Kinematics-02-OBJECTIVE SOLVED
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:
dv a dt
v2
dv a dt
v1
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 of curvature of the trajectory at the instant of projection will
(a) H/2 (b) H
(c) 2H (d) .
Solution:
Since, → →
. The radius
Radial acceleration a r g
We know a v2 / r
v2
r g
where r is the radius of curvature.
2gH g r
r 2H
( v
2gH )
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
(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
S 0
t 2v 0
g
v 0
t 1 gt 2 0
2
During the same time t, the body moves in absence of gravity through a distance
D v0t , because in absence of gravity g = 0
2v 0
2v 2
D v 0
0
...(i)
g g
In presence of gravity the total distance covered is
v 2 v 2
D 2H 2 0 0
...(ii)
2g g
(i) (ii) D 2D
Hence (c)
5. A particle is projected from point P with velocity 5
2ms1
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) sec.
(c)
Solution:
t
2 2 sec
u 5 2
2 1sec.
(d) 2 sec.
g sin 10
Hence (a) x
6. The relative velocity of a car ‘A’ with respect to car B is 30 2 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 10 2 m/s due
North-West. The speed of car C and the direction (in terms of angle it makes with the east).
(a)
20 m/s, 45º (b)
20 2m / s, 135º
(c) 10
Solution:
m / s, 45º
(d)
10 2 m / s, 135º .
Given → 30 2ms1
| vAB |
| vB | 20 ms
→ 1
| vCA | 10 2 ms
Now
→ → → ˆ ˆ
E(x)
vAB vA vB 30 2(cos 45º i sin 45º j)
or,
→ →
vA vB
(30iˆ 30ˆj)ms1
...(i) vB
and,
→ (20ˆj)ms1
...(ii)
and
→ → → ˆ ˆ
vCA vC vA 10 2 (cos 45º i sin 45º j)
or vC
→ (10iˆ 10ˆj)ms1
...(iii)
solving equation (i) (ii) and (iii) we’ll get →
20ˆi 20ˆj (a)
7. Two particles 1 and 2 are allowed to descend on the two frictionless chord OA and O
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 1
(OB is the diameter) 2
(a)
sin
(b)
tan
(c)
Solution:
cos
OA d cos , a OA g cos
(d) none of these. A
B
Along OA O
2 2g cos .d cos 1
Along OB
v 2 2gd
2
d
g cos
v A
v B
A
cos B
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) (b)
(c) (d) .
Solution:
As the ball falls, at height h the velocity of the ball is zero and at any height with decreasing h, v increases.
When h = 0, v = velocity is maximum.
h, v 2 u 2 2g(d h) ,
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.
9. A particle moves along a straight line according to the law S2 at 2 2bt c . The acceleration of the particle varies as
(a)
(c)
Solution:
S3 S2
(b)
(d)
S2 / 3
S5 / 2 .
S (at 2 2bt c)1/ 2
Differentiating,
dS 1 (at 2 2bt c)1/ 2 (2at 2b) at b
dt 2
at 2 2bt c a (at b)(at b)
d 2S
at 2 2bt c
dt 2
d 2S 1
(at 2 2bt c)
(ac b2 ) S S2
dt 2 S3
acceleration
S3
(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)
(c)
Solution:
V t
( )
V 2 t
( )
(b)
(d)
V t
( )
V 2 t
( )
From motion from A to B V t1 or
t V
1
V
From motion from B to C 0 V t 2 or
t 2
A B C
t t t
V V V( )
1 2
V t
( )
(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) 2
(c)
Solution:
Time of travel of each stone =
(d) .
Distance travelled by each stone h
2
For stone A, h 1 gt2
2 2
h 1
i.e.,
2
t
1 h
For stone B,
ut
gt u
g
or,
2 2
h u h
2 2
u h
2 g
u h g
h
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 S 1 gt 2 ;
2
For 2nd stone S u(t 2) 1 g(t 2)2 .
2
i.e.,
1 gt 2 ut 2u 1 gt 2 2gt 2g
2 2
0 (u 2g)t 2(u g); t 2(u g) 2(30 10) 4s
u 2g 30 20
Distance S at which they meet
1 gt 2 1 10 16
2 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)
(c)
u tan
u cos ec
(b)
(d)
u cot
u sec .
P
(90º-)
Solution:
v cos(90 ) v sin u cos ; v u cot
90º
u v
(b) O
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)
6 km/hr
(c) 2 km/hr (d) 2 km/hr
Solution:
→
VR
→
VRM
velcoity of rain VM
velocity of rain relative to man
→
VM velocity of man
Given, VM 6 kmh-1, VRM at angle 30º from vertical
→ → →
here VR VRM VM
from figure, VM
tan 30º VM
VR
6
VR
VR 6
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)
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