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Page 1
Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
w = Weight of plank,
f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
wall,
N R ( ) = Reaction to N i.e. normal force on
sphere B by sphere A,
w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N
Force acting on cylinder
Page 2
Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
w = Weight of plank,
f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
wall,
N R ( ) = Reaction to N i.e. normal force on
sphere B by sphere A,
w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
AB l =
( sin ) T l w
l
30
2
° =
Þ T w =
6. See figure (answer to question no. 3)
sin q =
+
OA
OB BC
=
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\ f = + ( ) 2.4 3 6
=10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Page 3
Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
w = Weight of plank,
f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
wall,
N R ( ) = Reaction to N i.e. normal force on
sphere B by sphere A,
w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
AB l =
( sin ) T l w
l
30
2
° =
Þ T w =
6. See figure (answer to question no. 3)
sin q =
+
OA
OB BC
=
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\ f = + ( ) 2.4 3 6
=10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
=10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\ Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
Net force on 2 kg block = ´ 2 a
= ´ 2 10
=20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ =
Þ a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
= ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e., T mg ma cos sin f = + q …(i)
and T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
= + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
)
=
7
4
2 2
m g
or T
mg
=
7
2
=5 7 N
Dividing Eq. (i) by Eq. (ii),
tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
Page 4
Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
w = Weight of plank,
f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
wall,
N R ( ) = Reaction to N i.e. normal force on
sphere B by sphere A,
w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
AB l =
( sin ) T l w
l
30
2
° =
Þ T w =
6. See figure (answer to question no. 3)
sin q =
+
OA
OB BC
=
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\ f = + ( ) 2.4 3 6
=10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
=10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\ Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
Net force on 2 kg block = ´ 2 a
= ´ 2 10
=20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ =
Þ a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
= ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e., T mg ma cos sin f = + q …(i)
and T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
= + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
)
=
7
4
2 2
m g
or T
mg
=
7
2
=5 7 N
Dividing Eq. (i) by Eq. (ii),
tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
=
+
a
g a
cos
sin
q
q
=
+
cos
sin
q
q 2
=
°
+ °
cos
sin
30
2 30
=
3
5
i.e., f =
é
ë
ê
ù
û
ú
-
tan
1
3
5
6.
a =
+ +
8
2 1 1
=2 m/s
2
Net force on 1 kg mass = - 8
2
T
\ 8 1 2
2
- = ´ T
Þ T
2
6 = N
Net force on 1 kg block = T
1
\ T a
1
2 = = ´ = 2 2 4 N
Introductory Exercise 5.3
1. F g g = ° = 2 30 sin
For the system to remain at rest
T g
2
2 = …(i)
T F T
2 1
+ = …(ii)
or T g T
2 1
+ = …[ii (a)]
T mg
1
= …(iii)
Substituting the values of T
1
and T
2
from
Eqs. (iii) and (i) in Eq. [ii(a)]
2g g mg + =
i.e., m =3 kg
2. As net downward force on the system is
zero, the system will be in equilibrium
\ T g
1
4 =
and T g
2
1 =
\
T
T
1
2
4 =
3. 2 2 g T a - =
Laws of Motion 77
2kg 1kg
F = 8 N
A B
2kg 1kg 8 N
T
1
T
1
T
2
T
2
B
A
T
1
T
1
mg
T
2
T
2
2 g
T
1
T
2
T
2
F
T
1
T
1
T
1
T
2
T
2
T
1
4g
1g
3g
T
T T
T
1g
2g
a
a
Page 5
Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
w = Weight of plank,
f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
wall,
N R ( ) = Reaction to N i.e. normal force on
sphere B by sphere A,
w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
AB l =
( sin ) T l w
l
30
2
° =
Þ T w =
6. See figure (answer to question no. 3)
sin q =
+
OA
OB BC
=
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\ f = + ( ) 2.4 3 6
=10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
=10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\ Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
Net force on 2 kg block = ´ 2 a
= ´ 2 10
=20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ =
Þ a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
= ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e., T mg ma cos sin f = + q …(i)
and T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
= + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
)
=
7
4
2 2
m g
or T
mg
=
7
2
=5 7 N
Dividing Eq. (i) by Eq. (ii),
tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
=
+
a
g a
cos
sin
q
q
=
+
cos
sin
q
q 2
=
°
+ °
cos
sin
30
2 30
=
3
5
i.e., f =
é
ë
ê
ù
û
ú
-
tan
1
3
5
6.
a =
+ +
8
2 1 1
=2 m/s
2
Net force on 1 kg mass = - 8
2
T
\ 8 1 2
2
- = ´ T
Þ T
2
6 = N
Net force on 1 kg block = T
1
\ T a
1
2 = = ´ = 2 2 4 N
Introductory Exercise 5.3
1. F g g = ° = 2 30 sin
For the system to remain at rest
T g
2
2 = …(i)
T F T
2 1
+ = …(ii)
or T g T
2 1
+ = …[ii (a)]
T mg
1
= …(iii)
Substituting the values of T
1
and T
2
from
Eqs. (iii) and (i) in Eq. [ii(a)]
2g g mg + =
i.e., m =3 kg
2. As net downward force on the system is
zero, the system will be in equilibrium
\ T g
1
4 =
and T g
2
1 =
\
T
T
1
2
4 =
3. 2 2 g T a - =
Laws of Motion 77
2kg 1kg
F = 8 N
A B
2kg 1kg 8 N
T
1
T
1
T
2
T
2
B
A
T
1
T
1
mg
T
2
T
2
2 g
T
1
T
2
T
2
F
T
1
T
1
T
1
T
2
T
2
T
1
4g
1g
3g
T
T T
T
1g
2g
a
a
T g a - = 1 1
Adding above two equations
1 3 g a =
\ a
g
=
3
Velocity of 1kg block 1 section after the
system is set in motion
v at = + 0
= ×
g
3
1
=
g
3
(upward)
On stopping 2 kg, the block of 1kg will go
upwards with retardation g. Time ( ) t¢
taken by the 1 kg block to attain zero
velocity will be given by the equation.
0
3
=
æ
è
ç
ö
ø
÷ + - ¢
g
g t ( )
Þ t¢ =
1
3
s
If the 2 kg block is stopped just for a
moment (time being much-much less than
1
3
s), it will also start falling down when
the stopping time ends.
In t¢ =
æ
è
ç
ö
ø
÷
1
3
s time upward displacement of
1 kg block
= = =
u
a
g
g
g
2 2
2
3
2 18
( / )
Downward displacement of 2 kg block
= = ×
æ
è
ç
ö
ø
÷ =
1
2
1
2 3 18
2
2
at g
g g
As the two are just equal, the string will
again become taut after time
1
3
s.
4. F g T a + - = 1 1 …(i)
and T g a - = 2 2 …(ii)
Adding Eqs. (i) and (ii),
F g a - = 1 3
\ a =
- 20 10
3
=
10
3
ms
-2
Introductory Exercise 5.4
1. 2 2 T a = ´ …(i)
and 1 2 g T a - = …(ii)
Solving Eqs. (i) and (ii),
a
g
=
3
\ Acceleration of 1 kg block
2
2
3
20
3
a
g
= = ms
-2
Tension in the string
T
g
= =
3
10
3
N
2. Mg T Ma - = …(i)
78 | Mechanics-1
T
T T
T
F
2g
a
1g
a
T T
2 kg
2T
a
1g
T
T
T T
2a
2T
T T
mg
T
T
a
M
T
T
a
Read More
FAQs on DC Pandey Solutions: Laws of Motion- 1 - DC Pandey Solutions for JEE Physics
| 1. What are the laws of motion? |  |
Ans. The laws of motion, formulated by Sir Isaac Newton, are three fundamental principles that describe the relationship between the motion of an object and the forces acting upon it. These laws are widely used to understand and analyze the motion of objects in physics.
| 2. What is the first law of motion? | |
Ans. The first law of motion, also known as the law of inertia, states that an object at rest will remain at rest, and an object in motion will continue to move at a constant velocity, unless acted upon by an external force. This law emphasizes the natural tendency of objects to resist changes in their state of motion.
| 3. What is the second law of motion? | |
Ans. The second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as F = ma, where F is the net force, m is the mass of the object, and a is the acceleration produced.
| 4. What is the third law of motion? | |
Ans. The third law of motion states that for every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. These forces always occur in pairs, acting on different objects.
| 5. How do the laws of motion apply to real-life situations? | |
Ans. The laws of motion have numerous applications in everyday life. For example, they explain why we feel a backward push when a car accelerates, why a goalkeeper pulls their hands backward while catching a fast-moving ball, or even why astronauts experience weightlessness in space. These laws provide a foundation for understanding and predicting the behavior of objects in motion.
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