(a) Earth alone as a system: Earth orbits the Sun due to gravitational attraction of the Sun. If we consider Earth as a system, then Sun's gravitational force is an external force. If we take the Moon into account, it also exerts an external force on Earth.

(b) (Earth + Sun) as a system: In this case, there are two internal forces which form an action and reaction pair-the gravitational force exerted by the Sun on Earth and gravitational force exerted by the Earth on the Sun.

(c) Our body as a system: While walking, we exert a force on the Earth and Earth exerts an equal and opposite force on our body. If our body alone is considered as a system, then the force exerted by the earth on our body is external.

(d) (Our body + Earth) as a system: In this case, there are two internal forces present in the system. One is the force exerted by our body on the Earth and the other is the equal and opposite force exerted by the Earth on our body.

(i) Tension T₁ of 24N pulls the masses (3m + 5m) with acceleration a.
∴ 24 = (3m + 5m)a or a = \(\frac { 3 }{ m } \)
(ii) Tension T₂ pulls mass 5m with acceleration \(\frac { 3 }{ m } \)
∴ T₂ = 5m\(\times\) \(\frac { 3 }{ m } \) = 15N
(iii) F = (m + 3m + 5m)a = 9m\(\times\)\(\frac { 3 }{ m } \) = 27N

Given: m = 6 kg
length = 2m
force = 50N
g = 10 ms^-2
As shown in the diagram, there are three forces acting on the midpoint p of the rope. Suppose the rope makes an angle ፀ with the vertical in equilibrium, resolving the forces horizontally and vertically, we get
T₁ sinፀ = T₃ = 50N   .....(1)
T₁ cosፀ = T₂ = 6kg wt = 60N .......(2)
Dividing (1) by (2) we get tanፀ = \(\frac { 5 }{ 6 } \) (or) \(\theta =tan^{ -1 }\left( \frac { 5 }{ 6 } \right) \) = 398

Formula:
(i) Impulse = Area ABC
= \(\frac { 1 }{ 2 } \) \(\times\)18000\(\times\)(2.5 -1)
=1.35\(\times\)104 kg ms^-1
(ii) Force = \(\frac { impulsw }{ time } =\frac { 1.35\times { 10 }^{ 4 } }{ (2.5-1) } \) = 9000 N
(iii) Maximum force = 18000 N

This is due to inertia of motion that the blades of an electric fan continue to rotate for sometime even after the fan has been switched off.

The variation of both static and kinetic frictional forces with external applied force is graphically shown in the figure.

Variation of static and kinetic frictional forces with external applied force
The Figure shows that static friction increases linearly with external applied force till it reaches the maximum. If the object begins to move then the kinetic friction is slightly lesser than the maximum static friction. Note that the kinetic friction is constant and it is independent of applied force.

Earth is not really an inertial frame since it has self-rotation and orbital motion. But these rotational effects of Earth can be ignored for the motion involved in our day-to-day life. Eg: when an object is thrown, or the time period of a simple pendulum is measured in the physics laboratory, the Earth's self rotation has very negligible effect on 'it. In this sense, Earth can be treated as an inertial frame. But at the same time, to analyse the motion of satellites and wind patterns around the Earth, we cannot treat Earth as an inertial frame since its self-rotation has a strong influence on wind patterns and satellite motion.

Jumping on a concrete cemented floor is more dangerous than jumping on the sand. Sand brings the body to rest slowly than the concrete floor, so that the average force experienced by the body will be lesser.

Mass (m) = 500 g = 0.5 kg
Angle (\(\theta\)) = 60°
Force (f) = 20 N
Horizontal component of force = F sin \(\theta\)
Acceleration of a body 'a = \(\frac { Fsin\theta }{ m } \)
\(\frac { 20sin{ 60 }^{ 0 } }{ 0.5 } \)
Acceleration of body "a'' = 17.32m.s²

The force F has to act downward, because the body is to move up with an acceleration greater than 'g' acceleration due to gravity.
According to Newton's II law,
In a force acting an a downward} mg + F = ma
F = m(g+a)
= 20(9.8 + 4)
= 20(13.8)
= 276N

1. Equal mass balancing each side so reading shows Zero.
2. Mass of string acting downward direction in the inclined plane.
mg sin θ = T
T = 2kg x 9.8m/s² x sin(30^o)
T = 9.8N

When a track goes round a level curve, the force of friction between the tyres and the road provides the necessary centripetal force. If the frictional force, which acts as centripetal force and keeps the body moving along the circular road is not enough to provide the necessary centripetal force, the car will skid. In order to avoid skidding while going round a curved path the outer edge of the road is raised above the level of the inner edge. This is known as banking of curved roads or tracks.
Let the surface of the road makes angle ፀ with horizontal surface. Then the normal force makes the same angle d with the vertical. When the car takes a turn, there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface).
We can resolve the normal force into two components. N cos ፀ and N sin ፀ as shown in Figure. The component N cos ፀ balances the downward gravitational force 'mg' and component N sin ፀ will, provide the necessary centripetal acceleration. By using Newton second law
N cos ፀ = mg .....(1)
N sin ፀ = \(\frac { { mv }^{ 2 } }{ r } \)  ....(2)
Eq (2) / (1) ⇒ tan ፀ = \(\frac { { v }^{ 2 } }{ rg } \)
v= \(\sqrt { rg\tan\theta } \)
Need Banking of Tracks
1. The banking angle ፀ and radius of curvature of the road or track determines the safe speed of the car at the turning.
2. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.
3. At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. 
4. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

The forces acting on the mango are
(i) The gravitational force exerted by the Earth on the mango acting downward along the negative y-axis
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis. The free body diagram for the mango is shown in the figure

(iii) Here, mg is the magnitude of the gravitational force and  -\(\hat { j } \) represents the unit vector in negative y-direction
\(\vec { T } -T\hat { j } \)
(iv) Here T is the magnitude of the tension force and \(\hat { j } \) represents the unit vector in positive y-direction
\(\vec { { F }_{ net } } -\vec { F } _{ g }+\vec { T } =-mg\hat { j } +\vec { T } \hat { j=(T-mg) } \hat { j } \)
(v) From Newton's second law  \({ \vec { F } }_{ net }=m\vec { a } \) Since the mango is at rest with respect to us (inertial coordinate system) the acceleration
is Zero \((\vec { a } =0)So{ \vec { F } }_{ net }=m\vec { a } =0\)
\((T-mg\hat { j } )=0\)
(vi) By comparing the components on both sides of the above equation  We get, T - mg  = 0, So the tension Force acting on the mango is Given By T = mg
Mass of the mango m = 400 g and
g = 9.8 m S^-2
Tension acting on the mango is 
T= 0 .4\(\times\)9. 8 = 3.92

Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called inertia.
Inertia of rest: It is the inability of a body to change by itself, its state of rest. This means a body at rest remains at rest and cannot start moving by its own.
(i) When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper part of the body as explained above.
(ii) Passengers experience a backward push due to inertia of rest.
Inertia of motion: It is the inability of a body to change itself its state of uniform motion i.e., a body in uniform motion can neither accelerate nor retard by its own.
(i) Passengers experience a forward push due to inertia of motion.
(ii) A person jumping out of a moving train may fall forward.
Inertia of direction: It is the inability of a body to change by itself direction of motion.
(i) A stone moves tangential to circle due to inertia of direction.
(ii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards.

Mass m = 100 kg
Acceleration a = 50 cm s^-2 = 0.5 ms^-2
Using Newton's second law,
F= ma
F=100 kg\(\times\)0.5 ms^-2 = 50N

v =\(\sqrt { (rg\quad tan\theta ) } =\sqrt { 20\times 9.8\times tan{ 15 }^{ 0 } } =\sqrt { 20\times 9.8\times 0.26 } \) = 7.1 ms^-1
The safe speed for the car on this road is 7.1 ms^-1

Change in momentum = (-mv cos45°) - (mv cos45°)
dp = - 2mv cos45°
\(|\vec { dp } |=2\times 0.2\times 25\times \frac { 1 }{ \sqrt { 2 } } \Rightarrow dp=5\sqrt { 2 } \).

No, the momentum remain conserved if some external force acts on the System.

A frame of references in which if an object experience force, depending on the acceleration of the frame.
Eg: An object remains at rest on it smooth table kept inside the train, and if the train suddenly accelerates (whiche may not sense), the object appears to accelerate backwards even without any force acting on it.

(i) Newton's second law is cause and effect relation. Force is the cause and acceleration is the effect.
(ii) Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing
Newton's second law is \(m\overset { \rightarrow }{ a } =\overset { \rightarrow }{ F } (or)\frac { d\overset { \rightarrow }{ p } }{ dt } =\overset { \rightarrow }{ F } \)

(i) If multiple forces \(\overset { \rightarrow }{ { F } } _{ 1 },\overset { \rightarrow }{ { F } } _{ 2 },\overset { \rightarrow }{ { F } } _{ 3 }...\overset { \rightarrow }{ { F } } _{ n }\) act on the same body, then the total force \(\left( \overset { \rightarrow }{ { F }_{ net } } \right) \) is equivalent to the vectorial sum of the individual forces
(ii) Their net force provides the acceleration. \(\overset { \rightarrow }{ { F } } _{ net }=\overset { \rightarrow }{ { F } } _{ 1 },\overset { \rightarrow }{ { F } } _{ 2 },\overset { \rightarrow }{ { F } } _{ 3 }...\overset { \rightarrow }{ { F } } _{ n }\)
(iii) Newton's second law for this case is \(\overset { \rightarrow }{ { F } } _{ net }=m\overset { \rightarrow }{ a } \)
In this case the direction of acceleration is in the direction of net force.

(i) The car is moving with uniform velocity v with respect to a person standing (at rest) on the ground.
(ii) As the car is moving with constant velocity with respect to ground to the person is at rest on the ground, both frames (with respect to the car and to the ground) are inertial frames.

The centrifugal force appears to act on the particle, only when we analyses the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the string. For this reason centrifugal force is called as a 'pseudo force'. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered.

Given:
F = 50N
m₁ = 5 kg
m₂ = 10 kg
m₃ = 15 kg
All the blocks move with common acceleration a under the force F = 50N
∴ F = (m₁ + m₂ + m₃)a
or a = \(\frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } =\frac { 50 }{ 5+10+15 } =\frac { 5 }{ 3 } \) ms^-2
To determine T₁ Refer to the free-body diagram for m₁ shown in the diagram. Clearly, the tension T₁ produces acceleration a in mass m₁.
∴ T₁ = m₁a = \(5\times \frac { 5 }{ 3 } =\frac { 25 }{ 3 } \) = 8.33 N

To determine T₂ Refer to the free-body diagram for m₃ shown in the diagram (b). Force F acts towards right and tension T₂ acts towards left.
∴ F - T₂ = m₃ a (or) 50 - T₂ = 15 x \(\frac { 5 }{ 3 } \) or T₂ =25N

When an object of mass m slides on a frictionless surface inclined at an angle e as shown in the figure, the forces acting on it decides the
(a) acceleration of the object 
(b) speed of the object when it reaches the bottom.
The force acting on the object is
(i) Downward gravitational force (mg)
(ii) Normal force perpendicular to inclined surface (N)

To draw the free body diagram, the block is assumed to be a point mass [in figure (a)]. Since the motion is on the inclined surface, we have to choose the coordinate system parallel to the inclined surface as shown
in the figure (b).
The gravitational force mg is resolved in to parallel component mg sine along the inclined plane and perpendicular component mg coss perpendicular to the inclined surface [figure (b)].
Note that the angle made by the gravitational force (mg) with the perpendicular to the surface is equal to the angle of inclination e' as shown in figure (c).
There is no motion (acceleration) along the y axis. Applying Newton's second law in the y direction
\(-mg\ cos\theta \hat { j } +N\hat { j } \) =0 (No acceleration)
By comparing the components on both sides, N-mg cos ፀ=0
N=mg cosፀ
The magnitude ~fnormal force (N) exerted by the surface is equivalent to mg cosፀ, Tlie object slides (with an acceleration) along the x direction. Applying Newton's second law in the x direction.
\(mg\ sin\theta\ \hat { i } =ma\hat { i } \)
By comparing the components on both sides, we can equate
mg sinፀ = ma
The acceleration of the sliding object is
a = g sinፀ

Note that the acceleration depends on the angle of inclination ፀ.

(i) If the object is at rest and no external force is applied on the object, the static friction acting on the object is zero (f_s = 0).
(ii) If the object is at rest, and there is an external force applied parallel to the surface, then the force of static friction acting on the object is exactly equal to the external force applied on the object (f_s = F_ext). But still the static friction Is is less than μ_sN.
(iii) When object begins to slide, the static friction (f_s) acting on the object attains maximum.

Static Friction Kinetic Friction It opposes the starting of motion It opposes the relative motion of the object with respect to the surface Independent of surface of contact Independent of surface of contact u_s depends on the nature of materials in mutual contact U_k depends on nature of materials and temperature of the surface Depends on the magnitude of applied force  Independent of magnitude of applied force It can take values from zero to It can never be zero and always equals to whatever be the speed (true <10 m/s) \({ f }_{ s }^{ max }>{ f }_{ k }\) It is less than maximal value of static friction   \({ u }_{ s }>{ u }_{ k }\) Coefficient of kinetic friction is less than coefficient of static friction

(i) Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest).
(ii) If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.
(iii) This implies that in addition to the inward centripetal force - mω²r there must be an equal and opposite force that acts on the stone outward with value + mω²r.
(iv) So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω²r=0).
(v) This outward force + mω²r is called the centrifugal force.