1. No category

FYC5_CON 4_ Rocket propulsion _SYNOPSIS
Rocket propulsion
Rocket is an example of variable mass system.
In a rocket, the fuel burns and produces gases at high temperatures. These gases are
ejected out of the rocket from the nozzle at the back of the rocket and hence a forward
force (thrust) is exerted on the rocket.
This is a consequence of Newton’s third law of motion or conservation of linear
momentum.
Motion parameters of rocket
Let
m0 = initial mass of rocket
m = mass of rocket at any instant ‘t’ (instantaneous mass)
u = velocity of exhaust gases
v = velocity of rocket at any instant ‘t’ (instantaneous velocity)
rate of change of mass of rocket = rate of ejection of the fuel.
Thrust on rocket
If the rocket is to just leave the ground without acceleration then
If the rocket is to move up with acceleration ‘a’ then
Velocity of the rocket when the mass of the rocket is reduced from 2.303 FYC5_CON 6_ Friction, Coefficient of friction, Angle of friction, Motion on a rough horizontal surface,
Motion along curved track_SYNOPSIS
PAGE 4
Sliding of a chain on a horizontal table
Consider a uniform chain of mass ‘m’ and length ‘L’ lying on a horizontal table. Suppose
the chain is about to slide from the table, when of its length is hanging from the edge of
the table.
Under limiting equilibrium i.e., when the chain just tends to slip down,
|"#$ %&&'#(#) $(| |%*$)
$((
#) $((
#
+#|
∴ -.-/ 01 2345 ⇒ 01 If 01 is coefficient of limiting friction then the maximum fractional length of chain hanging
from the edge of the table in equilibrium is ________
ℓ:;:<=>?=?
@
@
A
A:
: B
If 0- is coefficient of limiting friction then minimum fractional length of chain that can be on
the table is ________
ℓC>DEFG=
-.-/ -.-/
ℓ-.-/
2345
2345
8 9 ℓ-.-/
A
: B
Equilibrium condition for body placed on a rotating disc
Consider a body of mass ‘m’ placed at a distance ‘x’ from the centre of a circular disc. The
disc is rotating in a horizontal plane with a constant angular velocity ‘H’ about a vertical
axis passing through its centre.
For the body to be under equilibrium,
IJKLM.N35 I |$$
$(%*$)
$(|
∴ OHP 0- ⟹ H R
0- O
,T$#'#*$%*
$( T 2UV
O
0- Equilibrium condition for block pressed against a vertical wall
A body of mass ‘m’ is pressed against a vertical wall with a horizontal force ‘F’.
If the coefficient of static friction is 0- , then the equilibrium condition is
|weight of block| = |limiting friction|
∴ 0- Equilibrium condition for book pressed between two hands
Suppose a book is pressed between two hands, by each hand with a force F.
If the coefficient of static friction is 0- , then the equilibrium condition is
|weight of book| = |limiting friction|
∴ 0- 01 2
Equilibrium condition for block on front surface of truck
A block of mass ‘m’ is placed on the front surface of a truck.
The minimum acceleration with which the truck should travel, so that the body on its front
surface may not slide down is _________
Under equilibrium condition
|weight of block| = |limiting friction|
∴ 01 W 01 N
⇒ A
:
Body placed in contact with inner wall of rotating hollow Cylinder
Consider a hollow cylinder of radius ‘r’ rotating about a vertical axis passing through its
centre and parallel to its length with a constant angular velocity ‘H’. A block of mass ‘m’ is
in contact with the inner wall of the cylinder.
For equilibrium condition of block
|weight of block| = |limiting friction|
∴ 01 W 01 *HP N
⇒ H RA K ,T$#'#*$$$
T 2UR
:
A: K
N
A block of mass ‘m’ is placed in contact with vertical side of a body of mass ‘M’ as shown
in the figure. The coefficient of static friction between m and M is X. The least horizontal
force with which m is to be pushed so that the two blocks move together is _______
(neglect friction between M and ground)
acceleration of system = a = BZ
Y
ZY
normal reaction between the blocks = N = [ ZB
If ‘m’ is in equilibrium then,
mg =0N
ZY
∴ 0 ZB
⇒ NZB
AZ
Contd…….
FYC7_ SYSTEM OF PARTICLES AND ROTATIONAL MOTION_FORMULAE
Page 2
‘n’ number of particles of masses m, 2m, 3m, ....nm are at distances x1= 1, x2 = 4, x3 = 9, ......
xn = n2 units respectively from origin on the x-axis. The distance of their centre of mass
from origin is________
………. Position vector of centre of mass
……….
……… …….
n-particles having masses m1, m2, .....mn are with position vectors , , … … … . respectively. The position vector ( ) of their C.M is
………. ………
∑
!
Position vector of Centre of mass of continuous mass distribution.
A rigid body may be treated as a continuous distribution of matter.
The ‘particles’ then becomes differential mass elements ‘dm’ and the summation Σ
becomes integration #.
The integrals are evaluated for all mass elements in the object such that
# $%
!
& ! # &$%
' ! # '$%
In case of continuous distribution of mass to locate the position vector of centre of mass we
use the formula
# $% 1
* $% )
# $%
Few important applications for location of position of centre of mass
If two circular discs of radii ‘r1‘ and ‘r2‘ of same material, same thickness are kept in contact
then the distance of centre of mass of system from centre of a disc of radius ‘r1‘ is given by
+
, - + , - If two solid spheres of radii ‘r1‘ and ‘r2‘ of same material are kept in contact, then the
distance of centre of mass of the system from centre of a sphere of radius ‘r1’ is given by
+
, - +
, - For an equilateral triangle the centre of mass is at its centroid.
Distance of centre of mass of a uniform cone of height ‘h’ and base radius R, from the vertex
.
on the line of symmetry is
.
Formula to find shift in centre of mass when a small portion of mass is removed from a
uniform body.
Suppose a small portion of mass is removed from a uniform body as shown in the figure.
98694=6>
? >
9869:17>68
Where
> @@A = distance between original centre and centre of removed mass.
Negative sign implies that the shift of centre of mass is away from the side about which
mass is removed.
/012342567386429:// ; <
9
|Shift of Centre of mass| = +9 8694=6> , >
869:17>68
A circular portion of radius ‘r’ is removed from one edge of a circular plate of radius ‘R’
with uniform thickness, as shown in the figure. The shift in centre of mass is __________
LMNOPMQ
BCDEFGEHIJFIGE%KBB ; +
LMNRQML
,$
Here,
%STUSV W CX , %SYZVS W[ CX ; W CX W[ ; CX
⇒ ^CDEFGEHIJFIGE%KBB ; +_ ` , [ ; $ \\ [ ; Negative sign implies that the shift of centre of mass is away from the side about which
mass is removed.
Page 6
Moments of inertia of different bodies about different axes –Table
S.NO. Body
1.
2.
Thin rod (of mass
M and length l)
Circular ring (of
mass M and radius
R)
Axis
Moment of
Inertia
About an axis passing through the centre of
mass and perpendicular to its length
abc
About an axis passing through one end of the
rod and perpendicular to its length.
abc
About an axis passing through centre of ring
and perpendicular to its plane (generator axis
or transverse axis)
aec
About a tangent perpendicular to plane of
ring.
About a diameter of ring
Ac
d
caec
aec
c
,
daec
c
About a tangent in the plane of ring
3.
4.
Circular disc (of
mass M and radius
R)
Rectangular Lamina
(of mass ‘M’,
Length ‘l’, Breadth
‘b’)
About an axis passing through centre of disc
and perpendicular to its plane(generator axis
or Transverse axis).
About a tangent perpendicular to plane of
disc
aec
c
daec
c
About a diameter of disc
aec
About a tangent in the plane of disc
gaec
About an axis through its centre and parallel
to the length in its plane
ahc
Ac
About an axis through its centre and parallel
to the breadth in its plane
abc
Ac
About an axis through its centre and
perpendicular to its plane
a+
About an axis through its edge and parallel to
its length in its plane
!i About an axis through its edge and parallel to
its breadth in its plane.
abc
About an axis through its corner and
perpendicular to its plane
a
f
f
bc hc
,
Ac
d
d
bc - hc S.NO.
5.
Body
Axis
Square lamina (of
About an axis through its centre and parallel
mass M and side a)
to any side in its plane
About an axis through its centre and
Moment of
Inertia
a:c
Ac
a:c
j
perpendicular to its plane
About an axis through its edge in its plane
a:c
d
About an axis through its diagonal in its plane a:c
Ac
About an axis through its corner and
ca:c
d
perpendicular to its plane
6.
c
aec
About a tangent
k
Hollow sphere (or)
About an axis passing through diameter
c
aec
Spherical shell (of
(generator axis)
Solid Sphere (of
About an axis passing through diameter
mass M and radius
(generator axis)
R)
7.
d
aec
g
aec
d
Solid cylinder (of
About the axis of symmetry (an axis passing
aec
mass M , radius R
through the centre along the length)
and length l)
About an axis perpendicular to length and
R)
passing through the centre
9.
g
About a tangent
mass M and radius
8.
g
Hollow cylinder (of
About the axis of symmetry (an axis passing
mass M , radius R
through the centre along the length)
and length l)
About an axis perpendicular to length and
passing through the centre
c
bc
a +Ac -
ec
,
ec
,
f
aec
lc
a +Ac -
c
Page 11
Comparison of translational and rotational motions
S.No.
1.
Translational motion
Rotational motion
Liner displacement = /
Angular displacement = m
2.
Linear velocity = =
3.
Linear acceleration = a
Angular acceleration = o
4.
Mass = m
Moment of inertia = I
5.
q 9=
q
Linear momentum = qp
qqq
Angular momentum = qr sn
6.
7.
Angular velocity = n
q
>u
Force = qt q Torque = v
>3
Work w # qt. qqqq
>/
A
uc
q
>r
>3
Work w # v
q. qqqqq
>m
A
8.
Translational KE = c 9=c c9
9.
Work –energy theorem
A
A
x t/ 9=c2 ; 9=c1
c
c
q
Power = p qt. =
Work – energy theorem
A
A
x vm snc2 ; snc1
c
c
qqq
Power = p v
q. n
= z - :3
n2 n1 - o3
10.
11.
12.
13.
14.
15.
Linear impulse y # t >3
{ z3 -
:3c z - =3
c
c
=c ; zc c:/
A
{7 z - : <7 ; ?
c
Rotational KE = c snc rc
cs
Angular impulse y # v >3
n1 - n2 A
m n1 3 - o3c 2
c
c
nc2 ; nc1 com
A
m7 n1 - o <7 ; ?
c
Rolling motion
Rolling
If a body rotates about a fixed axis, and at the same time if the axis of rotation translates then
the motion is called combined translatory and rotatory motion (or) rolling motion .
Contd……
FYC7_ SYSTEM OF PARTICLES & ROTATIONAL MOTION_CONCEPTUAL MCQS
Page 4
22.
23.
Out of the given bodies (of same mass) for which the moment of inertia will be
maximum about the axis passing through its centre of gravity and perpendicular to its
plane
(1) Disc of radius a
(2) Ring of radius a
(3) Square lamina of side 2a
(4) Four rods each of length 2a making a square
be the force acting on a particle having position vector . If be the torque of this
Let force about the origin, then
(1) . 0 . 0
(2) . 0 . 0
(3) . 0 . 0
(4) . 0 . 0
24.
If the torque acting upon a system is zero, then which of the following will be constant
(1) Force
(2) Linear momentum
(3) Angular momentum
(4) Linear impulse
____________________________________________________________________________________
Key : 22. (4)
23. (1)
24. (3)
__________________________________________________________________________________
Explanations:
22.
M.I of disc = M.I of ring = M.I of square lamina of side 2 2 2 M.I of square system of side length 2a = 4 23.
! #$ % Here, moment of inertia is maximum for square system of four rods.
' ∴ )* + ,-.-,/
So, . 0 . 0
24.
!"
01
02
.
If 0 then L = constant
&
25.
A constant torque of 1000 Nm, turns a wheel of moment of inertia 200 kgm2 about an
axis through the centre. Angular velocity of the wheel after 3 s will be
(2) 10* 5
(1)15* 5
26.
(4) 1* 5
The instantaneous angular position of a point on a rotating wheel is given by the
equation 6 789 : ;87. The torque on the wheel becomes zero at
(1) t = 2s
27.
(3) 5* 5
(2) t = 1s
(3) t = 0.2s
(4) t = 0.25s
A thin circular disc of mass M and radius R is rotating in a horizontal plane about an axis
passing through its centre and perpendicular to its plane with an angular velocity <. If
another disc of same dimensions but of mass
=
>
is placed gently on the first disc co-
axially, then the new angular velocity of the system is
?
(1) @
(2) @
(3) ? @
(4) @
__________________________________________________________________________________
Key : 25. (1)
26. (2)
27. (3)
__________________________________________________________________________________
Explanations:
25.
Step 1
B
A
C
DDD
DD
5* 5
Step 2
@E @F A,
∴ @E 0 530 15* 5
26.
Torque zero means, the angular acceleration A is zero.
Here,
∴@
H 2, : 6, 0J
02
6, : 12,,A 0L
02
For the given condition, A 0
12, : 12
∴ 12, : 12 0 ⇒t = 1 second.
27.
By conservation of angular momentum
@ @ ∴ N @ O N # $ N P @ ⇒ @ ? @
28.
29.
What remains constant in the field of central force
(1) Potential energy
(2) Kinetic energy
(3) Angular momentum
(4) Linear momentum
Two bodies have their moments of inertia I and 2I respectively about their axis
of rotation. If their kinetic energies of rotation are equal, then their angular momenta
will be in the ratio
(2)√2: 1
(1) 1:2
30.
31.
(4) 1 ∶ √2
(3) 2 : 1
Before jumping in water from above a swimmer bends his body to
(1) Increase moment of inertia
(2) Decrease moment of inertia
(3) Decrease the angular momentum
(4) Reduce the angular velocity
A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with angular velocity
of 7TUVW5X . Its kinetic energy is
(1) 10 J
(2) 100 J
(3) 500 J
(4) 250 J
____________________________________________________________________________________
Key : 28. (3)
29. (4)
30. (2)
31. (4)
____________________________________________________________________________________
Explanations:
28.
29.
In the field central force, Torque = 0
* ∴ angular momentum remains constant.
Y
1"
C
∴ Z [2Y
If E is constant then,
1\
1"
C
C
]C\ ]C "
√
30.
In doing so moment of inertia is decreased and hence angular velocity is increased.
31.
Rotational kinetic energy = @ # N $ @ O ' 10 ' 0.5 P 20 250^
32.
Two discs of same thickness but of different radii are made of two different materials
such that their masses are same. The densities of the materials are in the ratio 1:3.The
moments of inertia of these discs about the respective axes passing through their centres
and perpendicular to their planes will be in the ratio
(1) 1 : 3
33.
(3) 1 : 9
(4) 9 : 1
A thin circular ring of mass M and radius r is rotating about its axis with a constant
angular velocity <. Two objects each of mass m are attached gently to the opposite ends
of a diameter of the ring. The ring will now rotate with an angular velocity
(1)
34.
(2) 3 : 1
L5_
L
(2)
`_
(3)
`_
L
`_
(4)
L`_
A child is standing with folded hands at the centre of a platform rotating about its central
axis. The kinetic energy of the system is K. The child now stretches his arms so that the
moment of inertia of the system doubles. The kinetic energy of the system now is
(1) 2K
a
(2)
(3)
a
(4) 4K
____________________________________________________________________________________
Key : 32. (2)
33. (2)
34. (2)
____________________________________________________________________________________
Explanations:
32.
"
M.I. of disc = N #b2c$ b2c
∵ eN ,f
If mass and thickness are same then,
C
c
∝ c ∴ C\ c" 33.
"
\
Initial angular momentum of ring = Z @ N @
Final angular momentum of system (Ring + Two particles) = N 2N @′
As there is no external torque on the system, angular momentum is constant
∴ N @ N 2N @i
L
∴ @i `_
34.
1"
j)k,)lYkmn j C o
Here L = constant ∴ j ∝ C
a
C
C
∴ a" C\ C\ ⇒ j \
"
\
a
Contd………
PREVIOUS AIPMT QUESTIONS DEMO
A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about horizontal axis. A
massless string is wound round the cylinder with one end attached to it and other hanging
freely. Tension in the string required to produce an angular acceleration of 2 revolutions
is _________
(AIPMT_2014)
(1) 25 N
(2) 50 N
(3) 78.5 N
(4) 157 N
PREVIOUS AIPMT QUESTIONS DEMO
The ratio of the acceleration for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an
incline of angle ′′ without slipping and slipping down the incline without rolling is
_________
(AIPMT_2014)
(1) 5 : 7
(2) 2 : 3
(3) 2 : 5
(4) 7 : 5
PREVIOUS AIPMT QUESTIONS DEMO
A rod PQ of mass ‘M’ and length ‘L’ is hinged at end ‘P’. The rod is kept horizontal by a
massless string tied to point Q as shown in figure. When string is cut, the initial angular
acceleration of the rod is _________
(NEET_2013)
(1)
(2)
(3)
(4)
PREVIOUS AIPMT QUESTIONS DEMO
A particle moves a distance x in time t according to equation . The
acceleration of particle is proportional to
(a) /
(b)
(c)
(d) /
PREVIOUS AIPMT QUESTIONS DEMO
A projectile is fired from the surface of the earth with a velocity of and angle with the horizontal. Another projectile fired from another planet with a velocity of
at the same angle follows a trajectory which is identical with the trajectory of the
projectile fired from the earth. The value of the acceleration due to gravity on the planet
is _____ ms-2 (given !"# = 9.8ms-2)
(AIPMT 2014)
1) 3.5
2) 5.9
3) 16.3
4) 110.8
PREVIOUS AIPMT QUESTIONS DEMO
A uniform force of $̂ &̂ newton acts on a particle of mass 2kg. Hence the particle is
( meter to position )$̂ &̂ * '
( meter. The work done
displaced from position $̂ '
by the force on the particle is :
(a) 6 J
(b) 13 J
(c) 15 J
(d) 9 J
PREVIOUS AIPMT QUESTIONS DEMO
A body of mass ‘m’ is taken from the earth’s surface to the height equal to twice the radius
(R) of the earth. The change in potential energy of body will be ________
(NEET_2013)
(1) +
(2) 3 mgR
(3) +
(4) 2 mgR
PREVIOUS EAMCET QUESTIONS DEMO
A particle of mass ‘m’ is attached to a thin uniform rod of length ‘a’ at a distance
of
from the mid point C as shown in the figure. The mass of the rod is ‘4m’. The
moment of inertia of the combined system about an axis passing through ‘O’ and
perpendicular to the rod is __________
(EAMCET_2013)
(1)
(2)
(3)
(4)
PREVIOUS EAMCET QUESTIONS DEMO
A stone tied to a rope is rotated in a vertical circle with uniform speed.
If the difference between the maximum and minimum tensions in the rope is 20N, mass
of the stone in kg is ___________ (EAMCET_2013)
(1) 0.5
(2) 0.75
(3) 1.0
(4) 1.5
PREVIOUS EAMCET QUESTIONS DEMO
Three uniform circular discs, each of mass ‘M’ and radius ‘R’ are kept in contact with
each other as shown in figure. Moment of inertia of the system about the axis AB
is ________
(EAMCET_2014)
(1)
(2)
(3)
(4)
PREVIOUS EAMCET QUESTIONS DEMO
A horizontal force F is applied to a block of mass m on a smooth fixed inclined plane of
inclination to the horizontal as shown in the figure. Resultant force on the block up the
plane is
(EAMCET 2014)
1) 2) 3) 4) PREVIOUS EAMCET QUESTIONS DEMO
The linear momentum of a particle varies with time t as P = a + bt + ct2. Then which of
the following is correct?
(EAMCET 2014)
1) Velocity of particle is inversely proportional to time
2) Displacement of the particle is independent of time
3) Force varies with time in a quadratic manner
4) Force is dependent linearly on time
PREVIOUS EAMCET QUESTIONS DEMO
A body is projected with an angle . The maximum height reached is h. If the time of
flight is 4seconds and g = 10ms-2, then the value of h is ______
(EAMCET 2014)
1) 40m
2) 20m
3) 5m
4) 10m
PREVIOUS EAMCET QUESTIONS DEMO
A bullet moving with a velocity of ms-1 is fired into a fixed target. It penetrated
into the target to the extent of S meters. If the same bullet is fired into a target of
!
thickness meters and of the same material with the same velocity, the bullet comes out
of the target with velocity
1) 20ms-1
2) 30ms-1
3) 4) ms-1
PREVIOUS EAMCET QUESTIONS DEMO
Keeping the mass of earth as constant, if its radius is reduced to th of its initial value,
then the period of revolution of earth about its own axis, and passing through the centre,
in hours, is __________ (Assume earth to be a solid sphere and its initial period of
rotation as 24 hrs).
(EAMCET_2014)
(1) 12
(2) 3
(3) 6
(4) 1.5
PREVIOUS EAMCET QUESTIONS DEMO
An artificial satellite of mass ‘m’ is revolving around in a circular orbit of radius ‘r’. If the
mass of earth is M, then angular momentum of the satellite with respect to the centre of
earth is _______ (G = universal gravitational constant)
(EAMCET_2012)
(1)
"
#
(2) " #
(3) "#
(4) "#
CHAPTER LEVEL TESTS – SOLUTONS -DEMO
FYC5_LAWS OF MOTION & FRICTION_TEST 1
__________________________________________________________________________________________
2.
A parachute of mass ‘m’ starts coming down with a constant acceleration a. Determine the
ballast mass to be released for the parachute to have an upward acceleration of same
magnitude. Neglect air drag.
(1)
(2)
(3)
(4)
__________________________________________________________________________________________
3.
Figure shows a bead of mass ‘m’ moving with uniform speed v through a U-shaped smooth
wire. The wire has a semi-circular bending between A and B. The average force exerted by
the bead on the part AB of the wire is
(3) (4) (1) 2 mv
(2) __________________________________________________________________________________________
4.
A water jet of area A hits a partition at an angle and reflects with same speed V at same
angle. Taking density of water as D, the normal thrust on partition is ________
(1) 2ADV sinα
(2) 2ADV cosα
(3) ADV sinα
(4) ADV cosα
__________________________________________________________________________________________
6.
The linear momentum P of a body moving in one dimension varies with time according to
the equation where a and b are positive constants. The net force acting on the
body is
(1) a constant
(2) proportional to t (3) inversely proportional to t
(4) proportional to t
__________________________________________________________________________________________
CHAPTER LEVEL TESTS DEMO_1
FYC5_LAWS OF MOTION & FRICTION_TEST 1_(KEY) DEMO
2
3
4
6
1
4
1
4
27
3
33
34
1
1
FYC5_LAWS OF MOTION & FRICTION_TEST 1_SOLUTIONS - DEMO
__________________________________________________________________________________________
2.
A parachute of mass ‘m’ starts coming down with a constant acceleration a. Determine the
ballast mass to be released for the parachute to have an upward acceleration of same
magnitude. Neglect air drag.
(1)
(2)
(3)
(4)
Ans. (1)
Sol. Step 1
If ‘B’ is the upward thrust (buoyancy) on parachute, then
ma = mg – B ………(i)
Step 2
If m’ is the mass released from the system so that the parachute moves up with acceleration ‘a’
then
(m – m’) a = B – (m – m’) g …….…(ii)
Adding (i) and (ii) we get
2ma – m’a = m’g
⇒ m" __________________________________________________________________________________________
CHAPTER LEVEL TESTS DEMO_2
3.
Figure shows a bead of mass ‘m’ moving with uniform speed v through a U-shaped smooth
wire. The wire has a semi-circular bending between A and B. The average force exerted by
the bead on the part AB of the wire is
(1) 2 mv
(2)
Ans. (4)
Sol. As the bead travels from A to B,
(3)
-
|Changeofmomentum| 2mv, Time of travel = =
∴Average force 3456758659:
9768;9<
6=
.
/0
1
>
(4)
.
/0
1
__________________________________________________________________________________________
4.
A water jet of area A hits a partition at an angle and reflects with same speed V at same
angle. Taking density of water as D, the normal thrust on partition is ________
(2) 2ADV cosα
(3) ADV sinα
(4) ADV cosα
(1) 2ADV sinα
Ans. (1)
Sol. Step 1
Change in momentum of water jet = MVsinα @ MA@VsinαB 2MVsinα
Step 2
Thrust E456758659:8;F96<G69
976
A<6L=6594L65J79MBIJ75K
HIJ75K
9
A
8=:6L65J79MBIJ75K
9
∴ Thurst 2AV Dsinα
9
__________________________________________________________________________________________
6.
The linear momentum P of a body moving in one dimension varies with time according to
the equation where a and b are positive constants. The net force acting on the
body is
(1) a constant
(2) proportional to t (3) inversely proportional to t
(4) proportional to t
Ans. (4)
Sol. F O
9
9 Aa bt B 2bt
∴F∝t
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SP1_FYC6_WORK, ENERGY ,POWER & COLLISIONS_CONTENT
SYLLABUS
01.
Introduction, Notions of work and kinetic energy : The work-energy theorem, Work,
Kinetic theory, Work done by a variable force, The work-energy theorem for a variable
force, The concept of potential energy, The conservation of mechanical energy, The
potential energy of a spring, Various forms of energy, The law of conservation of
energy, Power. Collisions
02.
NCERT TEXT e BOOK
→
CONCEPTS DIVISION, SYNOPSIS, FORMULAE, SOLVED EXAMPLES
Synopsis
03.
CONCEPT 1:
Work
Formulae
Solved examples_(18Q)
04.
CONCEPT 2:
Conservative force, Non conservative force,
Concept of potential energy
Synopsis
Formulae
Solved examples_(12Q)
Synopsis
05.
CONCEPT 3:
Kinetic Energy, Work – energy theorem, Law of
conservation of mechanical energy
Formulae
Solved examples_(40Q)
Synopsis
06.
CONCEPT 4:
Power
Formulae
Solved examples_(30Q)
Synopsis
07.
CONCEPT 5:
Collisions, Coefficient of restitution
Formulae
Solved examples_(14Q)
Synopsis
08.
CONCEPT 6:
Elastic collision
Formulae
Solved examples_(12Q)
Synopsis
09.
CONCEPT 7:
Semi elastic, Inelastic collisions
Formulae
Solved examples_(8)
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
→
CONTD……..
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OBJECTIVE ESSENCE OF NCERT TEXT & CBSE SYLLABUS
10.
Formulae for Quick review
11.
Conceptual MCQS_(102Q)
12.
Numerical MCQS_(52Q)
→
→
→
PREVIOUS EXAMINATIONS QUESTIONS
13.
PREVIOUS CBSE_AIPMT / NEET MCQS (15+ YEARS)_(36Q)
→
14.
PREVIOUS AP_EAMCET MCQS (15+ YEARS) _(31Q)
→
CHAPTER LEVEL TESTS
15.
TEST 1_(IN PRINT_40 QUESTIONS)
→
16.
TEST 2_(IN TABLET_40 QUESTIONS)
→
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01. SYLLABUS →→
02. NCERT TEXT e BOOK →→
03. CONCEPT 1 : WORK_SEP (18Q)
1.
A 50kg man with 20kg load on his head climbs up 20 steps of 0.25m height each. The
work done in climbing is
1) 5 J
2.
2) 350 J
2) +7
2) 270
4) 135
2) 360 J
3) 420 J
4) 100 J
A uniform chain of mass 'm' and length ‘ l ' is kept on a horizontal table with half of its
length hanging from the edge of the table. Work done in pulling the chain on to the table
so that only th of its length now hangs from the edge is,
2)
3)
!"
4)
""
A uniform chain of length 2m is kept on a table such that a length of 60cm hangs freely
from the edge of the table. The total mass of the chain is 4kg. The work done in pulling
the entire chain on to the table is ________(g = 10ms-2)
1) 7.2 J
8.
3) 35
̂ in displacing a particle from x = 4m to
The work done by a force x = –2m is
1)
7.
4) 2
A position dependent force F = (7 – 2x + 3x2) newton acts on a small body of mass 2 kg and
displaces it from x = 0 to x = 5m. The work done in joules is
1) – 240 J
6.
4) +13
3) 6
2) 1) 70
5.
3) +10
acts on a particle and produces a displacement of ̂ ̂ If the
̂ ̂ A force work done is zero, then the value of x is
1) – 2
4.
4) 3430 J
is applied over a particle which displaces it from its origin
̂ ̂ A force to the point ̂ ̂. The work done on the particle in joules is
1) – 7
3.
3) 100 J
2) 3.6 J
3) 120 J
4) 1200 J
A cord is used to lower vertically a block of mass M by a distance d with constant downward
#
acceleration $ . Work done by the cord on the block is
%
1) Mg &
%
2) 3Mg &
%
3) -3 Mg &
4) Mgd
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A ball moving with speed v hits another identical ball at rest. The two balls stick together
after collision. If specific heat of the material of the balls is S, then the temperature rise
resulting from the collision is
(1)
35.
'(
(2)
)
'(
&)
(3)
'(
(4)
)
'(
)
A wooden block of mass ‘M’ rests on a horizontal surface. A bullet of mass ‘m’ moving in
the horizontal direction strikes and gets embedded in it. The combined system covers a
distance ‘x’ on the surface. If the coefficient of friction between wood and the surface is
′+′, then the speed of the bullet at the time of striking the block is
(1) ,
36.
-.
/0
(2) ,
/0.
-1
(3) 22μgx7
-80
0
9
/01
(4) , -80
Two identical blocks A and B, each of mass 'm' resting on smooth floor are connected by
a light spring of natural length L and spring constant K, with the spring at its natural
length. A third identical block 'C' (mass m) moving with a speed v along the line joining
A and B collides with A. The maximum compression in the spring is
0
(1) v,;
37.
'
(2) m,;
(3) ,
0'
0'
(4) ;
;
A hammer of mass M falls from a height h repeatedly to drive a pile of mass m into the
ground. The hammer makes the pile to penetrate in the ground to a distance d in single
blow. Opposition to penetration is given by the expression __________
0( .=
(1) -80>
-( .=
(2) -80> M mg
-( .=
(3) -80>
0( .=
(4) -80> M mg
(KEY AND DETAILED SOLUTIONS ARE GIVEN IN TABLET)
16. CHAPTER LEVEL_TEST 2_(IN TABLET_40 Q) →→
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