OBJECTIVE SOLVED
1. Pick up the correct statements:
(a) area under a – t graph gives velocity
(b) area under a – t graph gives change is velocity
(c) path of projectile as seen by another projectile is parabola
(d) a body, whatever be its motion, is always at rest in a frame of reference fixed to body itself.
Solution:
a
dt
dv

 dt
a
dv
2
1
v
v

 
 v
 = Area under a – t graph
where v
 = magnitude of change in velocity.
.
The path of a projectile as seen by another projectile is a straight line becomes, the relative velocity
between the particles remains constant.
 (b) and (d)
2. The angular acceleration of a particle moving along a circular path with uniform speed is
(a) uniform but non-zero
(b) zero
(c) variable
(d) such as cannot be predicted from the given information.
Solution:
As angular speed of the particle is constant and hence angular acceleration is zero.
 (b)
3. A particle is projected horizontally from the top of a cliff of height H with a speed gH
2 . The radius
of curvature of the trajectory at the instant of projection will
(a) H/2 (b) H
(c) 2H (d)  .
Solution:
Since, v
g



Radial acceleration g
ar 
We know 2
r
a v / r


2
v
g
r
 where r is the radius of curvature.
 g
r
gH
2
 )
gH
2
v
( 

 H
2
r 
Hence (C)
4. A body when projected vertically up, covers a total distance D. During the time of its flight t. If there
were no gravity, the distance covered by it during the same time is equal to

x
y
Q
P
90º
(a) 0 (b) D
(c) 2D (d) 4D.
Solution:
The displacement of the body during the time t as it attains the point of projection
 0
S   0
gt
2
1
t
v 2
0 

 g
v
2
t 0

During the same time t, the body moves in absence of gravity through a distance
0
D v t
  , because in absence of gravity g = 0
 g
v
2
g
v
2
v
D
2
0
0
0 









 ...(i)
In presence of gravity the total distance covered is
g
v
g
2
v
2
H
2
D
2
0
2
0



 ...(ii)
(i)  (ii)  D
2
D 

Hence (c)
5. A particle is projected from point P with velocity 1
ms
2
5  perpendicular to the surface of a hollow
right angle cone whose axis is vertical. It collides at Q normally. The time of the flight of the particle is
(a) 1 sec. (b) 2 sec.
(c) sec
2
2 (d) 2 sec.
Solution:
.
sec
1
10
2
2
5
sin
g
u
t 




Hence (a)
6. The relative velocity of a car ‘A’ with respect to car B is 2
30 m/s due North-East. The velocity of
car ‘B’ is 20 m/s due south. The relative velocity of car ‘C’ with respect to car ‘A’ is 2
10 m/s due
North-West. The speed of car C and the direction (in terms of angle it makes with the east).
(a) 2
20 m/s, 45º (b) º
135
,
s
/
m
2
20
(c) º
45
,
s
/
m
2
10 (d) º
135
,
s
/
m
2
10 .
Solution:
Given 1
AB
| v | 30 2ms


1
B
| v | 20ms


1
CA
| v | 10 2 ms


Now
AB A B
ˆ ˆ
v v v 30 2(cos45ºi sin 45º j)
   
  
or, 1
A B
ˆ ˆ
v v (30i 30j)ms
  
 
...(i)
and, 1
B
ˆ
v ( 20j)ms
 

...(ii)
vCA
N(Y)
vAB
E(x)
vB
45º
45º

and CA C A
ˆ ˆ
v v v 10 2( cos45ºi sin 45º j)
    
  
or 1
C A
ˆ ˆ
v v ( 10i 10j)ms
   
 
...(iii)
solving equation (i) (ii) and (iii) we’ll get C
ˆ ˆ
v 20i 20j
 

(a)
7. Two particles 1 and 2 are allowed to descend on the two frictionless chord OA and
OB of a vertical circle, at the same instant from point O. The ratio of the velocities
of the particles 1 and 2 respectively, when they reach on the circumference will be
(OB is the diameter)
(a) 
sin (b) 
tan
(c) 
cos (d) none of these.
Solution:



 cos
g
a
,
cos
d
OA OA
Along OA
 

 cos
d
.
cos
g
2
v2
A
Along OB
gd
2
v2
B 
 
 cos
v
v
B
A
Hence, C is correct.
8. A ball is dropped vertically from a height d above the ground. If it hits the ground and bounces up
vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the
height h above the ground as
(a)
d
v
h (b)
d
v
h
(c)
d h
v
(d)
d h
v
.
Solution:
As the ball falls, at height h the velocity of the ball is zero and at any height )
h
d
(
g
2
u
v
,
h 2
2


 ,
with decreasing h, v increases.
When h = 0, v = velocity is maximum.
After the ball collides the floor, its velocity changes in magnitude as wall as direction, as the body goes
to a smaller height in bouncing up .The change in velocity takes place within zero height and with no
change in time.
Hence, (a) is correct choice.
A
B

1
2
O
A
B

1
2
O
g cos
d

9. A particle moves along a straight line according to the law c
bt
2
at
S 2
2


 . The acceleration of the
particle varies as
(a) 3
S (b) 3
/
2
S
(c) 2
S (d) 2
/
5
S .
Solution:
2
/
1
2
)
c
bt
2
at
(
S 


Differentiating,
c
bt
2
at
b
at
)
b
2
at
2
(
)
c
bt
2
at
(
2
1
dt
dS
2
2
/
1
2








 
)
c
bt
2
at
(
c
bt
2
at
)
b
at
)(
b
at
(
a
c
bt
2
at
dt
S
d
2
2
2
2
2













 


 
c
bt
2
at
c
bt
2
at
)
b
at
(
)
c
bt
2
at
(
a
2
2
2
2










2
2
(ac b )
S S



 3
2
2
S
1
dt
S
d
  acceleration 3
S

 (a)
10. A car accelerates from rest at a constant rate  for some time, after which it decelerates at a constant
rate  to come to rest. If the total time elapsed is t, the maximum velocity acquired by car is
(a) t
)
(
V




 (b) t
)
(
V





(c) t
)
(
2
V




 (d) t
)
(
2
V




 .
Solution:
From motion from A to B 1
t
V 
 or


V
t1
From motion from B to C 2
t
V
0 

 or


V
t2
(t )
1
v
A B C
(t )
2












)
(
V
V
V
t
t
t 2
1
or t
)
(
V





 (a)
11. A stone A is dropped from rest from a height h above the ground. A second stone B it simultaneous y
thrown vertically up from a point on the ground with velocity v. The line of motion of both the stones is
same. The values of v which would enable the stone B to meet the stone A midway between their initial
positions is
(a) 2 gh (b) gh
2
(c) gh (d) gh
2 .
Solution:
Time of travel of each stone =

Distance travelled by each stone
2
h

For stone
2
h 1
A, gt
2 2
 i.e., g
h
t 
For stone B, 











g
h
g
2
1
g
h
u
gt
2
1
ut
2
h 2
h/2
h
A
B
A
B
 2
h
g
h
u
2
h


or, h
g
h
u 
 gh
h
g
h
u 

The correct option is (c)
12. A stone is dropped from rest from the top of a cliff. A second stone is thrown vertically down with a
velocity of 30 m/s two seconds later. At what distance from the top of a cliff do they meet?
(a) 60 m (b) 120 m
(c) 80 m (d) 44 m.
Solution:
The two stones meet at distance S from top of cliff t seconds after first stone is dropped.
For 1st
stone ;
gt
2
1
S 2
 For 2nd
stone
2
)
2
t
(
g
2
1
)
2
t
(
u
S 


 .
i.e., g
2
gt
2
gt
2
1
u
2
ut
gt
2
1 2
2





s
4
20
30
)
10
30
(
2
g
2
u
)
g
u
(
2
t
);
g
u
(
2
t
)
g
2
u
(
0 










Distance S at which they meet 16
10
2
1
gt
2
1 2





= 80 m from top of cliff
 (c)
13. A particle is projected from a point O with velocity u in a direction making an angle  upward with the
horizontal. At P, it is moving at right angles to its initial direction of projection. Its velocity at P is
(a) 
tan
u (b) 
cot
u
(c) 
ec
cos
u (d) 
sec
u .
Solution:







 cot
u
v
;
cos
u
sin
v
)
90
cos(
v
 (b)
14. A man running at 6 km/hr on a horizontal road observes that the rain hits him at an angle of 30º from the
vertical. The actual velocity of rain has magnitude
(a) 6 km/hr (b) 3
6 km/hr
(c) 3
2 km/hr (d) 2 km/hr

u
P
O
90º
(90º- )

v


Solution:
R
V

 velcoity of rain
RM
V

 velocity of rain relative to man
M
V

 velocity of man
Given, M
V 6
 kmh-1
, RM
V  at angle 30º from vertical
RM
V
M
V
R
V
M
V
here R RM M
V V V
 
  
from figure,
M
R
V
tan30º
V


R
1 6
V
3

 R
V 6 3
 km/hr
 (b)
15. The angular acceleration of a particle moving along a circular path with uniform speed is
(a) uniform but non-zero
(b) zero
(c) variable
(d) such as cannot be predicted from the given information.
Solution:
As angular speed of the particle is constant and hence angular acceleration is zero.
 (b)