CBSE Class 11 Physics Newton laws of Motion Exam Notes

Question. In figure:

(A) retardation is uniform
(B) velocity is decreasing with time
(C) beyond M, the body has negative velocity
(D) all the above are incorrect
Answer: D

Question. A player completes a circular path of radius r in 40 s. At the end of 2 mintures 20 seconds, displacement will be:
(A) 2r
(B) 2πr
(C) 7πr
(D) zero
Answer: A

Question. A body whose speed in a particular direction is constant:
(A) must be accelerating
(B) must be retarding
(C) has a constant velocity
(D) all the above
Answer: C

Question. A body moving along a circular path has:
(A) a constant speed
(B) a constant velocity
(C) non tangential velocity
(D) no radial acceleration
Answer: A

Question. In figure BC represents a body moving: 

(A) backward with uniform velocity
(B) forward with uniform velocity
(C) backward with non-uniform velocity
(D) forward with non-uniform velocity
Answer: A

Question. The velocity of a particle increases from u to v in a time t during which it covers a distances S. If the particle has a uniform accelration, which one of the following euqations does not apply to the motion?
(A) 2S = (v + u)t
(B) a = v-u/t
(C) v² = u² – 2 aS
(D) S = (u+1/2 at)t
Answer: C

Question. A body has an acceleration of –4 ms^–2. What is its retardation?
(A) –4 ms²
(B) 4 ms^–2
(C) Zero
(D) Nothing can be decided
Answer: B

Question. A cyclist moves from a certain point X and goes round a circle of radius ‘r’ and reaches Y, exactly at the other side of the point X, as shown in figure.
The displacement of the cyclist would be:

(A) πr
(B) 2πr
(C) 2r
(D) 2π/r
Answer: C

Question. In the above problem, the distance covered by the cyclist would be:
(A) πr
(B) 2πr
(C) 2r
(D) 2π/r
Answer: A

Question. In the above question, the velocity:
(A) increases between points O and A
(B) increases between points A and B
(C) decreases between points A and B
(D) is zero throughout
Answer: B

Question. When a graph between two physical quantities is a straight line, the two quantities are:
(A) both constant
(B) independent of each other
(C) directly proportional
(D) inversely proportional
Answer: C

Question. A man walks 8m towards East and then 6m towards North. His magnitude of displacement is:
(A) 10 m
(B) 14 m
(C) 2 m
(D) zero
Answer: A

Question. Area undera velocity-time graph gives:
(A) the time taken by a moving object
(B) the distance travelled by a moving object
(C) the acceleration of moving object
(D) the retardation of a moving object
Answer: B

Question. In figure, the velocity of the body at A is: 

(A) zero
(B) unity
(C) maximum
(D) infinite
Answer: A

Question. Which of the following relations represents the relationship between the average speed, time and distance correctly?
(A) Average speed = distance × time
(B) Average speed = total distance/total time
(C) Time = average speed + distance
(D) Distance = average speed – time
Answer: B

CBSE Class 12 Physics: Newton’s Laws of Motion Exam Notes

NEWTON’S LAWS OF MOTION

In this chapter we will study the consequences of the cause of motion, i.e., force, which is defined as a push or pull experienced by the body or the system. The equation of motion is affected by the choice of reference frame. So we will also study the equations of motion in different types of reference frames.

NEWTON’S LAWS

First law of motion: The state of the body cannot be changed until or unless an external force is applied on the body. This statement is applicable in an inertial frame of reference only.

This law is also called the law of inertia, where inertial mass is the quantity of matter in a body and inertia is the property of a body due to which a body resists any change in its state of rest or of uniform motion in a straight line. The larger the mass of the body, the larger is the tendency to resist any change in its state of rest or of uniform motion in a straight line, i.e., to resist any change in its velocity. Newton’s first law gives a qualitative definition of force. A force is a push or pull exerted by a body which can accelerate, decelerate or deform a body.

EQUILIBRIUM

If the net external concurrent force acting on the body is zero, the body is said to be in equilibrium. There are two types of equilibrium:

1. Static equilibrium: When the body is at rest under the influence of external forces acting on the body.
2. Dynamic equilibrium: If the net external force is zero but the velocity of the body is not zero, i.e., the body moves with constant velocity.

Second law of motion: The rate of change of momentum of a body is directly proportional to the net external force, and this rate of change takes place in the direction of the force.

If a particle of mass \(m\) is moving with velocity \(\vec v\), its momentum is \(\vec p=m\vec v\).

If \(\vec F\) is the force applied, \(\vec F\propto\dfrac{d\vec p}{dt}\) or \(\vec F=K\dfrac{d\vec p}{dt}\), where \(K\) is a constant.

If the constant \(K\) is chosen as unity (if \(F\) is in newton, \(p\) in kg m/s and time \(t\) in second, then \(K=1\)),

\[\vec F=\frac{d\vec p}{dt}\]

The force \(\vec F\) is a vector. If the mass \(m\) is constant and \(\vec a\) is the acceleration of the body,

\[\frac{d\vec p}{dt}=m\frac{d\vec v}{dt}=m\vec a,\qquad \vec F=m\vec a\]

Newton’s second law of motion gives a quantitative measure of force. The unit of force is the newton. A force of 1 newton produces an acceleration of 1 m/s² in a body of mass 1 kg.

INERTIAL AND NON-INERTIAL FRAMES OF REFERENCE

Non-accelerated frames of reference are called inertial frames of reference, while accelerated frames are called non-inertial frames of reference.

If one is travelling in a train which is accelerated forward, the body in the train is pushed backwards, and he is pushed forward when the brakes are applied, due to the inertia of the body. Such an accelerated frame is called a non-inertial frame of reference. In order to make Newton’s laws applicable to such a frame, a fictitious inertial force or pseudo force is applied on the body. The magnitude of this pseudo force is equal to the product of the mass \(m\) of the body and the acceleration \(a\) of the reference frame, and its direction is opposite to the acceleration of the frame.

\[\therefore\ \text{pseudo force, }\vec F=-m\vec a\]

Thus in a non-inertial frame of reference, a trolley moving with an acceleration \(a\) has a hanging bob of mass \(m\) which will be deflected through angle \(\theta\) due to a pseudo force acting in the backward direction. In the non-inertial frame of reference, the bob is in equilibrium under the action of the force due to tension \(T\), the weight \(mg\) and the pseudo force \(ma\), in a direction making an angle \(\theta\) with the vertical.

\[T\sin\theta=ma\]

\[T\cos\theta=mg\]

\[\tan\theta=\frac{a}{g}\quad\text{or}\quad\theta=\tan^{-1}\!\left(\frac{a}{g}\right)\]

Third law of motion: Every action has an equal and opposite reaction.

Thus if a body A exerts a force on another body B, the second body B exerts an equal and opposite force, i.e., reaction, on the first body A.

\[\therefore\ \vec F_{AB}=-\vec F_{BA}\]

FREE BODY DIAGRAM

When all the external forces are indicated on the body, it is said to be representing the free body diagram (F.B.D.) of the body.

NORMAL REACTION

Normal reaction is the pressing force between the two surfaces in contact. It is always directed normal to the contact surface.

CONTACT REACTION

It is the resultant of all the forces which arise due to contact of two surfaces. The contact reaction may not be perpendicular to the contact surfaces. In case the contact reaction is not perpendicular to the contact surfaces, the component of the contact reaction normal to the contact surfaces is known as the normal reaction.

CONSTRAINT MOTION

When the motion of the body can be controlled, the motion is said to be constraint motion. For example, when a body tied with a string is lowered under the effect of gravity, its motion is constraint motion. Also the motion of masses \(m_1\) and \(m_2\) (shown in the figure) can be controlled by choosing appropriate values of \(m_1\) and \(m_2\).

PROCEDURE

• Measure the distances of different bodies or between different positions in such a way that we can measure the length of the string, which remains constant during motion.
• Double differentiate the equation for the length of the string. The sign of \( \dfrac{d^2y}{dt^2} \) is taken positive if the acceleration is in the direction in which \(y\) increases, and negative if it is in the direction in which \(y\) decreases.

FRICTION

If there are two bodies in contact with each other, the force which opposes relative motion between the two bodies in contact is called the force of friction. The magnitude of the frictional force depends upon the nature of the two surfaces in contact.

The variation of the force of friction with the applied force is shown in the graph when any block is moving over another surface. When the block is at rest, the force of static friction is equal to the force applied. It then reaches a maximum value at A. Once the motion starts, a smaller force is required for maintaining uniform smooth motion. It is due to surface irregularities at molecular scales, with the result that even a highly polished surface has irregularities. It results in producing interlocking of uneven surfaces. Once there is smooth motion of a body, the friction is less than the maximum force of static friction or limiting friction.

STATIC FRICTION

If the force applied in the horizontal direction on a surface is small, the body does not move because an equal and opposite frictional force is present. As \(F\) is increased, the frictional force also increases continuously until a stage is reached when the body is just at the point of sliding. The force of friction at this stage is called the limiting friction or maximum static friction \(f_m\), corresponding to point A.

COEFFICIENT OF FRICTION

The coefficient of friction \(\mu\) is the ratio of the limiting or maximum value of the force of friction \(f_m\) to the normal reaction \(R\) between the surfaces when the body is just about to move.

\[\mu=\frac{f_{max}}{R}\]

\[f_{max}=f_{limiting}=\mu R\]

Thus if the body does not move, the static frictional force \(f_s\) and the external applied force parallel to the surface are equal in magnitude and opposite in direction, and \(F\) is directly proportional to \(f_s\). If the external force \(F\) exceeds \(f_{max}\), the body slides on the surface and the magnitude of the frictional force decreases to less than \(f_{max}\). The frictional force for sliding is \(f_k=\mu_k R\), where \(\mu_k\) is the coefficient of kinetic friction. As \(f_k\) is less than \(f_{max}\), the coefficient of kinetic friction is less than the coefficient of static friction.

Angle of friction

The angle between the normal reaction \(R\) and the resultant of the limiting friction and the normal reaction is called the angle of friction and is denoted by \(\lambda\).

\[\tan\lambda=\frac{f_{max}}{R}=\mu\]

\[\text{or}\quad\lambda=\tan^{-1}(\mu)\]

If a body of mass \(m\) is placed on an inclined plane which is gradually moved upwards, then the body just starts sliding down at a certain angle of inclination \(\theta\). The weight \(mg\) can be resolved into a component \(mg\sin\theta\), due to which the body is about to slide down against the maximum or limiting value of friction \(f_s\), and the second component \(mg\cos\theta\), which balances the normal reaction \(R\) perpendicular to the inclined plane.

\[\therefore\ f_s=mg\sin\theta\]

\[R=mg\cos\theta\]

\[\therefore\ \frac{f_s}{R}=\tan\theta=\mu\]

As \(\mu=\tan\lambda\),

\[\theta=\lambda=\tan^{-1}(\mu)\]

The angle \(\theta\) is called the angle of repose or angle of friction. The angle of friction \(\lambda\) is the minimum angle of inclination of the inclined plane at which a body placed at rest on the inclined plane just starts sliding down.

When \(\theta<\lambda\), the body is in equilibrium and does not slide. If \(\theta>\lambda\), the body starts sliding down with an acceleration.

IMPULSE DUE TO A FORCE

If a force \(F\) acts on a body for a time \(\Delta t\), impulse is defined as the product of the force and the time for which the force acts on the body.

\[\text{Impulse}=\text{Force}\times\text{time}=F\times\Delta t\]

From Newton’s second law of motion, \(F=\dfrac{dp}{dt}\) or \(F\,dt=dp\).

If the force \(F\) acts from time \(t_i\) to \(t_f\), and the initial momentum \(p_i\) changes to \(p_f\), then by integrating,

\[\int_{t_i}^{t_f}F\,dt=\int_{p_i}^{p_f}dp=p_f-p_i\]

\[\therefore\ \text{Impulse}=\int_{t_i}^{t_f}F\,dt=p_f-p_i\]

If the force is constant, Impulse = Force × time. The S.I. unit of impulse is newton-second.

There are four fundamental forces in nature, i.e., gravitational, electromagnetic, weak and strong (nuclear) forces. On a macroscopic scale, only gravitational and electromagnetic forces are significant. The commonly used forces in mechanics include weight, tension, normal reaction, friction, contact force, spring force, upthrust of fluid, etc. The bodies which restrict the motion of other bodies impose a constraint on the motion of those bodies.

IMPULSE

The area under the \(F\)–\(t\) curve represents impulse or change in momentum.

STEPS FOR SOLVING PROBLEMS USING NEWTON’S LAWS

1. Draw a diagram showing the physical position of the body or bodies under consideration.
2. Draw the free body diagram of all the bodies by identifying all the forces acting on each of the bodies and giving the correct direction of each force. If there is a single body, identify the forces and the directions of the forces on the body.
3. Resolve the forces into components which are parallel and perpendicular to the coordinate system defined. In the case of motion along an inclined plane, resolve the components along and perpendicular to the plane.
4. Write the equations for equilibrium of forces or motion of the body along each axis of the coordinate system.
5. Solve the equations by the process of elimination to find the unknown required quantities.

A system is in equilibrium if its state does not change due to the application of a number of forces. When a number of forces are applied and the body remains at rest, it is in static equilibrium. The examples are a block where the normal reaction \(N\) and the weight \(W\) of the block are in equilibrium, a block pulled by a string with tension \(T\), and a block of weight \(mg\) compressing a spring with the spring exerting an equal spring force \(kx\).

Note: If a string is connected to a rigid body, the direction of the tension \(T\) is away from the point where the string is fixed. A string suspended from the ceiling will have a tension directed away from the ceiling.

APPLICATION OF NEWTON’S LAWS

1. Two blocks in contact: If two blocks of masses \(m_1\) and \(m_2\) are in contact on a horizontal frictionless surface, so that a force \(F\) applied horizontally imparts an acceleration \(a\), and \(F_C\) is the contact force, which is equal and opposite for \(m_1\) and \(m_2\), Newton’s second law, when applied to the free body diagrams, gives the following equations:

\[F-F_C=m_1a\]

\[F_C=m_2a\]

Adding,

\[a=\frac{F}{m_1+m_2}\]

\[F_C=\frac{m_2F}{m_1+m_2}\]

2. Blocks connected by strings: If two blocks of masses \(m_1\) and \(m_2\) are connected by an inextensible string, so that a force \(F\) is applied to \(m_1\) and there is an equal and opposite tension \(T\) in the string, and \(a\) is the acceleration of the masses, then Newton’s law gives

\[F-T=m_1a\]

\[T=m_2a\]

Adding,

\[F=(m_1+m_2)a\quad\text{or}\quad a=\frac{F}{m_1+m_2}\]

\[T=\frac{m_2F}{m_1+m_2}\]

If three blocks of masses \(m_1\), \(m_2\) and \(m_3\) are connected by two strings with tensions \(T_1\) and \(T_2\), and a force \(F\) applied to \(m_1\) imparts an acceleration \(a\) to all these blocks, Newton’s law gives the following relations for these three blocks:

\[F-T_1=m_1a\]

\[T_1-T_2=m_2a\]

\[T_2=m_3a\]

Adding,

\[F=(m_1+m_2+m_3)a\quad\text{or}\quad a=\frac{F}{m_1+m_2+m_3}\]

\[T_2=\frac{m_3F}{m_1+m_2+m_3}\]

\[T_1=\frac{(m_2+m_3)F}{m_1+m_2+m_3}\]

3. Apparent weight in a lift: If a person is standing in a stationary lift, his weight \(mg\) acts downwards and the normal reaction of the floor of the lift on him acts upwards, so that \(N=mg\) as per Newton’s third law of motion. If the lift is moving with a constant velocity, \(N\) is equal and opposite to \(mg\) as the net force is zero.

\[N-mg=0\quad\text{or}\quad N=mg\]

Thus the apparent weight is equal to the true weight.

If the lift is moving upwards with an acceleration \(a\),

\[N-Mg=Ma\quad\text{or}\quad N=M(g+a)\]

The apparent weight is greater than the actual weight.

If the lift is accelerating downwards with \(a'\),

\[Mg-N=Ma'\quad\text{or}\quad N=m(g-a')\]

Thus the apparent weight is less than the actual weight.

4. Masses connected by a pulley and constraint relation: Consider blocks of masses \(m_1\) and \(m_2\) connected by a string and passing over the pulley as shown in the figure. Let \(a_1\) be the acceleration of \(m_1\) downwards and \(a_2\) be the acceleration of \(m_2\) upwards. If \(T\) is the tension in the string, so that the pulley moves clockwise,

For block \(m_1\),

\[m_1g-T=m_1a_1\]

For block \(m_2\),

\[T-m_2g=m_2a_2\]

As there are three unknown parameters, the following steps should be taken for writing the constraint relation and finding one of the parameters:

1. Assume the direction of acceleration of each body.
2. Locate the position of each block from any fixed point, like the centre of the pulley.
3. Identify the constraint and write the equation of constraint in terms of distances.
4. Write the equation of constraint and differentiate twice to find one of the parameters.

In this case, the string is inextensible, therefore the constraint or restriction is that the length of the string remains constant. If \(x_0\) is the length of the string passing over the pulley, and \(x_1\) and \(x_2\) are the lengths of the string from the pulley to \(m_1\) and \(m_2\) respectively, the total length \(L\) of the string remains constant.

\[\therefore\ x_1+x_2+x_0=L=\text{constant}\]

As \(x_0\) and \(L\) are constants, differentiating,

\[\frac{dx_1}{dt}+\frac{dx_2}{dt}=0\]

Differentiating again,

\[\frac{d^2x_1}{dt^2}+\frac{d^2x_2}{dt^2}=0\]

As \(a_1\) and \(a_2\) are in opposite directions, they are equal in magnitude:

\[a_1=a_2=a\]

\[\therefore\ m_1g-T=m_1a\]

\[T-m_2g=m_2a\]

Adding,

\[a=\left(\frac{m_1-m_2}{m_1+m_2}\right)g\]

\[T=m_2a+m_2g=\frac{m_2(m_1-m_2)g}{m_1+m_2}+m_2g\]

\[T=\left(\frac{2m_1m_2}{m_1+m_2}\right)g\]

If the pulley is pulled up with an acceleration \(a\),

\[T=\left(\frac{2m_1m_2}{m_1+m_2}\right)(g+a)\]

5. Motion of a block on a smooth inclined plane: Consider a block of mass \(m\) placed on a frictionless inclined plane inclined at angle \(\theta\) to the horizontal. The normal reaction \(N\) acts perpendicular to the plane, and the weight is resolved into the component \(mg\sin\theta\) along the plane, which slides the block downwards with acceleration \(a\), and the component \(mg\cos\theta\) perpendicular to the plane downwards, which is equal and opposite to the normal reaction.

\[\therefore\ mg\sin\theta=ma\]

\[mg\cos\theta=N\]

\[\text{or}\quad a=g\sin\theta\]

If the plane is given a horizontal acceleration \(a'\) as shown, the body lies in an accelerating frame of reference and a pseudo force \(ma'\) acts horizontally in a direction opposite to that of \(a'\), because an inertial force \(ma'\) acts on it opposite to the direction of \(a'\). Thus \(ma'\) can be resolved into a component \(ma'\cos\theta\) up the plane and a second component \(ma'\sin\theta\) perpendicular to the plane in the downward direction, as shown in the figure. From Newton’s second law of motion:

\[mg\sin\theta-ma'\cos\theta=ma\]

\[\text{or}\quad a=g\sin\theta-a'\cos\theta\]

\[N=m(g\cos\theta+a'\sin\theta)\]

If the body is at rest relative to the inclined plane, \(a=0\),

\[\text{or}\quad g\sin\theta=a'\cos\theta\]

\[\text{or}\quad a'=g\tan\theta\]

6. Motion of a block on a rough inclined plane: When a block of mass \(m\) is moving down an inclined plane with an acceleration \(a\), with the coefficient of friction between the block and the inclined plane equal to \(\mu\), the force of friction \(\mu N\) will be acting upwards along the plane as shown in the figure. The weight \(mg\) of the block is resolved into the component \(mg\cos\theta\) opposite to the normal reaction and the component \(mg\sin\theta\) downwards, opposite to \(\mu N\).

From Newton’s second law of motion,

\[mg\sin\theta-\mu N=ma\]

\[N=mg\cos\theta\]

\[\text{or}\quad mg\sin\theta-\mu mg\cos\theta=ma\]

\[\therefore\ a=g(\sin\theta-\mu\cos\theta)\]

If the block is moving upwards and its retardation is \(a\) when the frictional force acts downwards,

\[ma=mg\sin\theta+\mu mg\cos\theta\]

\[\therefore\ a=g\sin\theta+\mu g\cos\theta\]

7. Banking of tracks: If a vehicle travels along a circular path on a horizontal level road, it skids away from the centre of curvature of the road due to inertia. When the vehicle tends to skid, a frictional force between the road and the tyres of the vehicle is produced, which provides the necessary centripetal force directed towards the centre and prevents the vehicle from skidding.

If a vehicle of mass \(m\) moves along a circular path of radius \(r\) with maximum speed \(v\) and \(\mu\) is the coefficient of static friction between the road and the tyres of the vehicle, the maximum speed of the vehicle allowed without skidding is given by:

\[\frac{mv^2}{r}=\mu mg\]

\[v=\sqrt{\mu rg}\]

The skidding can be avoided by banking the road at a suitable angle to the horizontal by raising the outer edge of the curved track with respect to the centre. If a vehicle of mass \(m\) is moving along a circular track of radius \(r\) which is banked or raised through an angle \(\theta\) with respect to the horizontal, the vehicle will have a normal reaction \(N\) which can be resolved into two rectangular components, \(N\cos\theta\) and \(N\sin\theta\), as shown in the figure.

The centripetal force is provided by the horizontal component of the normal reaction.

\[N\sin\theta=\frac{mv^2}{r}\qquad\ldots\text{(i)}\]

The vertical component of the normal reaction balances the weight of the vehicle.

\[\therefore\ N\cos\theta=mg\qquad\ldots\text{(ii)}\]

The angle of banking required for speed \(v\) can be calculated by dividing (i) by (ii):

\[\tan\theta=\frac{v^2}{rg}\quad\text{or}\quad\theta=\tan^{-1}\!\left(\frac{v^2}{rg}\right)\]

Thus a cyclist bends his body at an angle from the vertical direction when moving along a circular track.

If friction is present, a heavy vehicle of mass \(m\) moves along a circular track which is banked at angle \(\theta\); the vehicle tends to move up the plane, and to avoid skidding the frictional force \(F=\mu N\) acts on the vehicle in the downward direction (down the plane).

The equations of balance of forces are given as follows:

\[\frac{mv^2}{r}=N\sin\theta+F\cos\theta\]

\[mg=N\cos\theta-F\sin\theta\]

\[F=\mu N\]

Eliminating \(N\) and \(F\),

\[\frac{v^2}{rg}=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\]

\[\text{or}\quad\frac{v^2}{rg}=\frac{\tan\theta+\mu}{1-\mu\tan\theta}\]

\[v^2=rg\left[\frac{\tan\theta+\mu}{1-\mu\tan\theta}\right]\]

8. The conical pendulum: Consider a pendulum of length \(l\) with its bob of mass \(m\) revolving horizontally along a circular path of radius \(r\) with angular velocity \(\omega\), as shown in the figure. As the bob revolves, the string sweeps over the surface of a cone, forming a conical pendulum.

If \(T\) is the tension in the string when it is inclined at an angle \(\theta\) with the vertical, the vertical component of \(T\), \(T\cos\theta\), supports the weight of the bob and the horizontal component \(T\sin\theta\) provides the centripetal force.

\[T\cos\theta=mg\qquad\ldots\text{(i)}\]

\[T\sin\theta=mr\omega^2\qquad\ldots\text{(ii)}\]

Dividing (ii) by (i),

\[\tan\theta=\frac{r\omega^2}{g}\]

\[\omega=\sqrt{\frac{g\tan\theta}{r}}\]

The time period of revolution \(T_p\) is

\[T_p=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{r}{g\tan\theta}}\]

\[T_p=2\pi\sqrt{\frac{l\sin\theta}{g\tan\theta}}=2\pi\sqrt{\frac{l\cos\theta}{g}}=2\pi\sqrt{\frac{h}{g}}\]

where \(h=l\cos\theta\) is the vertical height of the bob below the point of suspension.

• Force is a push or pull upon an object resulting from the object's interaction with another object. The various effects of force are:
• Force can move a body initially at rest.
• Force can bring a moving body to rest.
• Force can change the direction of a moving body.
• Force can change the speed of a moving body.
• Force can change the shape of a body.
• Force can change the size of a body.

Muscular force – It involves the action of muscles.
Animals make use of muscular force to carry out their physical activities and other tasks.

Friction – It is an opposing force that acts between surfaces in contact moving with respect to each other.

• Frictional force always acts between two moving objects, which are in contact with one another.
• Frictional force always acts opposite to the direction of motion.
• Frictional force depends on the nature of the surface in contact.

Tension Force - This force appears in a string, attached to a rigid support, when an object is suspended by it.

Mechanical Force - It involves the force generated by machines.
Force exerted during collision -Two objects push each other with an equal but opposite forces if collision occurs between them. These forces are known as the force of action and force of reaction.

Combined Forces - When two or more forces are acting on the same object.
Non-contact force come into play even when the bodies are not in contact.

Magnetic force – Force acting between two magnets or a magnet and a magnetic material (eg. iron, steel, nickel, cobalt etc.). It can be attractive and repulsive.

Electrostatic force – Force due to electric charges. It can be attractive and repulsive.

Gravitational force – It is a kind of attractive force that comes into play because of the mass of a body. (eg. earth's gravitational attraction).

First law of motion
A body at rest remains at rest and a body in uniform motion continues its uniform motion unless an external force is applied.

Direction of change in momentum is the same as the direction of unbalanced force

Third law of motion
For every action force there is an equal and opposite reaction force.
The horse and the cart: From the third law of motion the pull by the horse in the forward direction is equal to the pull by the cart in the backward direction. The sum of these forces is therefore zero. Why should then the cart accelerate forward? Apart from the pulls of the horse and the cart there is frictional force and the reaction of the ground on the horse and the cart is also present. The resultant force of this normal reaction and the friction together helps the horse to move the cart in the forward direction.

Universal law of gravitation

For a point mass

The point of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the cell is concentrated at the center of the shell.
The force of attraction due to a hollow spherical shell of uniform density, on a point mass inside it is zero.

Gravitation Constant, G = 6.67 × 10^–11 Nm²/kg² 

Free Fall: A body is said to be free falling if it begins to fall downward towards the earth due to earth’s gravity only.
mg=GMmR2g=GMR2
‘g = 9.8 m/s2’ is the acceleration due to gravity, it is the acceleration of a freely falling body.

Tips to solve numerical
For upward motion take g = –9.8m/s² and final velocity at the highest point as 0.
For downward motion take g = 9.8m/s² and for a freely falling body take initial velocity as 0.

Equations of motion of an object under the influence of the earth’s gravity for downward motion of the particle
v=u+gts=ut+12gt2v2=u2+2gs
for upward motion of the particle
v=u-gts=ut-12gt2v2=u2-2gs

Mass: It is the amount of matter contained in the body
Weight: It is the force exerted on a body due to the gravitational pull of another body such as the earth, the sun etc.