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Chapter

4

Motion in a Plane
Solutions
SECTION - A
Objective Type Questions
1
...
Answer (3)
Acceleration is a vector quantity
...


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
...

3
...
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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Motion in a Plane

Solution of Assignment (Set-2)

The resultant of two vectors at an angle 150° is 10 units and is perpendicular to one vector
...
Answer (2)

⇒ R 2  A2  B 2


...


Two vectors, each of magnitude A have a resultant of same magnitude A
...
Answer (3)




| A || B || R |
R =

A2  B 2  2 AB cos 

A2 = A2 + A2 + 2A2cos
–A2 = 2A2cos
cos  = 

6
...
Which of the following figures correctly represents the angle
?

B

A
(1)

B



(2)

B



A

(3)



B
A

(4)


A

Sol
...

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Motion in a Plane

83

A is a vector of magnitude 2
...
What is the magnitude and direction of vector 4 A ?

(1) 4 units due east

(2) 4 units due west

(3) 2
...
8 units due east

Sol
...
7 iˆ


Vector 4A
⇒ 4(2
...
8iˆ or 10
...

8
...
If the resultant force is 17 N, the angle
between the forces has to be
(1) 60°

(2) 45°

(3) 90°

(4) 30°

Sol
...


A particle is moving in a circle of radius r having centre at O, with a constant speed v
...
Answer (4)



 V   2V sin


  2  V  sin ⎛ 60º ⎞
1
⎜
⎟  2  V  ⇒ V | V |
⎝ 2 ⎠
2
2

10
...
The direction of the forces are unknown
...
Answer (1)
The resultant of two vectors always lie between (A + B) & (A – B)
...

So answer is 15 N
...
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84

Motion in a Plane

Solution of Assignment (Set-2)

11
...
Now it is rotated by 45° anticlockwise about O
...
Answer (1)

OA  2iˆ  2 jˆ

OA  4  4 ⇒ 2 2

(4) 2 2 iˆ

P(2, 2)

On rotating by an angle of 45º anticlockwise it will lie along y-axis
...
A car moves towards north at a speed of 54 km/h for 1 h
...
The average speed and velocity of car for complete journey is
(1) 54 km/h, 0

(2) 15 m/s,

15
2

m/s

(3) 0, 0

(4) 0,

54
2

km/h

Sol
...
54 2
5
15

⇒ 27 2 
⇒
m/s
time
2
18
2

13
...
Answer (2)


 
2
2
R  A  B  A  B  2 AB cos 



A  B  R 1
1 = 1 + 1 + 2 × 1 × 1 × cos 
cos  = 

1
⇒   120º
2


 
2
2
R  A  B  A  B  2 AB cos120º
 
⎛ 1⎞
 12  12  2  1 1 ⎜  ⎟  3  A  B
⎝ 2⎠
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A particle projected from origin moves in x-y plane with a velocity v  3iˆ  6 xˆj , where iˆ and ˆj are the unit
vectors along x and y axis
...
Answer (1)
Method 2:

Method 1:

V  3iˆ  6 xjˆ


Vx iˆ  Vy ˆj  V

 dx
dy ˆ
also V 
iˆ 
j
dt
dt


Vx  3

dx
3,
dt

Vy  6 x

∫ dx  ∫ 3dt

We know
dy

x = 3t

dx

dy

dy
 6x
dt

dx

 tan  



Vy
Vx

6x
3x

dy  6 x  dt

∫ dy  ∫ 2xdx

∫ dy  ∫ 6  3 tdt

y  x2

 18 ∫ tdt ⇒ 18 

0

0

t2
2

y  9t 2
 9

x2
9

y  x2
15
...

The magnitude of their relative velocity is
(1) 3 m/s

(4) 6 2 m/s

(3) 6 3 m/s

(2) 6 m/s

E

VShyam

30º
60º

VRam

Sh
y
6 am
m
s –1

Sol
...
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86

Motion in a Plane

Solution of Assignment (Set-2)

16
...
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
...
Answer (4)
VT = 90 Kmh–1 = 90 

5
 25 ms1
18

Vm = ?
d = speed × time
dnet = Vnet × t
1 = (Vm – 25) × 1
Vm = 26 ms–1
17
...
The ships move such that B always remains

VA
north of A
...
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
...
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
...
Answer (2)
T 

d
v rel

v rel  v  v cos 90º

P

d
v

v

=v–0
=v
d
T 
v

S

v
Q

v

R

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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
...
5 m/s

(2) 3 m/s

(3) 4 m/s

(4) 1
...
Answer (1)
For drift to be zero
u = v sin 30º
= 5

v

1
2

30º

v cos 30
120º

u

v sin 30

= 2
...
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
...
Answer (1)

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

u sin 

u = 50 ms

–1


p  2mu sin 


= 2 × 1 × 50 × sin 30º

u cos 


p = 50 Kg ms–1

21
...
The rain drops fall vertically
with constant speed of 20 m/s
...
Answer (1)

v
tan   m 
vr

20  5
9
20

Q

vm


⎛ 5⎞
  tan1 ⎜ ⎟
⎝ 9⎠

vr

⎛
⎞
⎛
⎞

22
...

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
...

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A shell is fired vertically upwards with a velocity v1 from a trolley moving horizontally with velocity v2
...
Answer (4)
There is no acceleration in the horizontal direction
...
(1)
V1

1
Sy  U y T  g y T 2
2

O  V1T 

1 2
gT
2

V2

1
⇒ V1T  gT
2

T

2V1
g

We know,
(R) range = (Horizontal velocity 4x) × flight + time (T)
i
...
, R = 4x × T

R  V2 

2V1
2V1V2
⇒
g
g

24
...
The
angle of projection of projectile with vertical is
(1) 30°

(2) 37°

(3) 45°

(4) 60°

Sol
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Solution of Assignment (Set-2)

Motion in a Plane

89

for t = 0, Vy = 4

Vy

tan  



Vx

4
3

 = 53º with horizontal
With vertical
 = 37º
25
...
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
...


⇒ 80 × cos 30º ⇒ 40 3 m/s
...
A stone projected from ground with certain speed at an angle  with horizontal attains maximum height h1
...
The horizontal range of
projectile is
(1)

h1  h2
2

(3) 4 h1h2

(2) 2h1h2

(4) h1 + h2

Sol
...
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90

Motion in a Plane

Solution of Assignment (Set-2)

27
...
If both objects attain same
vertical height, then the ratio of magnitude of velocities with which these are projected is
(1)

5
3

(2)

3
5

2
3

(3)

(4)

3
2

Sol
...
For an object projected from ground with speed u horizontal range is two times the maximum height attained
by it
...
Answer (4)
R = 24 also,

H 1
 tan 
R 4
2

H 1
1 1
 ⇒  tan 
R 2
2 4

tan   2 

2

2u 2 sin  cos 
g

R

2u 2 2
1

...
The velocity at the maximum height of a projectile is
on the horizontal plane is
(1)

3u 2
2g

1

5

2


1

P
B

R

+

=

(2)

3u 2
2g

3
times its initial velocity of projection (u)
...
Answer (1)
uh  u cos 

uh

u cos 

3
u  u cos 
2
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A projectile is thrown into space so as to have a maximum possible horizontal range of 400 metres
...
Answer (2)
Rmax = 400 m
The velocity is minimum at the highest point

⇒ H

R
2

(200, 100)
200 N
400 m

R = 4H
400 = 4 × H
H = 100 m
31
...
Answer (1)

T 

2u sin 
gT
⇒u 
g
2sin 

R

2u 2 sin  cos 
g

R

2u sin 
 u cos 
g

R = T × u cos 
R T 

R

gT cos 
2sin 

gT 2 1
2 tan 

tan  

gT 2
2R

2⎞
⎛
  tan1 ⎜ gT ⎟
⎝ 2R ⎠

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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

Sol
...
The equation of a projectile is y = ax – bx2
...
Answer (1)
y = ax – bx2
When the body lands then y = 0, x = R, 0 = aR – bR2
aR = bR
R

y=0

R

a
b

u
30°

34
...
Then the
horizontal range of projectile is
(1) 20 3 m

(2) 40 3 m

(3) 40 m

(4) 20 m

40 m

Sol
...
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Solution of Assignment (Set-2)

Motion in a Plane

93

35
...
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
(1) t1  t 2 

2R
g

(2) t1  t 2 

R
g

(3) t1t 2 

2R
g

(4) t1t 2 

R
g

Sol
...
e
...
A projectile is thrown with velocity v at an angle  with horizontal
...
Answer (3)
2g ⎛ u 2 sin  ⎞
⎜
⎟
2 ⎝ 2g ⎠

v sin 

v B2  v 2 sin2  

v 2 sin2 
v B2 
2
vB 

vB
v


v sin 

v cos 

2

37
...
Answer (2)
The equation of trajectory for point 'P' can be written as :
x1 ⎞
x⎞
⎛
⎛
⎛ x  x2  x1 ⎞
= x1 tan  ⎜ 1
y = x tan  ⎜ 1  ⎟ = x1 tan  ⎜ 1 
⎟
⎟
x1  x2 ⎠
⎝ R⎠
⎝
⎝ x1  x2 ⎠

x1x2
y = x  x tan 
1
2

u


P

x1

y

P
x2

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Two paper screens A and B are separated by distance 100 m
...
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
...
1 m
It is a case of horizontal projectile
...
1
u 2
⇒
 100
100
10

1000
 707 ms1
2

39
...
Another car is going round a circle of radius
R2 with constant speed
...
Answer (3)
The angular speed is given by


2
T




T
1
⇒ 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
...
The magnitude of its average acceleration
during motion between two points in diametrically opposite direction is
(1) Zero

(2)

v2
r

(3)

2v 2
r

(4)

v2
2r

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Answer (3)

aavg

aavg

⎛⎞
2v sin ⎜ ⎟
⎝2⎠

⎛ r ⎞
⎜v ⎟
⎝ ⎠

2v 2 sin ⎛⎜ ⎞⎟
2
⎝
⎠

r

180º

B

A

Here,  =  rad

aavg


2v 2 sin ⎛⎜ ⎞⎟
⎝2⎠

r 
2v 2
r

aavg 

41
...
The kinetic energy of object is
(1)

1
FR
2

(2) FR

(3) 2FR

(4)

1
FR
4

(4)


rad/s
1800

Sol
...
The angular speed of earth around its own axis is
(1)


rad/s
43200

(2)


rad/s
3600

(3)


rad/s
86400

Sol
...
A particle moves in a circle of radius 25 cm at two revolutions per second
...
Answer (3)
a = r2
a=

25
 2  2 2
100

a = 42 m/s2
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A particle is revolving in a circular path of radius 25 m with constant angular speed 12 rev/min
...
Answer (4)



Angular acceleration is the rate of change of angular speed or angular velocity if  remains constant then
0

45
...
Then the ratio of their
centripetal acceleration is
(1) 1 : 1

(2) r1 : r2

(4) r22 : r12

(3) r2 : r1

Sol
...
A particle P is moving in a circle of radius r with uniform speed v
...
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
...
A car is moving at a speed of 40 m/s on a circular track of radius 400 m
...
The acceleration of car is
(1) 4 m/s2

(2) 7 m/s2

(3) 5 m/s2

(4) 3 m/s2

Sol
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Solution of Assignment (Set-2)

Motion in a Plane

97

48
...
At the instant shown relative velocity of A
with respect to B, C and D are in directions
B

A

C

D
(1)

(2)

(3)

(4)

Sol
...
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
...
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98

Motion in a Plane

mh =

Solution of Assignment (Set-2)

2
2

rad s1
60m 60  60

2
2
1
hh = 12 h  12  60  60 rad s

mh

hh

2
60  60  1  12
2
1 1
12  60  60

mh : hh ⇒ 12 : 1
50
...
Answer (3)
V

ac
a
aT

 between v & Q is
90º <  < 180º
51
...
Answer (1)
ac = r2 = (r)()
ac = v

⎛⎞
ac = (2v) ⎜ ⎟  v   ac
⎝2⎠
52
...
the radius of curvature of its
trajectory at maximum height from ground is
u 2 sin 2
g

(1)

(2)

u 2 cos2 
g

(3)

u 2 sin2 
g

(4)

u 2 sin2 
2g

Sol
...
Ltd
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Solution of Assignment (Set-2)

Motion in a Plane

99

SECTION - B
Objective Type Questions
1
...
The relative velocity of B w
...
t
...
Answer (1)
vA = 20 m/s

v B  30 2 m/s along 45º with x-axis

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


vB

45º


v BA  10iˆ  30 jˆ

2
...
The slop (m) of the trajectory of the particle
varies with time (t) as
m

m

m

(1)

t

O

(2)

t

O

(3)
O

m

t

(4)
O

t

Sol
...
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100
3
...
Answer (3)
1 + 2 = 90º

u

u 2 sin2 1
H1 
2g
H2 

u

2

u sin(90º 1 )
2g

H1H2 

1

R2
16

∵ R

2

u 2 sin2 1
g

R  4 H1H2
4
...
Answer (2)
H

u 2 sin2 
⇒ sin  
2g

R

2u 2 sin  cos 
g

R

2u 2
g

2gH

R

2u 2
g

2gH

gR
2 2gH

u

2

u2

 1



2gH
u2

2gH
u2

u 2  2gH
u2

 u 2  2gH

Squaring both the sides,
gR 2
 u 2  2gH
4  2gH

 u 2  2gH 

9R 2
8H

u  2gH 

gR 2
8H

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Motion in a Plane

101

A particle projected from ground moves at angle 45º with horizontal one second after projection and speed is
minimum two seconds after the projection
...
Answer (2)
 = 45º, t = 1 s
tan  

Vy
Uy



u sin   gt
u cos 

u sin   g  1
⇒ u cos   u sin   g
u cos 
also, Vy = 0, after 1st (as speed is minimum)
tan 45º 

u sin   g  2  0  u sin   2g


...
(ii)

(i) u sin  2g


(ii) u cos 
g

 tan   2
  tan 1(2)

6
...
Answer (4)

º
45
=


d1


d2

tan   tan   tan 
y
y

 tan 45º
d1 d 2

⎛ dd ⎞
y ⎜ 1 2 ⎟
⎝ d1  d 2 ⎠

7
...
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) Zero

(2) u sin 

(3) u cos 

(4)

1
u cos 
2

Sol
...

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Motion in a Plane

Solution of Assignment (Set-2)

A particle is projected from ground at an angle  with horizontal with speed u
...
Answer (4)
At the point of projection
rA 

u2
g cos 
2

u

2

u cos 
rH 
g

A

H

u = cos



u2
rA
r
1
g cos 
 2

 A
2
3
Ratio, rH
cos  rH
u cos 
g
9
...
The rate
of change of momentum of object one second after projection in SI unit is [Take g = 9
...
Answer (2)
Force =

p
, force remains constant = mg
t

 10 × 9
...

10
...
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
...
A particle is moving on a circular path with constant speed v
...
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
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Solution of Assignment (Set-2)

Motion in a Plane

103

12
...
The magnitude of its angular velocity
about point O is
y

v

(0,b)

v

(1)

2

a b

(2)

2

x

(a,0)

O
v
b

(3)

vb
(a  b 2 )

Sol
...
A particle is moving in xy-plane in a circular path with centre at origin
...
Answer (4)

 1
1 ˆ
r  iˆ 
j
2
2
 
v  r  0 as velocity is always tangential to the path
...
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104

Motion in a Plane

Solution of Assignment (Set-2)

14
...
After 6 s, the particle is found to be moving with same speed
in a direction 60° north of east
...
Answer (3)



⎛ 60 ⎞
| v |  2v sin  2v sin ⎜ ⎟  2v sin30º  v
⎝ 2⎠
2

6 ms–1
60º
6 m/s


| v |  6 ms1

aav 


( v )
t

t = 6 s
so, aav 

6
 1 ms2
6

15
...
e
...
Answer (2)
The path will be parabolic
...
Two stones are thrown with same speed u at different angles from ground in air
...
Answer (2)
If range is same then, one angle is  and other angle is (90 – )
 h1 

h1 

u 2 sin2 
u 2 sin2 (90   )
, h2 
2g
2g
u 2 sin2 
u 2 cos2 
, h2 
2g
2g

So, h 1  h 2 ⇒

h1  h2 

u 2 sin2  u 2 cos2  u 2


(sin2   cos2 )
2g
2g
2g

u2
2g

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When a force F acts on a particle of mass m, the acceleration of particle becomes a
...
Answer (3)
F
a
m

Fnet  32  42  5 F

4F

So, Fnet = ma
5F = ma
 a 

90º

5F
m

m

3F

a   5a
18
...
In uniform circular motion  , v and a are always mutually perpendicular
 

B
...
Answer (3)

v

a



Only first statement is correct
...

19
...
Answer (4)
Speed remains constant
...
A projectile is projected with speed u at an angle  with the horizontal
...
Answer (1)
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A ball is thrown at an angle  with the horizontal
...
This is
possible only when the value of tan  is
(1) 4

(2) 2

(3) 1

(4) 0
...
Answer (1)

H 1
 tan 
R 4
 H = R, given,

tan   4


  tan 1(4)

22
...
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
...
A particle is thrown with a velocity of u m/s
...
The
value of u is (g = 10 m/s2)
y

u
30º
O

(1) 20 m/s

(2) 10 m/s

A

B
x

(3) 40 m/s

(4) 5 m/s

Sol
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Solution of Assignment (Set-2)

Motion in a Plane

107

20 × 2 = u


u  40 ms1

24
...
Answer (2)
R

u 2 sin 2
 R  u2
g

25
...
r
...
B is (nearly)
(1) 30 m/s

(2) 10 m/s

(3) 22 m/s

(4) 42 m/s

Sol
...
A projectile is thrown with speed 40 ms–1 at angle  from horizontal
...
What is the angle of projection?
–1 ⎛ 1 ⎞
⎟
(2) tan ⎜
⎝ 3⎠

–1 ⎛ 1 ⎞
⎟
(1) tan ⎜
⎝ 2⎠

–1
(3) tan

 3

–1
(4) tan

 2

Sol
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108

Motion in a Plane

Solution of Assignment (Set-2)

⎛ 1 ⎞
  tan1 ⎜
⎝ 3 ⎟⎠
27
...
The total displacement after 8 such turn is equal to
(1) 12 m

(2) 15 m

(3) 17
...
14 m

Sol
...

60º

Displacement = AB

60º

Two vectors are at 60º
102  102  2  102 



B
60º

1  10 3 m
2

60º

10 m
60º

17
...




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
...
Answer (4)
R  4 sin(2t )iˆ  4 cos(2t ) ˆj  xiˆ  yjˆ
Now, x2 + y2 = 42 which is equation of circle of radius R
...




Two particles A and B, move with constant velocities v 1 and v 2
...
The condition for particles A and B for their collision is
[Re-AIPMT-2015]
 
 
(1) r 1 − r 2 = v 1 − v 2

(2)

 


r1 − r 2
v 2 − v1
  = 

r1 − r 2
v 2 − v1

 
 
(3) r 1 ⋅ v 1 = r 2 ⋅ v 2

 


(4) r 1 × v 1 = r 2 × v 2

Sol
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Solution of Assignment (Set-2)

Motion in a Plane

109

 
 
  | r1  r2 |
r
r
(
v
v
)




2
1 
 1 2
|v v |
2

3
...
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
...


A projectile is fired from the surface of the earth with a velocity of 5 ms–1 and angle  with the horizontal
...
The value of the acceleration due to gravity on the planet
is (in ms–2) is (Given g = 9
...
5

(2) 5
...
3

(4) 110
...
Answer (1)
Since trajectory is same, so range and maximum height both will be identical from earth and planet
...
5
9

9
...
5 m/s2
5
...



Average velocity vector (v av ) from t = 0 to t = 5 s is
(1)



1
13iˆ  14 ˆj
5



(2)

7 ˆ ˆ
(i  j )
3

[AIPMT-2014]
(3) 2(iˆ  ˆj )

(4)

11 ˆ ˆ
(i  j )
5

Sol
...


The velocity of a projectile at the initial point A is (2iˆ  3 ˆj ) m/s
...
Answer (2)
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∵ ax  0

The horizontal range and the maximum height of a projectile are equal
...
Answer (4)
H=R
tan = 4

H 1
⎛
⎞
 tan ⎟
⎜⎝∵
⎠
R 4

  tan 1(4)

8
...
3i  0
...
The magnitude of velocity after



10 s will be
(1) 5 units



[AIPMT (Prelims)-2012]
(2) 9 units

(3) 9 2 units

(4) 5 2 units

Sol
...
3iˆ  0
...
3iˆ  0
...


A particle moves in a circle of radius 5 cm with constant speed and time period 0
...
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
...
2  s

T 

2
20
⇒ 
 10 rad s 1

0
...
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Solution of Assignment (Set-2)

Motion in a Plane

111

10
...
After 10 s its velocity becomes 40 m/s towards north
...
Answer (1)

| v |
aav 
t



v  (v 2  v1 )  v 22  v12

–1

40 ms

(∵   90º )

 5 302  402  50 ms1
so, aav 

–1

30 ms

50
 5 ms 2  aav
10

11
...
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
...
A particle of mass m is released from rest and follows a parabolic path as shown
...
Answer (2)
13
...
Elevation angle of the projectile at its highest point as
seen from the point of projection is
[AIPMT (Mains)-2011]

⎛ 3⎞
(1) tan ⎜⎜
⎟⎟
⎝ 2 ⎠
Sol
...
(i)

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Six vectors, aˆ through fˆ have the magnitudes and directions indicated in the figure
...
Answer (3)


d

f


e
  
d e  f

15
...
The angle of projection is
[AIPMT (Mains)-2010]
(1) 60º

(2) 15º

(3) 30º

(4) 45º

Sol
...
A particle moves in x-y plane according to rule x = a sin t and y = a cost
...
Answer (2)
x  a sin t ⇒ x 2  a 2 sin2 t
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17
...
4iˆ  0
...
Its speed after 10 s is
[AIPMT (Prelims)-2010]
(1) 7 units

(2) 7 2 units

(3) 8
...
Answer (2)
18
...
When the particle
lands on the level ground the magnitude of the change in its momentum will be
[AIPMT (Prelims)-2008]
(1) Zero

(2) 2 mv

(3)

mv
2

(4) mv 2

Sol
...
A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane
...
The path of the particle makes with the x-axis an angle of

[AIPMT (Prelims)-2007]

(1) 0°

(4) 60°

(2) 30°

(3) 45°

Sol
...


3

 
 
A and B are two vectors and  is the angle between them, if A  B  3( A  B ) , the value of  is

[AIPMT (Prelims)-2007]
(1) 90°

(2) 60°

(3) 45°

(4) 30°

Sol
...
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
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114

Motion in a Plane

Solution of Assignment (Set-2)

22
...
8 s for every circular lap
...
Answer (2)
T = 62
...
14  10

T
62
...
1 rad s1
v = r
v = 100 × 0
...
The vectors A and B are such that: | A  B |  | A – B |
...
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

24
...
Answer (3)
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A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed
...
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
26
...
The boy at B starts running in a
direction perpendicular to AB with velocity v1
...
Answer (2)
The distance travelled by body at B,

v

C
v1

= Speed × t

A

= v1t

B

a

C

So, BC = v1t, similarly, AC = vt
Applying pythagoras in ABC,
2 2

v t 

v12t 2

a

t

v1t

2

(v 2  v12 )t 2  a2
t2 

vt
A

B

a

a2
2

v  v12

a
2

v  v12



  
27
...
Answer (4)
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A boat is sent across a river with a velocity of 8 km/h
...
8 km/h

(4) 6 km/h

Sol
...
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
...

30
...
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
...
A body is whirled in a horizontal circle of radius 20 cm
...
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
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Solution of Assignment (Set-2)

Motion in a Plane

117

32
...

(1) Distance

(2) Angular momentum

(3) Heat

(4) Energy

Sol
...

33
...
The rod slides along perpendicular rails as shown here
...
What is the velocity of B when angle  = 60°?
B



(1) 10 m/s

(2) 9
...
8 m/s

(4) 17
...
Answer (3)

v A cos 60º  v B cos30º

vB = cos30º

vB
vB = sin30º

30º

1
3
10   v B 
2
2
vB 

30º

vA = cos60º

10

60º

3

vA
vA = sin60º

34
...
It crosses a river of width 1
...
The velocity of the river water (in km/h) is
(1) 3

(2) 1

(3) 4

(4) 5

Sol
...
0 km
t = 15 min
35
...
Their speeds are
such that each makes a complete circle in the same time t
...
Answer (3)
If time is same then,
1 :  2 ⇒ 1 : 1

2 ⎤
⎡
⎢∵   T ⎥
⎣
⎦

36
...
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118

Motion in a Plane

Solution of Assignment (Set-2)

0
...
The speed of water in the stream is
(1) 0
...
5 m/s

(3) 1
...
433 m/s

Sol
...
5 

1
 u ⇒ u  0
...
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
...

38
...
If their time periods
are same, then the ratio of their angular velocities will be

(1)

r
R

(2)

R
r

(3) 1

(4)

R
r

Sol
...
If | A  B |  | A |  | B | then angle between A and B will be
(1) 90°

(2) 120°

(3) 0°

(4) 60°

Sol
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Solution of Assignment (Set-2)

Motion in a Plane

119

⎛ 20 ⎞
⎟ m with constant tangential acceleration
...
A particle moves along a circle of radius ⎜
⎝  ⎠
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
...
The vector sum of two forces is perpendicular to their vector differences
...
Answer (2)
   
( A  B)  ( A  B)  0
A2  B 2  AB  BA  0

A2  B 2


AB



so, | A |  | B |
42
...
0 rad/s2 and an initial angular speed of 2
...
In a time of 2 s it
has rotated through an angle (in radian) of
(1) 10

(2) 12

(3) 4

(4) 6

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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
43
...
Answer (4)
 
rf – ri
13 – 2 iˆ  14 – 3 jˆ  11 iˆ  jˆ
Vav 
=
5
t
 5 – 0



(4)

11 ˆ ˆ
(i  j )
5



SECTION - D
Assertion - Reason Type Questions
1
...

 
 
 
R : If A  B , then ( A  B ) is perpendicular to A  B
...
Answer (3)
2
...

   
R : By triangle law of vector addition we can prove P  Q  Q  P
...
Answer (1)
3
...

R : A vector can be divided by a scalar
...
Answer (2)
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Motion in a Plane

121

A : At the highest point the velocity of projectile is zero
...


Sol
...


A : Horizontal range of a projectile is always same for angle of projection  with horizontal or  with vertical
...


Sol
...


A : Horizontal motion of projectile without effect of air is uniform motion
...


Sol
...


A : Path of a projectile with respect to another projectile is straight line
...


Sol
...


A : In the case of ground to ground projection of a projectile from ground the angle of projection with horizontal is
 = 30º
...

R : Maximum deviation of the projectile is 2 = 60º
...
Answer (1)
9
...

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
...
Answer (1)
10
...

R : Instantaneous velocity is always normal to instantaneous acceleration in uniform circular motion
...
Answer (1)
11
...

R : When the speed is increasing, its tangential acceleration is in the direction of instantaneous velocity
...
Answer (1)
12
...

R : The direction of acceleration of a particle in uniform circular motion changes continuously
...
Answer (1)
13
...

R : The direction of angular displacement is perpendicular to plane of rotation of object
...
Answer (2)

  

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