kinematics - einstein classes

kinematics - einstein classes
PK – 1
KINEMATICS
The part of mechanics that deals with the description of motion is called kinematics. There are two types of
motion :
1.
One dimensional motion or Motion in straight a line : In this motion the velocity vector and acceleration
vector are always along the same line.
2.
Two dimensional motion : In this motion the velocity vector and acceleration vector will be in the same
plane but they are inclined at some angle, this angle may change during the motion or may be constant. For
example : circular motion and parabolic motion.
C1
Displacement and Distance
Displacement is defined as the change in position vector of the particle during a time interval whereas
distance is defined as the length of actual path. Displacement is a vector quantity whereas distance is a
scalar quantity.
Note that : (i) distance  |displacement| (ii) distance and magnitude of displacement are equal during the
time interval in which the velocity of the particle should not be zero at any moment along the straight line
motion.
C2
Velocity and Speed
Average Velocity : The change in position vector i.e. displacement divided by time interval during which
this change occurs is known as average velocity. For example, a particle changes its position from xi to xf
along x - axis at time t i and t f respectively . Then average velocity along x-axis is given by :
vav 
xf  xi x . In general, for a particle moving on curved path :

t f  t i t
  
x
y
z

 r rf  ri

î 
ĵ 
k̂ . Here  r is the displacement during the time interval t.
v av 

=
t
t
t
t t f  t i
Instantaneous Velocity : The velocity of the particle at a particular point or at a particular instant of time
is called the instantaneous velocity of the particle. It is given by


dx
dy
dz

r dr
î 
ĵ 
k̂ = v x î  v y ĵ  v z k̂ .
v  lim

=
t 0 t
dt
dt
dt
dt
For constant velocity, displacement = (velocity) (time)
Average Speed
The average speed of a particle in a time interval is defined as the distance travelled by the particle divided
by the time interval.
Note that : (i) Average Speed  |Average Velocity| (ii) average speed & magnitude of average velocity are
equal during the time interval in which the velocity of the particle should not be zero at any moment along
the straight line motion.
Instantanous Speed : The instantaneous speed equals the magnitude of the instantaneous velocity. The
s ds
where s is the distance travel during time t. Also the

t  0 t
dt
instantaneous speed is given by v  lim
speed is given by
v 2x  v 2y  v 2z .
For constant speed, distance = (speed) (time)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 2
Practice Problems :
1.
2.
Which of the following statement is true ?
(a)
|displacement|  distance
(b)
|Average velocity|  Average speed
(c)
distance and average speed never be zero or negative
(d)
all the above
A train travels from one station to another at a speed of v1 and returns to the first station at the speed
of v2. The average speed and average velocity of the train is respectively
(a)
3.
2 v1 v 2
,0
v1  v 2
0,
(b)
2v1 v 2
v1  v 2
0, 0
(c)
2v1 v 2 2 v1 v 2
,
v1  v 2 v1  v 2
(d)
A particle covers one quarter of a circular path of radius R. It takes time T. The average speed and
the magnitude of average velocity are given by
respectively.
(a)
R 2R
,
2T T
R R
,
2T 2T
(b)
2R 2 R
,
T
T
(c)
(d)
2R R
,
T 2T
4.
A particle starts from one point to another point along the straight path. It covers this path in n
equal distance with speed v1, v2.....vn. Find the average speed for the complete journey.
5.
A particle starts from one point to another point along the straight path. It covers this path in n
equal time interval with speed v1, v2.....vn. Find the average speed for the complete journey.
6.
A particle is moving along a circular path of radius r. Find magnitude of displacement and distance
for (a) one quarter of circle (b) half circle (c) three quarter circle (d) complete one circle (e) 2.5 circle.
n
[Answers : (1) d (2) a (3) a (4)
v
n
n
(5)
1
v
i 1
i 1
n
i
(6) (a)
2R,
3R (d) 0, 2R
R (b) 2R, R (c)
2R,
2
2
i
(e) 2R, 5R]
C2
Acceleration
Average Acceleration : Average acceleration is defined as the ratio of change in velocity to the time taken.
  




v vf  vi
where v f and v i are the velocity of the particle at tf (final time) and ti
 a  a av 

t t f  t i
(initial time) respectively.
For straight line motion (i.e. along x-axis) a av 
v xf  v xi v x

.
tf  ti
t

Instantaneous Acceleration : Instantaneous acceleration is defined as a  Lim
t  0
For straight line motion (i.e. along x-axis) a x 
Acceleration can also be expressed as a x 
Einstein Classes,


v d v

 t dt
dv x
dt
dv x dx
dv

.
 v x x . For uniform velocity a  0 .
dx dt
dx
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
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

Uniform acceleration means that the acceleration of the particle is constant and in this case < a > = a . If
acceleration is in same direction to the velocity then speed of the particle increases. If acceleration is in
opposite direction to the vleocity then speed decreases. This situation is called retardation. Note that
negative acceleration does not mean that motion is retardation.
C3
Flow chart to find displacement, velocity & acceleration :
Practice Problems :
1.
A particle moves along a straight line such that its displacement at any time t is given by
(t3 – 3t2 + 2)m. The displacement when the acceleration is zero
(a)
2.
2m
(c)
3m
(d)
–2 m
v=u
(b)
v = u + at
(c)
v = u + at2
(d)
v=u+
1 2
at
2
The displacement x of a particle moving in one dimension under constant acceleration is related to
the time t as t = x + 3. The displacement of the particle when its velocity is zero is
(a)
4.
(b)
The initial velocity of a particle is u and the acceleration at time t is at, a being a constant. Then the
velocity v at time t is given by
(a)
3.
0m
zero
(b)
3 units
(c)
3 units
(d)
9 units
2
The velocity of a particle moving on the x-axis is given by v = x + x where v in m/s and x is in m.
Its acceleration in m/s2 when passing through the point x = 2m.
(a)
0
(b)
5
(c)
11
(d)
30
5.
A particle is moving along the x-axis such that its velocity, v = ax where a is a constant quantity.
Prove that the acceleration of the particle is constant.
6.
A particle is moving along a straight path such that acceleration a = –v, where  is the constant and
v is the instant velocity. If initial velocity is u then (i) find velocity at any time t (ii) velocity after
covering the distance x (iii) draw velocity-time graph and velocity-distance graph. Also find the
maximum distance covered.
[Answers : (1) a (2) d (3) a (4) d (6) (i) v = ue–t (ii) v = u – x (iii) max. distance = u/]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 4
C4
GRAPHICAL REPRESENTATION
1.
The average velocity between two points A and B is the slope of line AB, whereas the instantaneous
velocity of the particle at P is the slope of tangent drawn at this point.
2.
Consider the velocity time graph for a particle moving along the straight line as shown in figure. Let the
magnitude of area of the triangle OAB is A1 and BCD is A2 then
Distance = A1 + A2
Magnitude of displacement = |A1 – A2|
3.
The average acceleration between two points A and B is the slope of line AB, whereas th e instantaneous
acceleration of the particle at P is the slope of tangent drawn at this point.
4.
On an acceleration (a) versus time (t) graph, the change in velocity in velocity
is the area bounded as shown in figure :
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 5
C5
Some typical graph : In the following graphs time is on the horizontal axis whereas displacement or
velocity on the vertical axis
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 6
Practice Problems :
1.
The velocity-time graph for straight line motion is shown in figure.
Find (a) total distance (b) total displacement (c) average velocity between 5s to 40s (d) total average
speed (e) total average velocity (f) average acceleration between 15s to 25s (g) acceleration at
t = 0 & 10s(h) draw the acceleration-time graph, distance-time graph and displacement-time graph.
2.
The velocity of a car moving along straight road is changing with time as shown in figure
Then :
(a)
The maximum acceleration of the car is between 40s to 50s.
(b)
The total distance covered by the car is 650 m
(c)
The total displacement covered by the car is 320 m
(d)
During the journey there is always non-uniform motion.
[Answers : (1) (a) 90 m (b) 20 m (c) 85/35 m/sec (d) 9/4 m/sec (e) 1/2 m/sec (f) –0.4 m/s2 (g) 2/5 m/s2,
1/5 m/s2 (2) a]
C6
MOTION WITH CONSTANT ACCELERATION, ALONG STRAIGHT LINE OR RECTELINEAR
MOTION
For a uniformly accelerated motion along a straight line (sav x-axis) the following equations can be used.
x = x0 + ux(t – t0) + ½ ax(t – t0)2
vx = ux + ax (t – t0)
x = x0 +
u x  vx
(t  t 0 )
2
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 7
2
x
2
x
v = u + 2ax (x – x0)
The symbols used above have following meaning;
x0  Initial position of the particle on x-axis at initial time t0.
ux  Initial velocity of the particle along x-axis.
vx  Velocity of the particle at any position x and any time t.
ax  Constant acceleration of the particle along x-axis.
NOTE :
we must decide at the beginning of a problem where the origin of co-ordinates is and which direction is
positive. The choices of frame of reference are usually a matter of convenience.
Practice Problems :
1.
A particle starts with velocity u along a straight line path with constant acceleration. It ends its
journey with velocity v. The velocity of the particle at the mid point of the journey is
(a)
2.
2v 2 u 2
v2  u 2
(d)
10 cm/s
(b)
12 cm/s
(c)
14 cm/s
(d)
16 cm/s
10 m
(b)
20 m
(c)
30 m
(d)
40 m
22 m
(b)
18 m
(c)
15 m
(d)
13 m
s2 = s1
(b)
s2 = 2s1
(c)
s2 = 3s1
(d)
s2 = 4s1
–2
A body moving in a straight line with constant acceleration of 10 ms covers a distance of 40 m in the
4th second. How much distance will it cover in the 6th second ?
(a)
7.
2 vu
vu
A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval
t during which it travels a distance s1. It continues to move with the same acceleration for the next
time interval t during which it travels a distance s2. The relation between s1 and s2 is
(a)
6.
(c)
A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time to
the driver. If he is driving a car at a speed of 54 km/h and the brakes causes a deceleration of
6.0 m/s2, the distance travelled by the car after he sees the need to put the brakes on.
(a)
5.
v2  u 2
2
The speed of a train is reduced from 60 km/h to
15 km/h while it travels a distance of 450 m. If
the retardation is uniform, how much further it will travel before coming to rest ?
(a)
4.
(b)
A body travels 200 cm in the first two seconds and 220 cm in the next four seconds. The velocity at
the end of the seventh second from the start is
(a)
3.
vu
2
50 m
(b)
60 m
(c)
70 m
(d)
80 m
A car, starting from rest, is accelerated at a constant rate  until it attains a speed v. It is then
retarded at a constant rate  until it comes to rest. The average speed of the car during its entire
journey is
(a)
zero
(b)
v
2
(c)
v
2
(d)
v
2
[Answers : (1) b (2) a (3) c (4) a (5) c (6) b (7) d]
C7
Vertical Motion Under Gravity
If a body is moving vertically downwards or upwards, it experiences a downward acceleration due to the
gravitational force of the earth. This is called acceleration due to gravity and is denoted by the symbol g.
Strictly speaking g is not a constant, but varies form place to place on the surface of the earth and also with
height. However the variation of g is so small that it can be neglected and g can be considered a constant
unless very large heights are involved. Therefore, we can use the above equations of motion for constant
acceleration.
For solving problems of vertical motion under gravity, either the upward or the downward direction is taken
as positive. If the upward direction is taken as positive, then g becomes negative and vice-versa. The signs
of other quantities like initial velocity, initial position will be decided according to the frame of reference.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 8
Practice Problems :
1.
2.
A stone is dropped from the top of a 30 m high cliff. At the same instant another stone is projected
vertically upwards from the ground with a speed of 30 m/s. The two stones will cross each other after
a time t and the height it which they cross each other is h then (g = 10 m/s2)
(a)
t = 2s, h = 25 m
(b)
t = 1s, h = 25 m
(c)
t = 1s, h = 15 m
(d)
t = 2s, h = 15 m
A particle, dropped from a height h, travels a distance 9h/25 in the last second. If g = 9.8 m/s2, then h
is
(a)
3.
100 m
(b)
122.5 m
(c)
145 m
(d)
167.5 m
A stone is dropped from a height h, simultaneously, another stone is thrown up from the ground
which reaches a height 4h. The two stones cross each other after time
(a)
h
2g
h
8g
(b)
(c)
(d)
8hg
2hg
[Answers : (1) b (2) b (3) b]
C8A MOTION IN A PLANE OR 2D MOTION
If a particle is moving in a plane, its motion can be split into two rectilinear motions along two
perpendicular directions. These two motions can be treated independently of each other and then the results
can be combined according to the rules of vector addition & requirement of the problem.
Now, if the acceleration is constant, then the motions along the two axes are governed by the following two
sets of equations :
X-direction
Y-direction
x = x0 + ux(t – t0) + ½ ax(t – t0)
2
y = y0 + uy(t – t0) + ½ ay(t – t0)2
vx = ux + ax (t – t0)
x = x0 +
vy = uy + ay (t – t0)
u x  vx
(t  t 0 )
2
y = y0 +
vx2 = ux2 + 2ax (x – x0)
C8B
u y  vy
2
(t  t 0 )
vy2 = uy2 + 2ay (y – y0)
Horizontal projection
Suppose a body is projected horizontally from a certain height h with a speed u then
time of flight =
T
2h
2h
and the horizontal range = R  uT  u
g
g
C8C Oblique Projection
Suppose a body is projected with initial velocity u at an angle  with the horizontal.
(i)
The equation of the trajectory of the projectile is y  (tan ) x 
g
x2
2
2u cos 
2
which represents a parabola.
(ii)
Maximum Height H 
u 2 sin 2 
(iii)
2g
(iv)
Horizontal Range R 
u 2 sin 2
g
Time of Flight T 
2u sin 
g
Two important points to be noted concerning horizontal range R :
(i)
For a given velocity of projection, R is maximum when  = 450.
Einstein Classes,
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(ii)
For a given velocity, there are two angles of projection for which the range is the same.
If one of these angles is , the other is

.
2
Practice Problems :
1.
2.
The x and y coordinates of a particle at any time t are given by x = 3t + 4t2 and y = 4t where x and y
are in m and t in s. Then
(a)
The initial speed of the particle is 5 m/s.
(b)
The acceleration of the particle is constant.
(c)
The path of the particle is parabolic.
(d)
All are correct
A particle is projected with speed u at an angle of  with the horizontal. Another particle of different
mass is projected with same speed from the same point. Both the particles has same horizontal
range. Let the time of flight and maximum height attained by the first particle and second particle
are t1, h1 and t2, h2 respectively. Then t1/t2 and h1/h2 are given by respectively
(a)
3.
5.
(b)
cot, cot2
(c)
cot, tan2
(d)
tan, cot2
Let the maximum height attained by the projectile is n times the horizontal range. Then the angle of
projection with the horizontal is given by
(a)
4.
tan, tan2
tan–1n
(b)
tan–12n
(c)
tan–13n
(d)
tan–14n
Two projectiles are projected from the same point with the same speed but at different angles of
projection. Neglect the air resistance. They land at the same point on the ground. Which of the
following angle of projections is possible ?
(a)


 ,  
4
4
(c)
,


2
(b)


 ,  
3
6
(d)
all are possible
If y = ax – bx2 is the path of a projectile, then which of the following is correct
(a)
Range = a/b
(b)
Maximum height = a2/4b
(c)
Angle of projection = tan–1a
(d)
all are correct
[Answers : (1) d (2) a (3) d (4) d (5) d]
C9
RELATIVE MOTION
The position, velocity and acceleration of a particle are relative terms and are defined with respect to
certain frame of reference. This frame of reference may be stationary, moving with constant velocity or
have some acceleration.
If xAB is is position of A with respect to B then xAB = xA – xB where xA and xB are the position of A and B with
respect to some common frame of reference. In the similar way for relative velocity vAB = vA – vB. In vector
form
rAB  rA  rB
v AB  v A  v B
a AB  a A  a B
where rAB is the position of A with respect to B, rA and rB are the position of A and position of B with
respect to some common frame of reference, v AB is the velocity of A with respect to B, v A and v B are
the velocity of A and velocity of B with respect to some common frame of reference, a AB is the
acceleration of A with respect to B, a A and a B are the acceleration of A and velocity of B with respect to
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 10
some common frame of reference.
The equations of motion for constant relative acceleration are written as :
xrel = (x0)rel + urel(t – t0) + ½ arel(t – t0)2
vrel = urel + arel (t – t0)
x = (x0)rel +
u rel  vrel
(t  t 0 )
2
vrel2 = urel2 + 2arel (xrel – (x0)rel)
In general,
v rel 
dx rel
dv
, a rel  rel
dt
dt
Practice Problems :
1.
2.
Rain is falling with a speed of 4 m/s in a direction making an angle of 300 with vertical towards south.
What should be the magnitude and direction of velocity of cyclist to hold his umbrella exactly
vertical, so that rain does not wet him
(a)
2 m/s towards north
(b)
4 m/s towards south
(c)
2 m/s towards south
(d)
4 m/s towards north
A motorboat covers the distance between two stations on the river t1 = 8h and t2 = 12h downstream
and upstream respectively. The time taken by the boat to cover this distance in still water is
(a)
3.
g + a, g – a
6.
4.3 h
(c)
2.2 h
(d)
6.7 h
(b)
g – a, g + a
(c)
both
(d)
none
A railway carriage moves over a straight level track with an acceleration a. A passenger in the
carriage drops a stone. The acceleration of the stone with respect to the carriage and the Earth are
respectively
a2  g2 ,g
(a)
5.
(b)
A lift moves with an acceleration a. A passenger in the lift drops a book. The acceleration of the book
with respect to the lift floor if the lift is going up and if the lift is going down is respectively
(a)
4.
9.6 h
(b)
g, a 2  g 2
a2  g2 ,g
(c)
(d)
g, a 2  g 2
A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s, a stone is
released from the balloon. The stone will
(a)
cover a distance of 40 m
(b)
have a displacement of 50 m
(c)
reach the ground in 4 sec
(d)
begin to move down after being
released
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest
possible path in 15 min. The velocity of the river water in km/h is
(a)
1
(b)
3
(c)
4
(d)
41
[Answers : (1) c (2) a (3) a (4) a (5) c (6) b]
C10
Circular Motion :
When an object follows a circular path at constant speed, the motion of the object is called uniform circular
motion. The magnitude of its acceleration is ac = v2/R. The direction of ac is always towards the centre of the
circle. The angular speed , is the rate of change of angular distance. It is related to velocity v by v = R.
The acceleration is ac = 2R. If T is the time period of revolution of the object in circular motion and f is its
frequency, we have  = 2f, v = 2fR, ac = 42f2R
Practice Problems :
1.
A point moves in x-y plane according to the law x = 4 sin 6t and y = 4(1 – cos 6t). The distance
traversed by the particle in 4 seconds is (x and y are in meters)
(a)
2.
96 m
(b)
48 m
(c)
24 m
(d)
108 m
The modulus of the acceleration vector is constant. The trajectory of the particle is a/an
(a)
parabola
(b)
ellipse
(c)
hyperbola
(d)
circle
[Answers : (1) a (2) d]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 11
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
A projectile has a maximum range of 500 m. If the
projectile is now thrown up an inclined plane of 300
with the same velocity, the distance covered by it
along the inclined plane will be about
(a)
250 m
(b)
500 m
(c)
750 m
(d)
1000 m
Three particles starts from the origin at the same
time, one with a velocity u1 along the x-axis, the
second along the y-axis with a velocity u2 and the
third along the x = y line. The velocity of the third
so that the three may always lie on the same line is
(a)
u1  u 2
2
(b)
u 1u 2
(c)
u 1u 2
u1  u 2
(d)
2 u 1u 2
u1  u 2
6.
7.
(a)
(c)
4.
5.
1 s
2 a
2s
a
(b)
(d)
2
(a)
variable
(c)

2
(b)
2

(d)

2
Two particles A and B are initially 40 m apart. A
behind B. Particle A starts moving with a uniform
velocity of 10 m/s towards B. Particle B starting
from the rest has an acceleration of 2 m/s2 in the
direction of velocity of A. The minimum distance
between the two is
Einstein Classes,
(b)
15 m
(c)
25 m
(d)
30 m
A particle P is projected from the origin (0, 0) with
(a)
5
î  20 ĵ
4
(b)
4
î  20 ĵ
5
(c)
4
î  10 ĵ
5
(d)
none
The distance x covered by a body moving in a
straight line in time t is given by x2 = t2 + 2t + 3. The
acceleration of the body will vary as
(a)
s
a
A particle starts from the origin of coordinates at
time t = 0 and moves in the xy plane with a
constant acceleration  in the y-direction. Its
equation of motion is y =  x 2 . Its velocity
component in the x-direction is
20 m
a velocity of 20 ˆj at t = 0 and another particle Q is
projected with a velocity from the origin at t = 5s.
There is a uniform acceleration on both particle
along the negative y-axis. The particle P crosses the
origin once again at t = 4s where as the particle Q
crosses the point (–5, 0) at t = 9s. The velocity of
the particle Q at t = 5s is
The greatest acceleration or deceleration that a train
may have is a. The minimum time in which the train
can get from one station to the next at a distance s
is
s
a
(a)
(c)
8.
9.
1
x
(b)
1
x
3
(d)
1
x2
1
x4
A projectile is projected with speed u at an angle 
with the horizontal. The time after which the
velocity vector of the particle become perpendicular to the initial velocity of projection
(a)
u
g sin 
(b)
u
g cos 
(c)
u sin 
g
(d)
2u sin 
g
A stone is projected from the ground with a
velocity of 50 m/s at an angle 300. It crosses the wall
after 4s. The distance beyond the wall at which the
stone strikes the ground is
(a)
25 m
(b)
253 m
(c)
50 m
(d)
25/3 m
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 12
10.
11.
12.
13.
14.
A car A is travelling on a straight level road with a
uniform speed of 60 km/h. It is followed by another
car B which is moving with a speed of 70 km/h.
When the distance between them is 2.5 km, the car
B is given a deceleration of 20 km/h2. The distance
after which B catch up with A
(a)
33.5 km
(b)
22.5 km
(c)
11.5 km
(d)
44.5 km
15.
A body is in straight line motion with an
acceleration given by a = 32 – 4v. At t = 0 the
velocity of the particle is 4 unit. The velocity when
t = ln 2 is
(a)
15/2
(b)
17/2
(c)
23/4
(d)
31/4
A body is projected vertically upwards with
velocity ‘u’. If t1 and t2 be the times at which it is at
height h above the point of projection while
ascending and descending respectively, then
(a)
h  gt 1t 2 , u  g(t 1  t 2 )
(b)
h
(c)
h  gt 1t 2 , u 
(d)
h
1
gt 1 t 2 , u  g(t 1  t 2 )
2
1
g( t 1  t 2 )
2
1
1
gt 1 t 2 , u  g(t 1  t 2 )
2
2
A parachutist drops freely from an aeroplane for
10 s before the parachute opens out. Then he
descends with a net retardation of 2.5 ms–2. If he
bails out of the plane at a height of 2495 m and
g = 10 ms–2, his velocity on reaching the ground will
be
(a)
2.5 ms–1
(b)
7.5 ms–1
(c)
5 ms–1
(d)
10 ms–1
Water drops fall at regular intervals from a roof.
At an instant when a drop is about to leave the roof,
the separations between successive drops below the
roof are in the ratio
(a)
1:2:3:4
(b)
(c)
1:3:5:7
(d)
1 : 4 : 9 : 16
1 : 5 : 13 : 21
A baloon starts rising from the ground with an
acceleration of 1.25 m/s2. After 8s, a stone is released
from the baloon. Take the ground as origin and
vertical upward direction as negative. The velocity
(v) time (t) graph for the particle during the
interval t = 8 s to t = 10 s is given by
(a)
(c)
Einstein Classes,
(b)
(d)
SINGLE CORRECT CHOICE TYPE
(ANSWERS)
1.
a
2.
d
3.
d
4.
d
5.
b
6.
d
7.
c
8.
a
9.
b
10.
a
11.
d
12.
c
13.
c
14.
a
15.
d
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 13
EXERCISE BASED ON NEW PATTERN
5.
COMPREHENSION TYPE
Comprehension-1
A particle accelerates from rest at a constant rate 
for some time after which it decelerates at a
constant rate  to come to rest. Its total time elapsed
is T then
1.
The displacement time graph of the particle is given
by
The value of H in terms of v0 and g so that at the
instant when the balls collide, the first ball is at the
highest point of its motion is
(a)
v02
2g
(c)
v02
4g
(b)
v02
3g
(d)
v02
g
Comprehension-3
(a)
(b)
A particle moves in the x-y plane with constant
acceleration  directed along the negative direction
of the y-axis. The equation of the trajectory of the
particle is y = ax – bx2 where a and b are constants.
6.
(c)
2.
3.
(d)
The maximum velocity acquired by the particle is
7.
The angle made by the velocity vector with the
x-axis at the origin is
(a)
tan–1a
(c)
tan–1
  

T
 
(b)
   T


  2
(a)
(c)
   T


 4
(d)
   T


 8
(c)
8.
(a)
  

T
 
(c)
   T


 4
tan–1b
(d)
tan–1
b
a
The speed of the particle at the origin is
(a)
The average velocity is
a
b
(b)

2b
2a

2b
(b)
(d)

2b
a
(1  a 2 )
The position along the y-axis at which vy is zero.
(b)
   T


  2
(a)
a2
b
(b)
a2
4b
(c)
a2
2b
(d)
(d)
   T


 8
a2
8b
Comprehension-4
Comprehension-2
A ball is thrown straight up from the ground with
speed v 0. At the same instant a second ball is
dropped from rest from a height H, directly above
the point where the first ball was thrown upward.
There is no air resistance.
4.
The time at which the two ball collide is
(a)
H
2v 0
(b)
H
4v 0
(c)
H
v0
(d)
H
3v 0
Einstein Classes,

2b
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 14
Figure shows a demonstration which involves a
blow gun G, Gun G projects ball as a projectile. The
target is a can suspended from a magnet M and the
tube of the blow Gun is aimed directly at the can.
Neglect the dimension of the can, that means can
be treated as particle. In one experiment the ball is
projected from the gun but the can is not released.
In another experiment the magnet releases the can
just as the ball leaves the blow gun.
9.
10.
If the free fall acceleration ‘g’ is assumed to be zero
then the ball will follow
12.
13.
18.
(a)
2.75 m
(b)
3.00 m
(c)
3.20 m
(d)
3.50 m
The time of flight of this ball
(a)
1s
(b)
1.2 s
(c)
1.4 s
(d)
1.6 s
The horizontal range of this ball is
(a)
4.8 m
(b)
7.2 m
(b)
projectile path
(c)
9.6 m
(d)
10.8 m
(c)
circular path
MATRIX-MATCH TYPE
(d)
hyperbolic path
Matching-1
If g is assumed to be zero then in which experiment
the ball hits the can ?
(a)
In the first experiment
(b)
In the second experiment
(c)
In both experiment
(d)
None of these
Column - A
Column - B
(A)
One dimensional
motion
(P)
Circular
motion
(B)
Two dimensional
motion
(Q)
Parabolic
motion
(C)
Vector quantity
(R)
Straight line
(D)
Scalar quantity
(S)
Velocity
(T)
Speed
In which experiment the ball hits the can ?
Matching-2
(a)
In the first experiment
(b)
In the second experiment
(c)
In both experiment
(d)
None of these
What is the path followed by the balls in each
experiment ?
(a)
Straight line path
(b)
Parabolic path
(c)
Circular path
(d)
Hyperbolic path
Column - A
Column - B
(A)
Less than one
(P)
Average
velocity
(B)
May be zero
(Q)
Average speed
(C)
Never be zero
(R)
Ratio of
magnitude of
average
velocity &
average speed
(D)
Always positive
(S)
Displacement
(T)
Distance
Let if the ball hits the can at a height of 2.75 m above
the ground in the second experiment then
Matching-3
The horizontal speed of the ball at any moment
before the hitting
The position of a particle moving on x-axis is given
by x = t3 + 4t2 – 2t + 4
(c)
15.
17.
The maximum height attained by the ball
straight line path
(a)
14.
16.
(a)
for the following questions ‘g’ equals to 10 m/s2.
11.
In the first experiment, if the ball has the same
projection speed as the ball in second experiment then
3 m/s
6 m/s
(b)
(d)
Column-A
Column-B
(A)
The velocity of the
particle at t = 4s
(P)
78
(B)
The acceleration of
the particle at t = 4s
(Q)
32
(C)
The average velocity of
the particle at t = 0 to
t = 4s
(R)
30
(D)
The average acceleration (S)
during the interval t = 0
to 4s
10
4.5 m/s
7.5 m/s
The speed of projection of the ball is
(a)
10 m/s
(b)
11 m/s
(c)
12 m/s
(d)
15 m/s
The time after which the ball hits the can
(a)
1
s
4
(b)
1
s
2
(c)
1s
(d)
1.25 s
(T)
Einstein Classes,
none
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 15
Matching-4
Column-A
Column-B
(A)
For a moving particle
(P)
the average velocity may be
Positive
(B)
For a moving particle the (Q)
average speed must be
Negative
(C)
The speed of a particle
(R)
at the highest point
projected at certain angle
with the vertical
Zero
(D)
The ratio of
displaceme nt
(S)
dis tan ce
(D)
(T)
MULTIPLE CORRECT CHOICE TYPE
1.
less then unity
must be
(T)
(S)
more than
unity
A bird flies for 4 sec with a velocity of (t – 2) m/s in
a straight line, where t = time in seconds. Choose
the correct statements
(a)
The average speed is 1m/s
(b)
The average velocity is 0
(c)
The average speed is ½ m/s
(d)
The distance travelled during 1s to 3s is
1m
Matching-5
A projectile is projected with certain speed at some
angle  with the horizontal.
Column-A
2.
Column-B
(A)
The maximum range for (P)
 is equal to
tan–14
(B)
The range and maximum (Q)
height are equal then  =
/4
(C)
The angle between
(R)
velocity and acceleration
at highest point
/2
(D)
The angle between
(S)
velocity and acceleration
at the point of projection
/3
(T)
3.
more than /2
Matching-6
Column-A
(A)
Column-B
(P)
4.
(B)
(Q)
(C)
(R)
Einstein Classes,
Four particles A, B, C and D are thrown from the
top of a tower. A is thrown straight up with speed
u, B is thrown straight down with the same speed
u, C is thrown horizontally with the same speed u
and D is released from rest. They hit the ground
with speed vA, vB, vC and vD respectively and time of
flight are tA, tB, tC and tD respectively. Choose the
correct statements from the following
(a)
v A = v B = vC
(b)
tD  tAtB
(c)
tD = tC
(d)
vD < vA
Two particle are projected simultaneously in the
same vertical plane from the same point, with
different speeds u1 and u2, making angles 1 and 2
respectively with the horizontal. The path followed
by one, as seen by the other (as long as both are in
flight) is
(a)
a vertical straight line if u1cos1 = u2cos2
(b)
a straight line if u1cos1  u2cos2
(c)
a parabola
(d)
a hyperbola
A particle is thrown with a speed u at an angle 
with the horizontal. When the particle makes an
angle  with the horizontal, its speed becomes v.
Then :
(a)
the speed at the highest point is ucos
(b)
the speed at the highest point is vcos
(c)
v = u cos .sec 
(d)
v = u sec .cos 
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 16
5.
A river is flowing from west to east at a speed of u.
A man on the south bank of the river, capable of
swimming at v with respect to river. The width of
the river is l. Choose the correct statement.
(a)
9.
x = t2 – t3.
If the man wants to swim across the river
in the shortest time, he should swim
due north.
(b)
The particle will come to rest after time
2/3.
(c)
(c)
If the man wants to swim across the river
in the shortest distance, he should
The initial velocity of the particle was zero
but its initial acceleration was not
zero.
(d)
none of these
(d)
10.
all are correct
If y = ax – bx2 is the path of a projectile, then which
of the following is correct
(a)
Range = a/b
(b)
Maximum height = a2/4b
At a certain moment a particle moves towards north
at a speed of 7 m/s. whereas its acceleration
2.8 m/s2 acting towards south. Choose the incorrect statement from the following
(a)
The velocity of the particle at t = 5s is
7m/s towards south.
(b)
The distance covered by the particle in
third second is 3.5 m.
(c)
The average speed of the particle is 7m/s
during t = 0 to t = 5s.
(d)
The average velocity of the particle from
t = 0 to t = 5s is negative.
–1
(c)
Angle of projection = tan a
(d)
at x = a/2b, vy = 0
Choose the correct statement from the following for
a projectile projected from the ground at certain
angle with the horizontal in the vertical plane.
(a)
8.
The particle will return to its starting
point after time /
If the man wants to swim across the river
in the shortest distance, he should
swim due north.
u
swim sin   north of west.
v
7.
(a)
(b)
1
6.
The displacement (x) of a particle depends on time
(t) as
11.
The angle between the velocity vector and
acceleration vector at the highest point
is /2.
(b)
The minimum speed at the highest point
equals to the initial horizontal speed.
(c)
The maximum horizontal range for the
projectile is at the angle of projection of
/4.
(d)
The horizontal speed is constant
The graph between the displacement x and time t
for a particle moving in a straight line is shown in
the diagram. During the intervals OA, AB, BC and
CD the acceleration of the particle is
12.
A particle starts with a velocity of 200 cm/s and
moves in a straight line with a retardation of
10 cm/s 2 . The time it takes to describe a
displacement of 1500 cm
(a)
10 s
(b)
15 s
(c)
25 s
(d)
30 s
Two masses A and B are moving in the same straight
line. A moves with a uniform velocity of 11 m/s ; B
starts from rest at the instant when it is 52.5 m ahead
of A and moves with a uniform acceleration of
1 m/s2. The time afterwhich, they will catch each
other
(a)
7s
(b)
9s
(c)
12s
(d)
15s
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(a)
The acceleration in OA is negative
(b)
The acceleration in AB is zero
(C)
Statement-1 is True, Statement-2 is False
(c)
The acceleration in BC is positive
(D)
Statement-1 is False, Statement-2 is True
(d)
The acceleration in CD is zero
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 17
1.
STATEMENT-1 : Consider a particle initially
moving with velocity of 5 m/s, starts decelerating
at a constant r ate of 2m/s2 along x-axis. The
distance travelled in the 3rd second is 0.5 m.
5.
STATEMENT-1 : The driver of a train moving at
a speed v1 sights a goods train a distance d ahead
of him on the same track moving in the same
direction with a slower speed v2. He puts on brakes
and gives his train a constant retardation a. There
STATEMENT-2 : The magnitude of displacement
is always equal to distance.
2.
will be no collision if d 
STATEMENT-1 : A particle P is projected with
velocity u from the origin at an angle  with xaxis. Another particle is projected with the same
velocity from the position (l, 0) at an angle  with
the x-axis. The locus of P is straight line as seen
from Q.
STATEMENT-2 : v2 = u2 + 2ax is valid only for
constant acceleration.
6.
STATEMENT-1 : From the top of a building a
ball is dropped while another is thrown
horizontally at the same time. Both will reach the
ground simultaneously.
STATEMENT-2 : Both particles have parabolic
path when they are in motion w.r.t. ground.
3.
STATEMENT-2 : They will have different speed
with which they will strike the ground.
STATEMENT-1 : Two particles A and B start from
rest and move for equal time on a straight line.
The particle A has an acceleration a for the first
half of the total time and 2a for the second half.
The particle B has an acceleration 2a for the first
half and a for the second half. Particle A has
covered larger distance.
7.
STATEMENT-1 : A particle projected with speed
‘u’ at an angle of  with the horizontal in the
vertical plane. Another particle project with the
same speed at an angle of  with the vertical in
the same plane. They have the same horizontal
range.
STATEMENT-2 : The velocity time graph for the
constant acceleration is a straight line with
non-zero slope.
4.
(v1  v 2 )2
2a
STATEMENT-2 : They have the same time of
flight.


dv d v

STATEMENT-1 :
dt
dt
STATEMENT-2 : For straight line motion the
acceleration equals to zero for constant speed.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
a
2.
a
3.
b
4.
c
5.
d
6.
a
7.
d
8.
b
9.
a
10.
c
11.
b
12.
b
13.
c
14.
a
15.
b
16.
c
17.
d
18.
c
MATRIX-MATCH TYPE
1.
[A-R ; B-P, Q ; C-S ; D-T]
2.
[A-P, Q, R, S, T ; B-P, S, R ; C-Q, T ; D-Q, T]
3.
[A-P, B-Q, C-R, D-T]
4.
[A-P, Q, R, S, T ; B-P,S,T ; C-P, S, T ; D-S]
5.
[A-Q ; B-P ; C-R ; D-T]
6.
[A-P, R ; B-Q, S ; C-Q ; D-T]
MULTIPLE CORRECT CHOICE TYPE
1.
a, b, d
2.
a, b, c, d
3.
a, b
4.
a, b, c
5.
a, c
6.
a, b, c, d
7.
a, b, c, d 8.
a, b, c, d
9.
a, b, c
10.
b, c, d
11.
a, d
12.
a, d
D
4.
D
5.
D
6.
B
ASSERTION-REASON TYPE
1.
C
7.
C
2.
B
Einstein Classes,
3.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 18
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
A ball is dropped from a height of 19.6 m above the
ground. It rebounds from the ground and raises
itself up to the same height. Take the statring point
as the origin and vertically downward as the
positive x-axis. Draw the approximate plots of x
versus t, v versus t and a versus t. Neglect the small
interval during which the ball was in contact with
the ground.
7.
Two cyclists move towards each other. The first
cyclist, whose initial velocity 5.4 km/h. descends the
hill, gathering speed with an acceleration of
0.2 m/s2. The second cyclist whose initial velocity
18 km/h climbs the hill with an acceleration
–0.2 m/s2. How long does it take for the cyclist to
meet if the distance separating them at the initial
moment of time is 195 m ?
2.
A body transversed half the distance with velocity
v0. The remaining part of the distance was covered
with velocity v 1 for half of the time, and with
velocity v2 for the other half of the time. Find the
mean velocity of the body averaged over the whole
time of motion ?
8.
(a)
A boat can travel at 10m/s relative to
water. It starts at one bank of river that
is 100 m wide and flows with a velocity
7.5 m/s. If the boat points directly across,
find (a) its velocity relative to the bank.
(b) how far downstream it travels.
3.
A particle starts moving along positive x-axis with
constant acceleration  from origin. At a time to
after the beginning of motion, the acceleration
reverses its direction remaining the same in
magnitude. Find the time t from the beginning of
motion in which the displacement of the particle
becomes zero ?
(b)
In which direction the boat must point
so as to cross the river in a direction
perpendicular to the river currents. Also
find the time taken to cross the river.
4.
A basket ball player throws a ball with initial
velocity v0 at  above the horizontal to the hoop
which is located a horizontal distance L and at a
height h above the point of release.
(a)
A gun shoots bullets that leave the muzzle at 250
m/s. If the bullet is to hit a target 100 m away at the
level of the muzzle, the gun must be aimed at a point
above the target. How far above the target is this
point ?
10.
A projectile is fired with a velocity ‘u’ at right angle
to the slope which is inclined at an angle  with the
horizontal. Find the range along the inclined
plane ?
11.
To a man walking at the rate of 3km/h the rain
appears to fall vertically. When he increases his
speed to 6 km/h it appears to meet him at an angle
of 450 with vertical from the front. Find the actual
speed of the rain ?
12.
A particle is projected so as to graze the tops of two
walls each of height 20 m at distances of 30 m and
170 m respectively from the point of projection.
Find the angle of projection.
13.
Two bodies move towards each other in a straight
line at initial velocities v1 & v2 and with constant
acceleration a 1 and a 2 directed against the
corresponding velocities at the initial instant. What
must be the maximum initial separation l max
between the bodies for which they meet during the
motion ?
14.
A point moves in the xy plane according to the law
x = asint, y = a(1 – cost), where and  are
constants. Find (a) the distance transvered by the
point during the time . (b) the angle between the
point’s velocity and acceleration.
Prove that the initial speed required is
v 02 
(b)
9.
gL
.
h

2
2 cos   tan   
L

Prove that the angle  to the horizontal
at which it reaches the loop is
tan  
2h
 tan  .
L
5.
A projectile is projected with velocity v from the
origin at an angle  with x-axis. Another particle B
is projected with the same velocity from the
position (l, –h, 0) at an angle  with the negative
x-axis such that the two particles move towards each
other. Find the time after which the separation
between the particles is minimum. Also, find the
minimum distance between the particles.
6.
A car goes out of control and slides off a steep
embankment of height h at  to the horizontal It
lands in a ditch at distance R from the base. Find
the speed at which the car leaves the slope.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 19
15.
16.
17.
The dependence of a particle’s speed v on the
distance s it has travelled is determined by the
function v = v0 – bs, where b is the constant.
(i)
Find how s depends on the time t.
(ii)
Determine the dependence of v on t.
Six particles situated at the corners of a regular
hexagon of side l move at a constant speed v. Each
particle maintains a direction towards the particle
at the next corner. Find the time the particle will
take to meet each other.
A body falling freely from a given height H hits on
inclined plane in its path at a height h. As a result
of this impact the direction of the velocity becomes
(a)
23.
h
 the body will
H
horizontal. For what value of 
take maximum time to reach the ground ?
18.
Two guns, situated at the top of a hill of height 10
m, fire one shot each with the same speed 53 m/s.
at some interval of time. One gun fires horizontally
and other fires upwards at an angle of 600 with the
horizontal. The shots collide in air at a point P. Find
(a) the time interval between the firings, and (b)
the coordinates of the point P. Take origin of the
coordinates system at the foot of the hill right
below the muzzle and trajectories in X-Y plane.
24.
19.
A particle is projected with speed u at an angle 
with the x-axis in the xy plane. Find the average
velocity of the particle during its (a) total time of
flight (b) half time of flight, from time of
projection to the time at which particle attains
maximum y-coordinates.
25.
20.
A car starts moving rectilinearly, first with
acceleration a = 5.0 m/s2 (the initial velocity is equal
to zero), then uniformly, and finally, decelerating
at the same rate a, comes to stop. The total time of
motion equals t = 25 sec. The average velocity
during that time is equal to v = 72 km/hr. How long
does the car move uniformly.
26.
21.
A ball rolls down from the top of a staircase with
some horizontal speed u. If the height and width of
the steps are h and b respectively, then show that
ball will just strike the edge of nth step directly
(without hitting any step in between) if n 
22.
2hu 2
gb 2
.
A ball is projected from origin with an initial
velocity v0 = 700 cm/s in a direction 370 above the
horizontal as shown in figure. Another ball B 300
cm from origin on a line 370 above the horizontal is
released from rest at the instant A starts. (g = 9.8
m/s2)
Einstein Classes,
27.
28.
How far will B have fallen when it is hit
by A ?
(b)
In what direction is A moving when it hits
B?
A man is sitting on the shore of a river. He is in the
line of a 1.0 m long boat and is 5.5 m away from the
centre of the boat. He wished to throw and apple
into the boat. If he can throw the apple only with a
speed of 10 m/s, find the range of angle of
projection for successful shot. Assume that the point
of projection and the edge of the boat are in the
same horizontal level.
A river 400 m wide is flowing at a rate of 2.0 m/s. A
boat is sailing at a velocity of 10 m/s with respect to
the water, in a direction perpendicular to the river.
(a)
Find the time taken by the boat to reach
the opposite bank.
(b)
How far from the point directly opposite
to the starting point does the boat reaches
the opposite bank ?
A motor-boat going down stream overcome a float
at a point M. 60 minutes later it turned back & after
some time passed the float at a distance of 6 km
from the point M. Find the velocity of the stream
assuming a constant velocity for the motor-boat in
still water.
A very broad elevator is going down vertically with
a constant acceleration 1 ms–2. At the instant when
the velocity of the lift is 2 m/s, a stone is projected
from the floor of the lift with a speed of 2 m/s
relative to the floor at an elevation 300. Find
(a)
the time taken by the stone to return to
the floor.
(b)
the range of the stone over the floor of
the lift. [g = 10 ms–2]
An elevator car whose floor-to-ceiling distance is
equal to 2.7 m starts ascending with constant
acceleration 1.2 m/s2; 2.0 after start a bolt begins
falling from the ceiling of the car. Find :
(a)
free fall time of the bolt ;
(b)
the displacement and the distance
covered by the bolt during the free fall in
the reference frame fixed to the ground.
A projectile is fired from a point on a cliff to hit a
mark 10 m horizontally from the point and 10 m
vertically below it. The velocity of projection is
equal to that due to falling freely under gravity
through 5 m from rest. Show that the two possible
directions are at right angles and the time of flight
are approximately 2.6 sec & 1.08 sec. [g = 10 ms–2].
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 20
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
A stone is projected from a point on the ground in
such a direction so as to hit a bird on the top of a
telegraph post of height h and then attain a
maximum height 2h above the ground. If, at the
instant of projection, the bird were to fly away
horizontally with a uniform speed, find the ratio
between the horizontal velocities of the bird and
the stone, if the stone still hits the bird.
2.
The graph of velocity v/s time for a block as shown.
The area of OABC is 83 m.
(a) Prove that journey is covered in least time if it is
accelerated for a time 6s. (b) Hence or otherwise
find the value of minimum journey time.
3.
A particle is projected with a velocity
6.
A hunter is riding an elephant of height 4m moving
in straight line with uniform speed of 2m/sec. He
sights a deer running with a speed V in front at a
distance 45m moving perpendicular to the
direction of motion of the elephant. If hunter can
throw his spear with a speed of 10m/sec. relative to
the elephant, then at what angle  to it’s direction
of motion must he throw his spear horizontally for
a successful hit. Find also the speed ‘V’ of the deer.
7.
Two swimmers leave point A on one bank of the
river to reach point B lying across on the other bank.
One of them crosses the river along the straight line
AB while the other swims at right angles to the
stream and then walks the distance that he has been
carried away by the stream to get to point B. What
was the velocity u of his walking if both swimmers
reached the destination simultaneously ? The
stream velocity v0 = 2.0 km/hr and the velocity v’ of
each swimmer with respect to water equals
2.5 km/hr.
8.
Two boats A and B, moves away from a buoy
anchored at the middle of a river along the
mutually perpendicular straight lines; the boat A
along the river, and the boat B across the river.
Having moved off an equal distance from the buoy
the boats returned. Find the ratio of times of
motion of boats A/B if the velocity of each boat
with respect to water  = 1.2 times greater than the
stream velocity.
9.
A radius vector of a point A relative to the origin
varies with time t as r = ati – bt2j, where a and b are
positive constants, and i and j are the unit vectors
of the x and y axes. Find :
2 ag so that
it just clears two walls of equal height a which are
at a distance ‘2a’ apart. Prove that the time of
passing between the wall is 2 a .
g
4.
5.
The current velocity of a river grows in proportion
to the distance from its bank and reaches its
maximum value v0 in the middle. Near the bank,
the velocity is zero. A boat is so moving in the river
that its velocity u relative to the water is constant
and perpendicular to the current. Find the distance
through which the boat crossing the river will be
carried away by the current if the width of the river
is d. Also determine the trajectory of the boat ?
A large heavy box is sliding without friction down
a smooth plane of inclination .
From a point P on the bottom of a box, a particle is
projected inside the box. The initial speed of the
particle with respect to box is u and the direction of
projection makes on angle  with the bottom as
shown. (a) Find the distance along the bottom of
the box between the point of projection P and the
point Q where the particle lands. (b) If the
horizontal displacement of the particle as seen by
an observer on the ground is zero, find the speed of
the box with respect to the ground at the instant
when the particle was projected.
Einstein Classes,
10.
(a)
the equation of the point’s trajectory y(x);
plot this function;
(b)
the time dependence of the velocity v and
acceleration w vectors, as well as of
the moduli of these quantities ;
(c)
the time dependence of the angle 
between the vectors w and v;
(d)
the mean velocity vector averaged over
the first t seconds of motion, and the
modulus of this vector.
A particle is moving in a plane with velocity given

by : v  u 0 î  a cos t ĵ . If the particle is at the
origin at t = 0. (a) calculate the trajectory of the
particle; (b) find its distance from the origin at time
(3/2)
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 21
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
2.
2v 0 ( v1  v 2 )
2v 0  v1  v 2
5.
l
,h
2 v cos 
6.
R
cos 
7.
3.
2 t0 (1 + 2)
30 s
8.
(a)
(b)
9.
0.8 m
10.
2v 2
tan sec 
g
11.
32 km/h
12.
tan  
40
51
13.
( v1  v 2 ) 2
2(a1  a 2 )
14.
(a)
a
15.
(a)
(b)
v = v0e–bt
16.
2a
v
17.
½
18.
(a)
1s
(b)
(53 m, 5m)
19.
(a)
u cos 
(b)
u cos  î +
20.
 t  t 1  4 v / at  15 sec .
22.
24.
25.
(a)
90 cm, (b) horizontal
(a)
40 s,
(b) 80 m
3 km/hr
26.
(a)
0.22 sec, (b)
27.
(a)
0.7 s, (b) 0.7 & 1.3 m respectively
g
2(h  R tan )
s
v0
(1  e bt )
b
Einstein Classes,
2
(a)
(b)
12.5 m/s
6.61 m/s
(b)
(b)
75 m

2
u sin 
ĵ
2
23.
150 <  < 18.50 and 71.50 <  < 750
m
3 3
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
PK – 22
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
4.
2
2 1
2.
(a)
(b)
u cos(    )
cos 
7.
u
v 0d 2 ud
d
,y 
x; y 
2u
v0
2
u 2 sin 2
g cos 
5.
(a)
6.
 = 370, v = 6m/s

8.
A / B 
9.
(a)
y = –x2b/a2;
(c)
tan  = a/2bt (d)
(a)
y = a sin (x/u0)
(b)
[a 2  ( 3u 0 / 2) 2
10.
8 sec.
2  1
Einstein Classes,
v0
2 

1  v0 

v  2 

 3.0 km / hr
1 / 2
1
 1 .8
(b)


v = ai – 2btj, w = –2bj, v =
a 2  4b 2 t 2 , w  2b
a 2  b 2t 2
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
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