = 108 Km
108 5
Average speed=  54 Kmh1   15 ms1
2 18
disp. 54 2 5 15
Average velocity   ⇒ 27 2  ⇒ m/s
time 2 18 2
13. If the sum of two unit vectors is also a unit vector, then magnitude of their difference and angle between the
two given unit vectors is
(1) 3 , 60 (2) 3 , 120  (3) 2, 60 (4) 2 , 120 

Sol. Answer (2)
2 2
R  A  B  A  B  2 AB cos 

A  B  R 1
1 = 1 + 1 + 2 × 1 × 1 × cos 
1
cos  =  ⇒   120º
2
2 2
R  A  B  A  B  2 AB cos120º
⎛ 1⎞
 12  12  2  1 1 ⎜  ⎟  3  A  B
⎝ 2⎠
14. A particle projected from origin moves in x-y plane with a velocity v  3iˆ  6 xˆj , where iˆ and ĵ are the unit
vectors along x and y axis. Find the equation of path followed by the particle
1 1
(1) y = x2 (2) y  (3) y = 2x2 (4) y 
x2 x
Sol. Answer (1)
Method 1: Method 2:

V  3iˆ  6 xjˆ Vx iˆ  Vy ˆj  V
dx dy ˆ
also V  iˆ  j Vx  3
dt dt
dx
 3, Vy  6 x
dt
∫ dx  ∫ 3dt We know
dy Vy
x = 3t  tan  
dx Vx
dy dy 6x
 6x 
dt dx 3x
dy  6 x  dt ∫ dy  ∫ 2xdx
0 0
∫ dy  ∫ 6  3 tdt y  x2
t2
 18 ∫ tdt ⇒ 18 
2
y  9t 2
x2
 9
9
y  x2
15. Ram moves in east direction at a speed of 6 m/s and Shyam moves 30° east of north at a speed of 6 m/s.
The magnitude of their relative velocity is
(1) 3 m/s (2) 6 m/s (3) 6 3 m/s (4) 6 2 m/s

Sol. Answer (2)
6 am
E
s –1
y
m
VShyam
Sh
30º 30º
60º 60º
VRam 6 ms–1 N
Ram
2 2 30º
VRS  VR  VS  2VRVS cos 
W E
1
 62  62  2  62 
2
S
= 6 ms–1
16. A train is running at a constant speed of 90 km/h on a straight track. A person standing at the top of a boggey
moves in the direction of motion of the train such that he covers 1 meters on the train each second. The speed
of the person with respect to ground is
(1) 25 m/s (2) 91 km/h (3) 26 km/h (4) 26 m/s
Sol. Answer (4)
5
VT = 90 Kmh–1 = 90   25 ms1
18
Vm = ?
d = speed × time
dnet = Vnet × t
1 = (Vm – 25) × 1
Vm = 26 ms–1
17 . Figure shows two ships moving in x-y plane with velocities VA and VB. The ships move such that B always remains
VA
north of A. The ratio V is equal to
B
N
y
VB
W E

S
B
A VA x
(1) cos (2) sin (3) sec (4) cosec

Sol. Answer (1)
If ship B is always north of ship A then, their horizontal component should be equal, so,
VA = VBcos 
VA
⇒  cos 
VB
18. Four persons P, Q, R and S are initially at the four corners of a square of side d. Each person now moves
with a constant speed v in such a way that P always moves directly towards Q, Q towards R, R towards S,
and S towards P. The four persons will meet after time
d d 3d
(1) (2) (3) (4) They will never meet
2v v 2v
Sol. Answer (2)
d P d S
T  v
v rel
v rel  v  v cos 90º v

=v–0
v
=v
Q v R
d
T 
v
19. A person, reaches a point directly opposite on the other bank of a flowing river, while swimming at a speed
of 5 m/s at an angle of 120° with the flow. The speed of the flow must be
(1) 2.5 m/s (2) 3 m/s (3) 4 m/s (4) 1.5 m/s
Sol. Answer (1)
For drift to be zero
u = v sin 30º v cos 30
v 30º
1 u
= 5 120º
2
v sin 30
= 2.5 ms–1
20. A body of mass 1 kg is projected from ground at an angle 30º with horizontal on a level ground at a speed 50 m/s.
The magnitude of change in momentum of the body during its flight is (g = 10 m/s2)
(1) 50 kg ms–1 (2) 100 kg ms–1 (3) 25 kg ms–1 (4) Zero
Sol. Answer (1)
⇒ The change in momentum = 2mu sin ˆj u sin  –1

u = 50 ms

p  2mu sin 
= 2 × 1 × 50 × sin 30º 
u cos 

p = 50 Kg ms–1
21. A car with a vertical windshield moves in a rain storm at a speed of 40 km/hr. The rain drops fall vertically
with constant speed of 20 m/s. The angle at which rain drops strike the windshield is
5 9 3 2
(1) tan–1 (2) tan–1 (3) tan–1 (4) tan–1
9 5 2 3
Sol. Answer (1)
20  5 Q
v 9 vm
tan   m 
vr 20

⎛ 5⎞ vr
  tan1 ⎜ ⎟
⎝ 9⎠
⎛ ⎞ ⎛ ⎞ 
22. Two projectiles are projected at angles ⎜   ⎟ and ⎜   ⎟ with the horizontal, where   , with same speed.
⎝ 4 ⎠ ⎝ 4 ⎠ 4
The ratio of horizontal ranges described by them is
(1) tan  : 1 (2) 1 : tan2  (3) 1 : 1 (4) 1 : 3
Sol. Answer (3)

The horizontal range is same when the angles of projection are complimentary to each other.
23. A shell is fired vertically upwards with a velocity v1 from a trolley moving horizontally with velocity v2. A person
on the ground observes the motion of the shell as a parabola, whose horizontal range is
2v 12v 2 2v 12 2v 22 2v 1v 2
(1) (2) (3) (4)
g g g g
Sol. Answer (4)

There is no acceleration in the horizontal direction.
1
Sx  U xT  a0  T 2
2
R  U xT .....(1)
V1
1
Sy  U y T  g y T 2
2
1 2
O  V1T  gT
2
V2
1
⇒ V1T  gT
2
2V1
T
g
We know,
(R) range = (Horizontal velocity 4x) × flight + time (T)
i.e., R = 4x × T
2V1 2V1V2
R  V2  ⇒
g g
24. The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given
by y = (4t – 2t2)m and x = (3t) metre, where t is in second and point of projection is taken as origin. The
angle of projection of projectile with vertical is
(1) 30° (2) 37° (3) 45° (4) 60°
Sol. Answer (2)
y = 4t – 2t2
x = 3t
V  Vxiˆ  Vyjˆ
dx dy
Vx  , Vy 
dt dt
Vx = 3, Vy = 4 – 4t
for t = 0, Vy = 4
Vy 4
tan   
Vx 3
 = 53º with horizontal
With vertical
 = 37º
25. A particle is projected from ground with speed 80 m/s at an angle 30° with horizontal from ground. The magnitude
of average velocity of particle in time interval t = 2 s to t = 6 s is [Take g = 10 m/s2]
(1) 40 2 m/s (2) 40 m/s (3) Zero (4) 40 3 m/s
Sol. Answer (4)

Average velocity of the projectile when it is at the same vertical height is : u cos.
⇒ 80 × cos 30º ⇒ 40 3 m/s .
h
t=2 t=6
26. A stone projected from ground with certain speed at an angle  with horizontal attains maximum height h1.
When it is projected with same speed at an angle  with vertical attains height h2. The horizontal range of
projectile is
h1  h2
(1) (2) 2h1h2 (3) 4 h1h2 (4) h1 + h2
2
Sol. Answer (3)
When the angles are complimentary the range is same,
u 2 sin2 
h1  ,
2g u u
u 2 sin2 (90  )
h2  h1 h2
2g  (90–)
R R
2 2
u sin 
h1 
2g
u 2 cos2 
h2 
2g
2
u 4 sin2  cos2  ⎛ 2u sin  cos  ⎞ 1 1
h1h2  ⇒ ⎜ ⎟  4g  4
4g 2 ⎝ g ⎠
1
h1h2  R 2 ⇒ R 2  16 h1h2
16
R  4( h1h2 )
27. Two objects are thrown up at angles of 45° and 60° respectively, with the horizontal. If both objects attain same
vertical height, then the ratio of magnitude of velocities with which these are projected is
5 3 2 3
(1) (2) (3) (4)
3 5 3 2
Sol. Answer (4)
h1 = h2
u12 sin2 45º V22 sin2 60º


2g 2g
3 3
u12 
2 2 3
 
V22 1 2
2
V1 3

V2 2
28. For an object projected from ground with speed u horizontal range is two times the maximum height attained
by it. The horizontal range of object is
2u 2 3u 2 3u 2 4u 2
(1) (2) (3) (4)
3g 4g 2g 5g
Sol. Answer (4)

H 1
R = 24 also,  tan  5
R 4 =
1
2 + 2
H 1 1 1 2
 ⇒  tan  
R 2 2 4
1
P
tan   2 
B
2u 2 sin  cos 
R
g
2u 2 2 1
R . 
g 5 5
4u 2
R
5g
3
29. The velocity at the maximum height of a projectile is times its initial velocity of projection (u). Its range
on the horizontal plane is 2
3u 2 3u 2 3u 2 u2
(1) (2) (3) (4)
2g 2g g 2g
Sol. Answer (1)
uh  u cos  uh
u cos 
3
u  u cos 
2
3
⇒ cos  
2
 = 30º
u 2 sin 2
R =
g
u 2 sin 60º 3u 2
= ⇒ R
g 2g
30. A projectile is thrown into space so as to have a maximum possible horizontal range of 400 metres. Taking
the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum
are
(1) (400, 100) (2) (200, 100) (3) (400, 200) (4) (200, 200)
Sol. Answer (2)
Rmax = 400 m
The velocity is minimum at the highest point
(200, 100)
R
⇒ H 200 N
2
400 m
R = 4H
400 = 4 × H
H = 100 m
31. If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is
1 ⎛ 2R ⎞
2
⎛ gT 2 ⎞ ⎛ 2R ⎞ 1 ⎛ 2R ⎞
(1) tan 1 ⎜⎜ ⎟
⎟ (2) tan ⎜⎜ ⎟
⎟ (3) tan 1 ⎜⎜ 2 ⎟⎟ (4) tan ⎜⎜ ⎟⎟
⎝ 2R ⎠ ⎝ gT ⎠ ⎝g T ⎠ ⎝ gT ⎠
Sol. Answer (1)
2u sin  gT
T  ⇒u 
g 2sin 
R
g
2u sin 
R  u cos 
g
R = T × u cos 
gT cos 
R T 
2sin 
gT 2 1
R
2 tan 
gT 2
tan  
2R
⎛ 2⎞
  tan1 ⎜ gT ⎟
⎝ 2R ⎠
32. In the graph shown in figure, which quantity associated with projectile motion is plotted along
y-axis? y-axis
x-axis
t
(1) Kinetic energy (2) Momentum (3) Horizontal velocity (4) None of these
Sol. Answer (3)
y-axis
It is the horizontal component of
velocity that remains constant
throughout the motion as there is no
acceleration in that direction ax = 0,
x-axis
ux = constant t
33. The equation of a projectile is y = ax – bx2. Its horizontal range is

a b
(1) (2) (3) a + b (4) b – a
b a
Sol. Answer (1)
y = ax – bx2
When the body lands then y = 0, x = R, 0 = aR – bR2
y=0
aR = bR R
a
R
b
u
34. Figure shows a projectile thrown with speed u = 20 m/s at an angle
30° with horizontal from the top of a building 40 m high. Then the 30°
horizontal range of projectile is
40 m
(1) 20 3 m (2) 40 3 m
(3) 40 m (4) 20 m
Sol. Answer (2)
1
Sy  uyT  g yT 2
2
1 2
–40 = 4 sin30T  gT uy = 4 sin 30º
2
u = 20 ms–1
1 2
–40 = 20  T  5T
2 30º
–8 = 2T – T2 ux = u cos 30º
T2 – 2T – 8 = 0
T2 – 4T + 2T – 8 = 0 40 m
T = –2, 4
3
R  u cos T = 20  4
2
R  40 3 m
35. When a particle is projected at some angle to the horizontal, it has a range R and time of flight t1. If the same
particle is projected with the same speed at some other angle to have the same range, its time of flight is
t2, then
2R R 2R R
(1) t1  t 2  (2) t1  t 2  (3) t1t 2  (4) t1t 2 
g g g g
Sol. Answer (3)
The angles has to be complimentary i.e., if 1  , 2  (90  )
2u sin  2u sin(90  )
t1  , t2 
g g
2u cos 
t2 
g
2u sin  2u cos 
t1t2  
g g
2R
t1t2 
g
36. A projectile is thrown with velocity v at an angle  with horizontal. When the projectile is at a height equal to
half of the maximum height, the vertical component of the velocity of projectile is
v sin v sin v sin

(1) v sin  × 3 (2) (3) (4)
3 2 3
Sol. Answer (3)
2g ⎛ u 2 sin  ⎞
v B2  v 2 sin2   ⎜ ⎟
2 ⎝ 2g ⎠
vB
v sin 
v 2 sin2  v
v B2 
2 
v sin 
vB  v cos 
2
37. In the given figure for a projectile

u
P

x1 x2
⎡ x1x 2 ⎤ ⎡ x1x 2 ⎤ ⎡ 2 x1x 2 ⎤ ⎡ 2 x1x 2 ⎤

(1) y  ⎢ x  x ⎥ tan θ (2) y  ⎢ ⎥ tan θ (3) y  ⎢ ⎥ cos θ (4) y  ⎢ x  x ⎥ tan θ
⎣ 1 2⎦ ⎣ x1  x 2 ⎦ ⎣ x1  x 2 ⎦ ⎣ 1 2⎦
Sol. Answer (2)

The equation of trajectory for point 'P' can be written as :
⎛ x⎞ ⎛ x1 ⎞ ⎛ x  x2  x1 ⎞ u P
y = x tan  ⎜ 1  ⎟ = x1 tan  ⎜ 1  ⎟ = x1 tan  ⎜ 1 ⎟
⎝ R⎠ ⎝ x1  x2 ⎠ ⎝ x1  x2 ⎠
 y
x1x2  P
y = x  x tan  x1 x2
1 2
38. Two paper screens A and B are separated by distance 100 m. A bullet penetrates A and B, at points P and
Q respectively, where Q is 10 cm below P. If bullet is travelling horizontally at the time of hitting A, the velocity
of bullet at A is nearly
(1) 100 m/s (2) 200 m/s (3) 600 m/s (4) 700 m/s
Sol. Answer (4)
10 cm ⇒ 10 × 10–2 m ⇒ 10–1 ⇒ 0.1 m A B
It is a case of horizontal projectile.
P
So, ax = 0, ux = 4, uy = 0, ay = –g
2H 10 cm
R = 100m, T  ⇒ Time of flight
g
Q
R = uxT 100 m
2  0.1 u 2
100 = u ⇒  100
100 10
1000
u  707 ms1
2
39. A car is going round a circle of radius R1 with constant speed. Another car is going round a circle of radius
R2 with constant speed. If both of them take same time to complete the circles, the ratio of their angular
speeds and linear speeds will be
R1 R1 R1 R1
(1) , (2) 1, 1 (3) 1, (4) ,1
R2 R2 R2 R2
Sol. Answer (3)

The angular speed is given by
2

T
1  T
 ⇒ 1  2
T  2 T1
if T1 = T2 ⇒ 1 = 2
So, ratio ⇒ 1 : 1
and linear speed v = R
V R
V1 R1

V2 R2
40. A body revolves with constant speed v in a circular path of radius r. The magnitude of its average acceleration
during motion between two points in diametrically opposite direction is
v2 2v 2 v2
(1) Zero (2) (3) (4)
r r 2r
Sol. Answer (3)
⎛⎞
2v sin ⎜ ⎟
aavg  ⎝2⎠
⎛ r ⎞
⎜v ⎟
⎝ ⎠

2v 2 sin ⎛⎜ ⎞⎟
aavg  ⎝ 2 ⎠
180º
r B A
Here,  =  rad

2v 2 sin ⎛⎜ ⎞⎟
aavg  ⎝2⎠
r 
2v 2
aavg 
r
41. An object of mass m moves with constant speed in a circular path of radius R under the action of a force of
constant magnitude F. The kinetic energy of object is
1 1
(1) FR (2) FR (3) 2FR (4) FR
2 4
Sol. Answer (1)
1 1F 1 F v2 1
KE = mv 2 = v2 = = FR
2 2a 2 ⎛ v 2 ⎞ 2
⎜ ⎟
⎝R ⎠
42. The angular speed of earth around its own axis is
   
(1) rad/s (2) rad/s (3) rad/s (4) rad/s
43200 3600 86400 1800
Sol. Answer (1)
2
Angular speed =
T
T  Time period of earth = 24 h
2 
=  rad s1
24  60  60 43200
43. A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of the particle is
(in m/s2)
(1) 2 (2) 82 (3) 42 (4) 22
Sol. Answer (3)
a = r2
25
a=  2  2 2
100
a = 42 m/s2
44. A particle is revolving in a circular path of radius 25 m with constant angular speed 12 rev/min. Then the angular
acceleration of particle is
(1) 22 rad/s2 (2) 42 rad/s2 (3) 2 rad/s2 (4) Zero
Sol. Answer (4)

Angular acceleration is the rate of change of angular speed or angular velocity if  remains constant then
0
45. Two particles are moving in circular paths of radii r1 and r2 with same angular speeds. Then the ratio of their
centripetal acceleration is
(1) 1 : 1 (2) r1 : r2 (3) r2 : r1 (4) r22 : r12

Sol. Answer (2)
Centripetal acceleration is given by
v2
a  r 2
r
For same ''
a1 r1
ac  r ⇒ 
a2 r2
46. A particle P is moving in a circle of radius r with uniform speed v. C is the centre of the circle and AB is
diameter. The angular velocity of P about A and C is in the ratio
(1) 4 : 1 (2) 2 : 1 (3) 1 : 2 (4) 1 : 1
Sol. Answer (3)
P
d
P /C 
dt
1 d /2 
P / A  B
2 dt C
1
P / A  P /C
2
P / A 1
  1: 2
P /C 2
47. A car is moving at a speed of 40 m/s on a circular track of radius 400 m. This speed is increasing at the
rate of 3 m/s2. The acceleration of car is
(1) 4 m/s2 (2) 7 m/s2 (3) 5 m/s2 (4) 3 m/s2
Sol. Answer (3)
v = 40 ms–1
r = 400 m
aT = 3 ms–2
V 2 40  40
ac    4 ms2
r 400
a  aC2  aT2
a= 42  32  5 ms2
a = 5 ms–2
48. Four particles A, B, C and D are moving with constant speed v each. At the instant shown relative velocity of A
with respect to B, C and D are in directions
B
C A
(1) (2) (3) (4)
Sol. Answer (1)

vC
vA
(i)
vA –vC vA – vC
v A  vC
v AC  v A  vC ⇒
vB

(ii) vA ⇒ v AB  v A  v B ⇒ v A  ( v A )
v B
vA
v A  ( vB )
vA
(iii) ⇒ vAD  vA  vD  vA  (vD )
vD
v A  vD
vA
⇒
v D
49. The ratio of angular speeds of minute hand and hour hand of a watch is
(1) 6 : 1 (2) 12 : 1 (3) 60 : 1 (4) 1 : 60
Sol. Answer (2)
mh = Angular speed of minute hand
hh = Angular speed of hour hand
2 2
mh =  rad s1
60m 60  60
2 2 1
hh = 12 h  12  60  60 rad s
2
mh 60  60  1  12

hh 2 1 1
12  60  60
mh : hh ⇒ 12 : 1
50. If  is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed,
then
(1)  = 90° (2) 0° <  < 90° (3) 90° <  < 180° (4) 0°    180°
Sol. Answer (3)
V
ac
aT
 between v & Q is
90º <  < 180º
51. If speed of an object revolving in a circular path is doubled and angular speed is reduced to half of original
value, then centripetal acceleration will become/remain
(1) Same (2) Double (3) Half (4) Quadruple
Sol. Answer (1)
ac = r2 = (r)()
ac = v
⎛⎞
ac = (2v) ⎜ ⎟  v   ac
⎝2⎠
52. An object is projected from ground with speed u at angle  with horizontal. the radius of curvature of its
trajectory at maximum height from ground is
u 2 sin 2 u 2 cos2  u 2 sin2  u 2 sin2 

(1) (2) (3) (4)
g g g 2g
Sol. Answer (2)
v2
ac 
r
u cos 
v 2 u 2 cos2 
r  , = 90º
ac g g
u 2 cos2 
r 
g
SECTION - B
Objective Type Questions
1. Two particles A and B start moving with velocities 20 m/s and 30 2 m/s along x-axis and at an angle 45º with
x-axis respectively in xy-plane from origin. The relative velocity of B w.r.t. A
(1) (10iˆ  30 ˆj ) m/s (2) (30 iˆ  10 jˆ) m/s
(3) (30 iˆ  20 2 ˆj )m/s (4) (30 2 iˆ  10 2 ˆj ) m/s
Sol. Answer (1)

vA = 20 m/s
v B  30 2 m/s along 45º with x-axis 

vB
v B  v B cos 45º iˆ  v B sin 45º jˆ  30iˆ  30 jˆ

v BA  v B  v A  30iˆ  30 jˆ  20iˆ
45º

v BA  10iˆ  30 jˆ
2. A particle is projected at angle  with horizontal from ground. The slop (m) of the trajectory of the particle
varies with time (t) as
m m m
m
(1) t (2)
t
(3) (4)
O t t
O O O
Sol. Answer (1)

Slope of trajectory
u sin   gt
tan  
u cos 
u sin  gt
So, m  
u cos  u cos 
g y
m  tan   t
u cos 
 y = a – bx
Therefore, x
m
O t
3. If H1 and H2 be the greatest heights of a projectile in two paths for a given value of range, then the horizontal
range of projectile is given by
H1  H2 H1  H2
(1) (2) (3) 4 H1 H2 (4) 4[H1  H2 ]
2 4
Sol. Answer (3)
1 + 2 = 90º
u
u 2 sin2 1
H1 
2g
u
2
u sin(90º 1 )
H2  2
2g 1
R2 u 2 sin2 1
H1H2  ∵ R
16 g
R  4 H1H2
4. If R and H are the horizontal range and maximum height attained by a projectile, than its speed of projection is
4R 2 R 2g 8H R2
(1) 2gR  (2) 2gH  (3) 2gH  (4) 2gH 
gH 8H Rg H
Sol. Answer (2)
u 2 sin2  2gH
H ⇒ sin  
2g u2
R
g
2u 2 2gH 2gH
R 2
 1
g u u2
2u 2 2gH u 2  2gH
R 
g u2 u2
gR
 u 2  2gH
2 2gH
Squaring both the sides,
gR 2
 u 2  2gH
4  2gH
9R 2
 u 2  2gH 
8H
gR 2
u  2gH 
8H
5. A particle projected from ground moves at angle 45º with horizontal one second after projection and speed is
minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]
(1) tan–1(3) (2) tan–1(2) (3) tan1( 2) (4) tan–1(4)
Sol. Answer (2)

 = 45º, t = 1 s
Vy u sin   gt
tan   
Uy u cos 
u sin   g  1
tan 45º  ⇒ u cos   u sin   g
u cos 
also, Vy = 0, after 1st (as speed is minimum)
u sin   g  2  0  u sin   2g ...(i)
so, u cos   2g  g
u cos   g ...(ii)
(i) u sin  2g
so,  
(ii) u cos  g
 tan   2
  tan 1(2)
6. A ball is projected from ground at an angle 45º with horizontal from distance d1 from the foot of a pole and just
after touching the top of pole it the falls on ground at distance d2 from pole on other side, the height of pole is
d1  d 2 2 d1 d 2 d1 d 2
(1) 2 d1d 2 (2) (3) d  d (4) d1  d 2
4 1 2
Sol. Answer (4)
º
45
=
 
d1 d2
tan   tan   tan 
y y
  tan 45º
d1 d 2
⎛ dd ⎞
y ⎜ 1 2 ⎟
⎝ d1  d 2 ⎠
7. A particle is projected with speed u at angle  with horizontal from ground. If it is at same height from ground
at time t1 and t2, then its average velocity in time interval t1 to t2 is
1
(1) Zero (2) u sin  (3) u cos  (4) u cos 
2
Sol. Answer (3)
When projectile is at same height, average velocity = u cos.
8. A particle is projected from ground at an angle  with horizontal with speed u. The ratio of radius of curvature
of its trajectory at point of projection to radius of curvature at maximum height is
1 1 1
(1) 2 (2) cos2  (3) (4)
sin  cos  sin3  cos3 
Sol. Answer (4)
At the point of projection
u2
rA 
g cos 
u H
2 2
u = cos
u cos 
rH 
g 
A
u2
rA g cos  1 r
 2 2
 3
 A
Ratio, rH u cos  cos  rH
g
9. An object of mass 10 kg is projected from ground with speed 40 m/s at an angle 60º with horizontal. The rate
of change of momentum of object one second after projection in SI unit is [Take g = 9.8 m/s2]
(1) 73 (2) 98 (3) 176 (4) 140
Sol. Answer (2)
p
Force = , force remains constant = mg
t
 10 × 9.8  98 N
At t = 1, particle is at its maximum height.
10. An object is projected from ground with speed 20 m/s at angle 30º with horizontal. Its centripetal acceleration
one second after the projection is [Take g = 10 m/s2]
(1) 10 m/s2 (2) Zero (3) 5 m/s2 (4) 12 m/s2
Sol. Answer (1)
v2
Centripetal acceleration =  g  10 ms2
r
11. A particle is moving on a circular path with constant speed v. It moves between two points A and B, which
subtends an angle 60º at the centre of circle. The magnitude of change in its velocity and change in magnitude
of its velocity during motion from A to B are respectively
(1) Zero, Zero (2) v, 0 (3) 0, v (4) 2v, v
Sol. Answer (2)

v  2v sin Change in magnitude of velocity = 0
2
⎛ 60 ⎞
 2v  sin ⎜ ⎟
⎝ 2⎠
| v |  v
12. A particle is moving with constant speed v in xy plane as shown in figure. The magnitude of its angular velocity
about point O is
y
(0,b) v
x
O (a,0)
v v vb v
(1) (2) (3) (4)
2
a b 2
b (a  b 2 )
2
a
Sol. Answer (3)

v cos
v sin   r  r
(0, b)  v v  r 
v sin   a 2  b 2  
v sin
v b b
 r
2 2 2 2
a b a b 
O
(a, 0)
vb a

(a 2  b 2 )
13. A particle is moving in xy-plane in a circular path with centre at origin. If at an instant the position of particle is
1 ˆ ˆ
given by (i  j ), then velocity of particle is along
2
1 ˆ ˆ 1 1 ˆ ˆ
(1) (i  j ) (2) ( jˆ  iˆ) (3) (i  j ) (4) Either (1) or (2)
2 2 2
Sol. Answer (4)
1 1 ˆ
r  iˆ  j
2 2

v  r  0 as velocity is always tangential to the path.
1
(v x iˆ  v y jˆ)  (iˆ  ˆj )  0
2
vx  vy  0 ⇒ v x  v y
or v y  v x
v
v  v x2  v y2  2v x  v  vx 
2
v
vy  
2
v v
or vx   , vy 
2 2
v ˆ v ˆ v ˆ v ˆ
So, possible value of v  v x iˆ  v y ˆj  i  j or i j
2 2 2 2
14. A particle is moving eastwards with a speed of 6 m/s. After 6 s, the particle is found to be moving with same speed
in a direction 60° north of east. The magnitude of average acceleration in this interval of time is
N 6 m/s
60°
W E
6 m/s
S
(1) 6 m/s2 (2) 3 m/s2 (3) 1 m/s2 (4) Zero

Sol. Answer (3)
6 ms–1 
( v )
 ⎛ 60 ⎞ aav 
| v |  2v sin  2v sin ⎜ ⎟  2v sin30º  v t
2 ⎝ 2⎠ 60º
6 m/s

| v |  6 ms1
t = 6 s
6
so, aav   1 ms2
6
15. What is the path followed by a moving body, on which a constant force acts in a direction other than initial velocity
(i.e. excluding parallel and antiparallel direction)?
(1) Straight line (2) Parabolic (3) Circular (4) Elliptical
Sol. Answer (2)
The path will be parabolic.
16. Two stones are thrown with same speed u at different angles from ground in air. If both stones have same range
and height attained by them are h1 and h2, then h1 + h2 is equal to
u2 u2 u2 u2
(1) (2) (3) (4)
g 2g 3g 4g
Sol. Answer (2)

If range is same then, one angle is  and other angle is (90 – )
u 2 sin2  u 2 sin2 (90   )

 h1  , h2 
2g 2g
u 2 sin2  u 2 cos2 
h1  , h2 
2g 2g
u 2 sin2  u 2 cos2  u 2
So, h 1  h 2 ⇒   (sin2   cos2 )
2g 2g 2g
u2
h1  h2 
2g
17. When a force F acts on a particle of mass m, the acceleration of particle becomes a. Now if two forces of
magnitude 3F and 4F acts on the particle simultaneously as shown in figure, then the acceleration of the
particle is
4F
90º
3F
m
(1) a (2) 2a (3) 5a (4) 8a
Sol. Answer (3)
F 4F
Fnet  32  42  5 F a
m
So, Fnet = ma
5F = ma
5F 90º
 a  3F
m m
a   5a
18. Consider the two statements related to circular motion in usual notations

A. In uniform circular motion  , v and a are always mutually perpendicular

B. In non-uniform circular motion,  , v and a are always mutually perpendicular
(1) Both A and B are true (2) Both A and B are false
(3) A is true but B is false (4) A is false but B is true
Sol. Answer (3)
v

a
Only first statement is correct.

  mutually perpendicular to v and a.
19. Which of the following quantities remains constant during uniform circular motion?
(1) Centripetal acceleration (2) Velocity
(3) Momentum (4) Speed
Sol. Answer (4)
Speed remains constant.
20. A projectile is projected with speed u at an angle  with the horizontal. The average velocity of the projectile
between the instants it crosses the same level is
(1) u cos  (2) u sin  (3) u cot  (4) u tan 
Sol. Answer (1)
21. A ball is thrown at an angle  with the horizontal. Its horizontal range is equal to its maximum height. This is
possible only when the value of tan  is
(1) 4 (2) 2 (3) 1 (4) 0.5
Sol. Answer (1)
H 1
 tan 
R 4
 H = R, given,
tan   4
   tan 1(4)
22. A ball is projected from a point O as shown in figure. It will strike the ground after (g = 10 m/s2)
10 m/s
30°
O
60 m
(1) 4 s (2) 3 s (3) 2 s (4) 5 s

Sol. Answer (1)
1 10 ms
–1
sy  u xT  ayT 2
2
1 2 
60  10 sin30º T  gT
2
60  5T  5T 2 60 m
T 2 T  2  0
T 4s
23. A particle is thrown with a velocity of u m/s. It passes A and B as shown in figure at time t1 = 1 s and t2 = 3 s. The
value of u is (g = 10 m/s2)
y
u
A B
30º
O x
(1) 20 m/s (2) 10 m/s (3) 40 m/s (4) 5 m/s
Sol. Answer (3)
2u sin 
t1  t2 
g
2u  sin30º
1 3 
10
20 × 2 = u
 u  40 ms1
24. Which one of the following statements is not true about the motion of a projectile?
(1) The time of flight of a projectile is proportional to the speed with which it is projected at a given angle of
projection
(2) The horizontal range of a projectile is proportional to the square root of the speed with which it is projected
(3) For a given speed of projection, the angle of projection for maximum range is 45°
(4) At maximum height, the acceleration due to gravity is perpendicular to the velocity of the projectile
Sol. Answer (2)
u 2 sin 2
R  R  u2
g
25. Out of the two cars A and B, car A is moving towards east with a velocity of 10 m/s whereas B is moving towards
north with a velocity 20 m/s, then velocity of A w.r.t. B is (nearly)
(1) 30 m/s (2) 10 m/s (3) 22 m/s (4) 42 m/s
Sol. Answer (3)

v AB  v A  v B
–1
vB = 20 ms
v AB  v A2  v B2
vA = 10 ms–1
| v AB |  102  202  100  400  500 22 ms 1
26. A projectile is thrown with speed 40 ms–1 at angle  from horizontal. It is found that projectile is at same height
at 1 s and 3 s. What is the angle of projection?
–1 ⎛ 1 ⎞ –1 ⎛ 1 ⎞
(1) tan ⎜
⎝ 2⎠
⎟ (2) tan ⎜
⎝ 3⎠
⎟ (3) tan
–1
 3 (4) tan
–1
 2
Sol. Answer (2)
vy
tan  
vx
2u sin 
Also, t1  t2 
g
2  40  sin 
4
10
1
sin   ⇒   30º
2
1
So, tan   tan30º ⇒
3
⎛ 1 ⎞
  tan1 ⎜
⎝ 3 ⎟⎠
27. A man moves in an open field such that after moving 10 m on a straight line, he makes a sharp turn of 60º to
his left. The total displacement just at the start of 8th turn is equal to
(1) 12 m (2) 15 m (3) 17.32 m (4) 14.14 m
Sol. Answer (3)
After 8 such turns object is at 'B'. 60º
60º
Displacement = AB 60º
Two vectors are at 60º B
60º 60º
1  10 3 m
102  102  2  102 
2 10 m
 60º A
17.32 m  AB
10 m
SECTION - C
Previous Years Questions
1. In the given figure, a = 15 m/s2
represents the total acceleration of a particle moving in the clockwise direction
in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is [NEET (Phase-2) 2016]
30°
R
a
O
(1) 4.5 m/s (2) 5.0 m/s (3) 5.7 m/s (4) 6.2 m/s
Sol. Answer (3)
v2 3 v2
a cos 30   15   v = 5.7 m/s
r 2 2.5
2. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle
between these vectors is [NEET-2016]
(1) 180° (2) 0° (3) 90° (4) 45°
Sol. Answer (3)

| A  B |  | A  B |  cos = 0   = 90°

3. A particle moves so that its position vector is given by r  cos t x  sin t y, where  is a constant. Which
of the following is true? [NEET-2016]

(1) Velocity is perpendicular to r and acceleration is directed away from the origin

(2) Velocity and acceleration both are perpendicular to r

(3) Velocity and acceleration both are parallel to r

(4) Velocity is perpendicular to r and acceleration is directed towards the origin
Sol. Answer (4)

r  cos t x  sin t y,
⇒ v.r  0

dr
v    sin t x   cos t y
dt
2
a   cos t x   sin t y   r
2 2
4. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What
is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10–4 J by the
end of the second revolution after the beginning of the motion? [NEET-2016]
(1) 0.2 m/s2 (2) 0.1 m/s2 (3) 0.15 m/s2 (4) 0.18 m/s2
Sol. Answer (2)
m = 0.01 kg, r = 6.4 cm
1
mv 2  8  104 J
2
16  104
v2   16  102
0.01
Speed v2 = 2ats
v2 16  102
v2 = 2at 4r  at    0.1 m/s2
8r 8  3.14  6.4  102

5. The position vector of a particle R as a function of time is given by R  4 sin(2t )iˆ  4 cos(2t ) ˆj , where R is
in meters, t is in seconds and iˆ and jˆ denote unit vectors along x-and y-directions, respectively. Which one
of the following statements is wrong for the motion of particle? [Re-AIPMT-2015]
(1) Path of the particle is a circle of radius 4 meter

(2) Acceleration vector is along R
v2
(3) Magnitude of acceleration vector is , where v is the velocity of particle
R
(4) Magnitude of the velocity of particle is 8 meter/second
Sol. Answer (4)
R  4 sin(2t )iˆ  4 cos(2t ) jˆ  xiˆ  yjˆ

Now, x2 + y2 = 42 which is equation of circle of radius R.
So, the motion is UCM with speed
V  8 2 m/s

6. Two particles A and B, move with constant velocities v 1 and v 2 . At the initial moment their position vectors

are r 1 and r 2 respectively. The condition for particles A and B for their collision is [Re-AIPMT-2015]

r1 − r 2 v 2 − v1
(1) r 1 − r 2 = v 1 − v 2 (2) = (3) r 1 ⋅ v 1 = r 2 ⋅ v 2 (4) r 1 × v 1 = r 2 × v 2
r1 − r 2 v 2 − v1
Sol. Answer (2)
For collision final positions should be equal

 r1  v1t  r2  v 2 t

 r1  r2  (v 2  v1 )t

| r1  r2 |
 r1  r2  (v 2  v1 ) | v  v |
2 1
7. A ship A is moving Westwards with a speed of 10 km h–1 and a ship B 100 km South of A, is moving
Northwards with a speed of 10 km h–1. The time after which the distance between them becomes shortest,
is [AIPMT-2015]
(1) 10 2 h (2) 0 h (3) 5 h (4) 5 2 h
Sol. Answer (3)
8. A projectile is fired from the surface of the earth with a velocity of 5 ms–1 and angle  with the horizontal. Another
projectile fired from another planet with a velocity of 3 ms–1 at the same angle follows a trajectory which is identical
with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet
is (in ms–2) is (Given g = 9.8 ms–2) [AIPMT-2014]
(1) 3.5 (2) 5.9 (3) 16.3 (4) 110.8
Sol. Answer (1)
Since trajectory is same, so range and maximum height both will be identical from earth and planet. So
equating maximum height (Answer can be obtained by equating range also)
ue2 sin2  u p sin 

2 2

2ge 2g p
2.5 9

9.8 g p  gp = 3.5 m/s
2
9. A particle is moving such that its position coordinates (x, y) are
(2 m, 3 m) at time t = 0,
(6 m, 7 m) at time t = 2 s and
(13 m, 14 m) at time t = 5 s.

Average velocity vector (v av ) from t = 0 to t = 5 s is [AIPMT-2014]
(1)
1
5

13iˆ  14 ˆj  (2)
7 ˆ ˆ
3
(i  j ) (3) 2(iˆ  ˆj ) (4)
11 ˆ ˆ
5
(i  j )
Sol. Answer (4)
10. The velocity of a projectile at the initial point A is (2iˆ  3 jˆ) m/s. Its velocity (in m/s) at point B is
[NEET-2013]
Y
B
A X
(1) 2iˆ  3 jˆ (2) 2iˆ  3 jˆ (3) 2iˆ  3 jˆ (4) 2iˆ  3 jˆ
Sol. Answer (2)

The change is only in the y-component
So, v f  2iˆ  3 ˆj ∵ ax  0
11. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile
is [AIPMT (Prelims)-2012]
⎛ 1⎞
(1)  = tan–1(2) (2)  = 45º (3)   tan–1 ⎜ ⎟ (4)  = tan–1(4)
⎝4⎠
Sol. Answer (4)
H=R
⎛ H 1 ⎞
tan = 4 ⎜⎝∵  tan ⎟
R 4 ⎠
  tan 1(4)

12. A particle has initial velocity  2i  3 j   
and acceleration 0.3i  0.2 j . The magnitude of velocity after
10 s will be [AIPMT (Prelims)-2012]
(1) 5 units (2) 9 units (3) 9 2 units (4) 5 2 units
Sol. Answer (4)

u  3iˆ  3 ˆj , a  0.3iˆ  0.2 ˆj
t = 10 s

v  u  at

v  2iˆ  3 ˆj  (0.3iˆ  0.2 ˆj )  10
 2iˆ  3 ˆj  3iˆ  2 ˆj

v  5iˆ  5 ˆj
v  52  52  50
v ⇒ 5 2 ms1
13. A particle moves in a circle of radius 5 cm with constant speed and time period 0.2  s. The acceleration of
the particle is [AIPMT (Prelims)-2011]
(1) 5 m/s2 (2) 15 m/s2 (3) 25 m/s2 (4) 36 m/s2
Sol. Answer (1)
r = 5 cm, v = ?, T = 0.2  s
2 20
T  ⇒   10 rad s1
 0.2
a  r  2  5  10 2  100
a  5 ms2
14. A body is moving with velocity 30 m/s towards east. After 10 s its velocity becomes 40 m/s towards north.
The average acceleration of the body is [AIPMT (Prelims)-2011]
(1) 5 m/s2 (2) 1 m/s2
(3) 7 m/s2 (4) 7 m/s2

Sol. Answer (1)

| v |
aav 
t
–1
40 ms
v  (v 2  v1 )  v 22  v12 (∵   90º )
 5 302  402  50 ms1 30 ms

–1
50
so, aav   5 ms 2  aav
10
15. A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2, the range of the
missile is [AIPMT (Prelims)-2011]
(1) 20 m (2) 40 m (3) 50 m (4) 60 m
Sol. Answer (2)
For maximum range  = 45º
v = 20 ms–1
u 2 20  20
R  [∵   45º ]
g 10
R  40 m
16. A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the
displacement of the mass from the origin is small, which graph correctly depicts the position of the particle
as a function of time? [AIPMT (Prelims)-2011]
v(x) m
(x)
0
x(t) x(t) x(t) x(t)
0 t 0 t
(1) 0 t (2) 0 t (3) (4)
Sol. Answer (2)
17. A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as
seen from the point of projection is [AIPMT (Mains)-2011]
⎛ 3⎞
(1) tan ⎜⎜ ⎟⎟ (2) 45°
⎝ 2 ⎠
1 1
(3) 60° (4) tan
2
Sol. Answer (4)
1 H
B
tan  
4 R
4H = R ...(i) H
H 2H 
In ABC, tan    45º
R R A
2 R C
2H 1 2
1
tan       tan1 ⎛⎜ ⎞⎟
4H 2 ⎝ 2⎠
18. Six vectors, â through fˆ have the magnitudes and directions indicated in the figure. Which of the following
statements is true? [AIPMT (Prelims)-2010]
b
a c
f
d
e

(1) b  c  f (2) d  c  f (3) d  e  f (4) b  e  f
Sol. Answer (3)
 f
d

e

d e  f
19. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
[AIPMT (Mains)-2010]
(1) 60º (2) 15º (3) 30º (4) 45º
Sol. Answer (1)
uH  u cos 
u 1
 u  cos   cos  
2 2
  60º
20. A particle moves in x-y plane according to rule x = a sin t and y = a cost. The particle follows
[AIPMT (Mains)-2010]
(1) An elliptical path
(2) A circular path
(3) A parabolic path
(4) A straight line path inclined equally to x and y-axes
Sol. Answer (2)
x  a sin t ⇒ x 2  a 2 sin2 t
y  a cos t ⇒ y 2  a2 cos2 t
x 2  y 2  a2 (sin2 t  cos2 t )
x 2  y 2  a 2  equation of circle.
21. A particle has initial velocity (3iˆ  4 ˆj ) and has acceleration (0.4iˆ  0.3 jˆ) . Its speed after 10 s is
[AIPMT (Prelims)-2010]
(1) 7 units (2) 7 2 units (3) 8.5 units (4) 10 units
Sol. Answer (2)
22. A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle
lands on the level ground the magnitude of the change in its momentum will be [AIPMT (Prelims)-2008]
mv
(1) Zero (2) 2 mv (3) (4) mv 2
2
Sol. Answer (4)
p  2mv sin ˆj

1
| p |  2mv sin   2mv 
2

| p |  2 mv
23. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later
time are ( 3,3) . The path of the particle makes with the x-axis an angle of [AIPMT (Prelims)-2007]
(1) 0° (2) 30° (3) 45° (4) 60°

Sol. Answer (4)
P ( 3, 3)
P 3
tan   
B 3

  60º
3

24. A and B are two vectors and  is the angle between them, if A  B  3( A  B ) , the value of  is
(1) 90° (2) 60° (3) 45° (4) 30°
Sol. Answer (2)
25. For angles of projection of a projectile at angles (45°– ) and (45° + ), the horizontal ranges described by
the projectile are in the ratio of [AIPMT (Prelims)-2006]
(1) 1 : 1 (2) 2 : 3 (3) 1 : 2 (4) 2 : 1
Sol. Answer (1)
26. A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The
average velocity and average speed for each circular lap respectively is [AIPMT (Prelims)-2006]
(1) 0, 0 (2) 0, 10 m/s
(3) 10 m/s, 10 m/s (4) 10 m/s, 0
Sol. Answer (2)
T = 62.8 s
r = 100 m
2
T 

2 2  3.14  10 m
  0
T 62.8  100 10
  0.1 rad s1

v = r
v = 100 × 0.1
v  10 ms1
Average velocity = 0
Average speed = 10 ms 1
27. The vectors A and B are such that: | A  B |  | A – B | . The angle between the two vectors is
(1) 90° (2) 60°
(3) 75° (4) 45°
Sol. Answer (1)

| AB|  | AB|
A2  B 2  2 AB cos   A2  B 2  2 AB cos 
Squaring both the sides,
A2  B 2  2 AB cos   A2  B 2  2 AB cos 
4 AB cos   0  cos   0    90º
28. If a vector 2iˆ  3 ˆj  8kˆ is perpendicular to the vector 4 jˆ  4iˆ  kˆ , then the value of  is
1
(1) –1 (2)
2
1
(3)  (4) 1
2
Sol. Answer (3)
29. A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the
stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?
2
(1) ms–2 and direction along the radius towards the centre
4
(2) 2 ms–2 and direction along the radius away from centre
(3) 2 ms–2 and direction along the radius towards the centre
(4) 2 ms–2 and direction along the tangent to the circle
Sol. Answer (3)
22  2
= ⇒  rad s1
44 s
Centripetal acceleration, a = r2
a  1   2 ms2 along the radius towards the centre
30. Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a
direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and
catches the other boy in a time t, where t is [AIPMT (Prelims)-2005]
a a2 a a
(1)
v 2
 v12
(2) 2
v  v 12
(3)
 v  v1  (4)  v  v1 
C
Sol. Answer (2)
The distance travelled by body at B, v v1
= Speed × t
= v1t A a B
C
So, BC = v1t, similarly, AC = vt
Applying pythagoras in ABC, vt v1t
2 2
v t  v12t 2 a 2
A a B
(v 2  v12 )t 2  a2
a2
t2 
v 2  v12
a
t
v  v12
2

31. If the angle between the vectors A and B is , the value of the product (B  A)  A is equal to
(1) BA2 cos (2) BA2 sin
(3) BA2 sin cos (4) Zero
Sol. Answer (4)
32. A boat is sent across a river with a velocity of 8 km/h. If the resultant velocity of the boat is
10 km/h, then velocity of the river is
(1) 8 km/h (2) 10 km/h (3) 12.8 km/h (4) 6 km/h
Sol. Answer (4)
v  
vr  v  u
vr  v 2  u2
vr = 10 kmh–1, v = 8 kmh–1
u=?
100  82  uR2
 uR2  36  uR  6 km/h

33. Which of the following is correct relation between an arbitrary vector A and null vector 0 ?

(1) A  0  A  0  A (2) A  0  A  0  A (3) A  0  A  0  0 (4) None of these
Sol. Answer (1)
Knowledge based.
34. An object is being thrown at a speed of 20 m/s in a direction 45° above the horizontal. The time taken by
the object to return to the same level is
(1) 20/g (2) 20 g (3) 20 2/g (4) 20 2g
Sol. Answer (3)

u = 20 ms–1
 = 45º
2u sin 
T 
g
2u 1 2
T    u
g 2 10
2  20 1 20 2
T  ⇒ T
g 2 g
35. A body is whirled in a horizontal circle of radius 20 cm. It has an angular velocity of 10 rad/s. What is its linear
velocity at any point on circular path?
(1) 20 m/s (2) 2 m/s (3) 10 m/s (4) 2 m/s

Sol. Answer (4)
v  r   20  10 2  10
v  2 ms1
36. Identify the vector quantity among the following.

(1) Distance (2) Angular momentum (3) Heat (4) Energy
Sol. Answer (2)
Angular momentum is an axial vector.
37. Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here.
The velocity of A to the left is 10 m/s. What is the velocity of B when angle  = 60°?
B
 A
(1) 10 m/s (2) 9.8 m/s (3) 5.8 m/s (4) 17.3 m/s
Sol. Answer (3)
vB
vB = cos30º
v A cos 60º  v B cos30º 30º vB = sin30º
1 3
10   vB  30º
2 2
vA = cos60º
10 60º
vB 
3 vA
vA = sin60º
38. The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest possible
path in 15 minute. The velocity of the river water (in km/h) is
(1) 3 (2) 1
(3) 4 (4) 5
Sol. Answer (1)
v = 5 kmh–1
d = 1.0 km
t = 15 min
39. Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are
such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to
the second car is
(1) r1 : r2 (2) m1 : m2 (3) 1 : 1 (4) m1 m2 : r1 r2
Sol. Answer (3)
If time is same then,
⎡ 2 ⎤
1 :  2 ⇒ 1 : 1 ⎢∵   T ⎥
⎣ ⎦
40. A person aiming to reach exactly opposite point on the bank of a stream is swimming with a speed of
0.5 m/s at an angle of 120° with the direction of flow of water. The speed of water in the stream is
(1) 0.25 m/s (2) 0.5 m/s
(3) 1.0 m/s (4) 0.433 m/s

Sol. Answer (1)
30º
120º
v sin30º u
v sin30º = u
1
0.5   u ⇒ u  0.25 ms 1
2
41. Two projectiles of same mass and with same velocity are thrown at an angle 60° and 30° with the horizontal,
then which will remain same
(1) Time of flight (2) Range of projectile
(3) Maximum height acquired (4) All of these
Sol. Answer (2)
Range is same for complimentary angles.
42. Two particles having mass M and m are moving in a circular path having radius R and r. If their time periods
are same, then the ratio of their angular velocities will be
r R R
(1) (2) (3) 1 (4)
R r r
Sol. Answer (3)
43. If | A  B |  | A |  | B | then angle between A and B will be
(1) 90° (2) 120°

(3) 0° (4) 60°
Sol. Answer (2)

| AB|  | A|  |B|

| AB|  A2  B 2  2AB cos 
A2  A2  A2  2 A2 cos 
1
  cos  ⇒   120º
2
⎛ 20 ⎞
44. A particle moves along a circle of radius ⎜ ⎟ m with constant tangential acceleration. If the velocity of the
⎝  ⎠
particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is
(1) 40 m/s2 (2) 640 m/s2
(3) 160 m/s2 (4) 40 m/s2
Sol. Answer (1)
20
r  m

aT  constant
v = 80 ms–1,  = 4 rad
v = r
20
80    ⇒   4 rad s1

 = 0,
2  02  2
4  4  2    4
   2 rad s2
20
a  r   2

a  40 ms2
45. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
(1) Are equal to each other (2) Are equal to each other in magnitude
(3) Are not equal to each other in magnitude (4) Cannot be predicted
Sol. Answer (2)

( A  B)  ( A  B)  0
A2  B 2  AB  BA  0
A2  B 2  A  B

so, | A |  | B |
46. A wheel has angular acceleration of 3.0 rad/s2 and an initial angular speed of 2.00 rad/s. In a time of 2 s it
has rotated through an angle (in radian) of
(1) 10 (2) 12
(3) 4 (4) 6
Sol. Answer (1)

 = 3 rad s–2
0 = 2 rad s–1
t=2s
 = 0 + t
=2+3×2
  8 rad s1
2  02  2  
64  4  2  3  
60
  ⇒   10 rad s 1
6
47. A particle is moving such that its position coordinates (x, y) are
(2 m, 3 m) at time t = 0,
(6 m, 7 m) at time t = 2 s and
(13 m, 14 m) at time t = 5 s

Average velocity vector (Vav ) from t = 0 to t = 5 s is
1 7 ˆ ˆ 11 ˆ ˆ
(1) (13iˆ  14 ˆj ) (2) (i  j ) (3) 2(iˆ  ˆj ) (4) (i  j )
5 3 5
Sol. Answer (4)

13 – 2 iˆ  14 – 3 ˆj  11 iˆ  ˆj
Vav 
rf – ri
t
=
 5 – 0 5
 
SECTION - D
Assertion - Reason Type Questions

1. A : If A  B , then | A  B |  | A  B | .

R : If A  B , then ( A  B ) is perpendicular to A  B .
Sol. Answer (3)

2. A : The addition of two vectors P and Q is commutative.

R : By triangle law of vector addition we can prove P  Q  Q  P .
Sol. Answer (1)

3. A : A vector cannot be divided by other vector.
R : A vector can be divided by a scalar.
Sol. Answer (2)
4. A : At the highest point the velocity of projectile is zero.
R : At maximum height projectile comes to rest.
Sol. Answer (4)
5. A : Horizontal range of a projectile is always same for angle of projection  with horizontal or  with vertical.
R : Horizontal range depends only on angle of projection.
Sol. Answer (4)
6. A : Horizontal motion of projectile without effect of air is uniform motion.
R : Without air effect the horizontal acceleration of projectile is zero.
Sol. Answer (1)
7. A : Path of a projectile with respect to another projectile is straight line.

R : Acceleration of a projectile with respect to another projectile is zero.
Sol. Answer (1)
8. A : In the case of ground to ground projection of a projectile from ground the angle of projection with horizontal is
 = 30º. There is no point on its path such that instantaneous velocity is normal to the initial velocity.
R : Maximum deviation of the projectile is 2 = 60º.
Sol. Answer (1)
9. A : Three vectors having magnitudes 10, 10 and 25 cannot produce zero resultant.
R : If three vectors are producing zero resultant, then sum of magnitude of any two is more than or equal to
magnitude of third and difference is less than or equal to the magnitude of third.
Sol. Answer (1)
10. A : Uniform circular motion is accelerated motion still speed remains unchanged.
R : Instantaneous velocity is always normal to instantaneous acceleration in uniform circular motion.
Sol. Answer (1)
11. A : When a body moves on a curved path with increasing speed, then angle between instantaneous velocity
and acceleration is acute angle.
R : When the speed is increasing, its tangential acceleration is in the direction of instantaneous velocity.
Sol. Answer (1)
12. A : A uniform circular motion have non uniform acceleration.
R : The direction of acceleration of a particle in uniform circular motion changes continuously.
Sol. Answer (1)
13. A : Angular displacement is vector quantity only for small values.
R : The direction of angular displacement is perpendicular to plane of rotation of object.
Sol. Answer (2)


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CLS Aipmt 17 18 XI Phy Study Package 1 SET 2 Chapter 4

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Chapter 4

1. Which of the following is a vector?

(1) Current (2) Time (3) Acceleration (4) Volume

Sol. Answer (3)

Acceleration is a vector quantity.

2. The change in a vector may occur due to

(1) Rotation of frame of reference (2) Translation of frame of reference

(3) Rotation of vector (4) Both (1) & (3)

Sol. Answer (3)

(1) 6 N & 8 N (2) 7 N & 51 N (3) 6 2 N & 2 7 N (4) 9 N & 1 N

Sol. Answer (4)

The vector magnitude = Ax 2  Ay 2

But (4) option gives the magnitude

⇒ 92  12  82  10 [by trial method check options]

(1) 10 units (2) 10 3 units (3) 10 2 units (4) 5 3 units

Sol. Answer (2)

Sol. Answer (3)

⇒ 4(2.7iˆ)  10.8iˆ or 10.8 units due east.

cos = 0 ⇒ = 90º

(1) 2v (2) 0 (3) 3v (4) v

Sol. Answer (4)

(1) 2 2 ĵ (2) 2 ĵ (3) 2 iˆ (4) 2 2 iˆ

Sol. Answer (1)

(1) 3 , 60 (2) 3 , 120  (3) 2, 60 (4) 2 , 120 

(1) 3 m/s (2) 6 m/s (3) 6 3 m/s (4) 6 2 m/s

(1) cos (2) sin (3) sec (4) cosec

v rel  v  v cos 90º v

⇒ The change in momentum = 2mu sin ˆj u sin  –1

Sol. Answer (1)

(1) tan  : 1 (2) 1 : tan2  (3) 1 : 1 (4) 1 : 3

Sol. Answer (3)

Sol. Answer (4)

(R) range = (Horizontal velocity 4x) × flight + time (T)

(1) 30° (2) 37° (3) 45° (4) 60°

Sol. Answer (2)

(1) 40 2 m/s (2) 40 m/s (3) Zero (4) 40 3 m/s

Sol. Answer (4)

⇒ 80 × cos 30º ⇒ 40 3 m/s .

u12 sin2 45º V22 sin2 60º

Sol. Answer (4)

Sol. Answer (1)

Sol. Answer (1)

33. The equation of a projectile is y = ax – bx2. Its horizontal range is

v sin v sin v sin

37. In the given figure for a projectile

⎡ x1x 2 ⎤ ⎡ x1x 2 ⎤ ⎡ 2 x1x 2 ⎤ ⎡ 2 x1x 2 ⎤

Sol. Answer (2)

Sol. Answer (3)

42. The angular speed of earth around its own axis is

(1) 1 : 1 (2) r1 : r2 (3) r2 : r1 (4) r22 : r12

(1) (2) (3) (4)

Sol. Answer (1)

u 2 sin 2 u 2 cos2  u 2 sin2  u 2 sin2 

Sol. Answer (2)

(1) (10iˆ  30 ˆj ) m/s (2) (30 iˆ  10 jˆ) m/s

(3) (30 iˆ  20 2 ˆj )m/s (4) (30 2 iˆ  10 2 ˆj ) m/s

Sol. Answer (1)

v B  30 2 m/s along 45º with x-axis 

Sol. Answer (1)