## Revision Notes for CBSE Class 11 Physics Chapter 4 (Laws of Motion) - Free PDF Download

If students need reference by their side for exams, they must prefer Chapter 4 physics Class 11 notes. This chapter deals with Newton’s laws of motion and its real-time application. Students are provided with proper formulas that they can directly apply to solve their queries during exams. According to the CBSE board examination pattern, students should prepare well from notes readily available on Vedantu. These Newton’s laws of motion are thoroughly understood with notes of Physics Class 11 Chapter 4. These are highly helpful if it is about solving complex problems at different levels.

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## Laws of Motion Class 11 Notes Physics - Basic Subjective Questions

Section-A (1 Mark Questions)

1. Name the factor on which coefficient of friction depends?

Ans. The coefficient of friction will mainly depend upon two factors, they are as following:

1. The materials of the surfaces in contact.

2. The characteristics of the surfaces.

2. What provides the centripetal force to a car taking a turn on a level road?

Ans. The frictional force between the tyres and the road provides centripetal force.

3. Why is it desired to hold a gun tight to one's shoulder when it is being fired?

Ans. As the gun recoils after shooting, it must be held softly on the shoulder. Here the gun and the shoulder are one mass system, due to this the back kick will be reduced. A gunman must keep his weapon securely against his shoulder when shooting.

4. Why does a swimmer push the water backwards?

Ans. From the Newton's 3rd laws of motion, we know that "when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted". As a result, in order to swim ahead, the swimmer pushes water backward with his hands.

5. Friction is a self-adjusting force. Justify.

Ans. Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

6. A thief jumps from the roof of a house with a box of weight W on his head. What will be the weight of the box as experienced by the thief during jump?

Ans. The thief is in free fall during the jump. Both he/she and the box will be weightless during that time. So, the weight of the box experience by the thief during the jump will be zero. So, mathematically it can be written as: Weight of the box, W=m(g-a)=m(g-g)=0.

7. Action and reaction forces do not balance each other. Why?

Ans. Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

8. If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?

Ans. When a force acts in a perpendicular direction on a moving body, the work done by the force is zero. Since W = F.S cosθ, where S = 90° and cos90° = 0, therefore W = 0.

As a result, the magnitude of the body's velocity (or speed) will remain unchanged. The direction of motion of the body, however, will be altered.

9. The two ends of spring - balance are pulled each by a force of 10kg.wt. What will be the reading of the balance?

Ans. As the spring balancing is based on the tension in the spring, it reads weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg , and the reading will be 10kg .

10. A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Ans. There will be an increase in perceived weight. The apparent weight will stay the same as the true weight, if the lift moves at a constant pace.

### Section-B (2 Marks Questions)

11. Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed.

(b) A kite skillfully held stationary in the sky.

Ans.

(a) As the raindrop is falling with a constant speed, so its acceleration a will be 0. As the force acting on a particle is given by F = ma, so the net force acting on the rain drop will be 0.

(b) As the kite is held stationary, so by Newton's first laws of motion, the algebraic sum of forces acting on the kite is zero.

12. Two blocks of masses m1, m2 are connected by light spring on a smooth horizontal surface. The two masses are pulled apart and then released. Prove that the ratio of their acceleration is inversely proportional to their masses.

Ans. Due to inertia, the mass of the two bodies tries to expand, and the acceleration will act in the opposite direction as it shrinks. So let us assume that the F1 and F2 be the forces acting in opposite directions due to masses m1 and m2.

Thus $F_{1}+F_{2}=0$

$m_{1}+a_{2}=m_{2}a_{2}=0$

$m_{1}+a_{1}=m_{2}a_{2}=0$

$m_{1}+a_{1}=m_{2}a_{2}$

$\dfrac{a_{1}}{a_{2}}=-\dfrac{m_{1}}{m_{2}}$

Hence the above is proved.

13. Force of 16 N and 12 N are acting on a mass of 200 kg in mutually perpendicular directions. Find the magnitude of the acceleration produced?

Ans. In the given question, we have the force of 16 N and 12 N given and they are acting on a mass of 200 kg in mutually perpendicular directions. We need to find the magnitude of the acceleration produced.

$F=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\;cos\theta }$

Since, the forces are in mutually perpendicular directions. Therefore, (θ = 90°). Hence, the force will become:

$F=\sqrt{F_{1}^{2}+F_{2}^{2}}$

Now substituting the values, we get

$F=\left ( \sqrt{(16)^{2}+(12)^{2}} \right )$

=20 N

Hence, the magnitude of the acceleration will be:

$a=\dfrac{F}{m}$

$=\dfrac{20}{200}$

$=0\cdot 1ms^{-2}$

14. An elevator weighs 3000 kg. What is its acceleration when the tension supporting cable is 33000 N. Given that g = 9.8 ms−2.

Ans. From the question, we have an elevator having the weighs given as 3000 kg. We need to find the acceleration, if the tension in the supporting cable is given as 33000 N.

Net upward force on the Elevator F is equal to $=T-ms(\vec{}F=ma)$

$\Rightarrow ma=T-mg$

$\Rightarrow T=m(a+g)$

$\Rightarrow T=33000N=3000(a+9\cdot 8)$

$a=\dfrac{33000-3000\times 9\cdot 8}{3000}$

$=1\cdot 2ms^{-2}$

15. How does banking of roads reduce wear and tear of the tyres?

Ans. When a curving road is unbanked, the centripetal force is provided by friction between the tyres and the road. Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground's natural response supplies the necessary centripetal force, reducing tyre wear and tear.

16. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Ans. As from the question, we have the mass of each ball given as 0.05 kg

Initial velocity of each ball will be = 6 ms-1

Magnitude of the initial momentum of each ball given as, pi = 0.3 kg ms-1

The balls alter their directions of motion after colliding, but their velocity magnitudes do not change. Final momentum of each ball given as, $p_{f}=-0\cdot 3kg\;ms^{-1}$

Each ball's impulse equals a change in the system's momentum.

$=p_{f}-p_{i}$

$=0\cdot 3-0\cdot 3=-0\cdot 6\;kg\;ms^{-1}$

The negative indication implies that the balls are receiving opposite-direction shocks.

PDF Summary - Class 11 Physics Laws of Motion Notes (Chapter 4)

### Force:

A force is something that causes a body's rest or motion to alter. An interaction between two bodies is also referred to as force. Two bodies exert forces on each other even if they are not physically in contact, e.g., the electrostatic force between two charges or gravitational force between any two bodies.

It is a vector quantity having SI unit Newton (N) and dimension \[\left[ {ML{T^{--2}}} \right]\].

Superposition of Force: When many forces are acting on a body then their resultant is obtained by the laws of vector addition:

$\overrightarrow {{F_{res.}}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} + \overrightarrow {{F_3}} .............. + \overrightarrow {{F_n}}$

$\overrightarrow {{F_{res.}}} = \sqrt {{{\overrightarrow {{F_1}} }^2} + {{\overrightarrow {{F_2}} }^2} + 2{{\overrightarrow {{F_1}} }^{}}\overrightarrow {{F_2}} \cos {\theta ^{}}} $

$\tan \alpha = \dfrac{{{F_2}\sin \theta }}{{{F_1} + {F_2}\cos \theta }}$

The resultant of the two forces \[\overrightarrow {{F_1}} \] and \[\overrightarrow {{F_2}} \]acting at angle \[\theta \] is given by: \[\overrightarrow {{F_{res.}}} = \sqrt {{{\overrightarrow {{F_1}} }^2} + {{\overrightarrow {{F_2}} }^2} + 2{{\overrightarrow {{F_1}} }^{}}\overrightarrow {{F_2}} \cos {\theta ^{}}} \]

The resultant force is directed at an angle α with respect to force F1 where:

\[\tan \alpha = \dfrac{{{F_2}\sin \theta }}{{{F_1} + {F_2}\cos \theta }}\]

### Lami’s Theorem:

If three forces F1 , F2 and F3 are acting simultaneously on a body and the body is in equilibrium, then according to Lami’s theorem:

\[\dfrac{{{{\text{F}}_1}}}{{\sin (\pi - \alpha )}} = \dfrac{{{{\text{F}}_2}}}{{\sin (\pi - \beta )}} = \dfrac{{{{\text{F}}_3}}}{{\sin (\pi - \gamma )}}\] where α, β & γ are the angles opposite to the forces F1 , F2 & F3 respectively.

(Image will be uploaded soon)

### Basic Forces:

There are basically four types of forces: Weight, Contact force, Tension and Spring Force.

Weight: It is the force with which earth attracts a body toward itself. It is also called the gravitational force.

Contact Force: When two bodies come in contact, they exert forces on each other that are called contact forces.

Normal Force (N): This is the contact force component that is normal to the surface. It determines how tightly two surfaces are forced together.

Frictional Force (f): It is the component of contact force parallel to the surface. It opposes the relative motion (or attempted motion) of the two in contact surfaces.

Tension: It refers to the force exerted by the end of a taut string, rope, or chain. The direction of strain is pulling the body, whereas the natural reaction is pushing it.

Spring Force: It is the resists to change its length; the more you alter its length the harder it resists. The force exerted by a spring is given by \[F{\text{ }} = {\text{ }}--k.x\], where x is the change in length and k is the spring constant (unit Nm-1)

### Newton's Laws of Motion:

### Newton's First Laws:

If a body is at rest or in motion in a straight line, it will remain at rest or in motion unless it is acted by any external force. This is known as Law of Inertia or First laws of Newton

Inertia is the property of the inability of a body to change its position of rest or uniform motion in a straight line unless some external force acts on it.

Newton’s first laws is valid only in a frame of reference of the inertial frame, i.e., if a frame of reference is at rest or in uniform motion it is called inertial, otherwise non-inertial.

### Newton's Second Laws:

According to second laws, rate of change of momentum of a body is proportional to the resultant force acting on the body, i.e.,

$F \propto \dfrac{{dp}}{{dt}}$ or $F \propto ma$

$P = mv$

$\Rightarrow \dfrac{{dp}}{{dt}} = m\dfrac{{dv}}{{dt}}$.

The applied resultant force causes a change in momentum in the direction of the applied resultant force is a measure of the total amount of motion in the body.

\[\vec F=\dfrac {d{\vec v}}{dt}\]

\[\Rightarrow m {\vec a}=\dfrac {{\vec P_2}-{\vec P_1}}{t}\]

The external force acting on a body can accelerate it, either by changing the magnitude of velocity or direction of velocity or both.

Special Cases:

Case 1:

If the force is parallel or antiparallel to the motion of the body, then it changes only the magnitude of $\overrightarrow v $ but not the direction. Therefore, the path will be a straight line.

Case 2:

If the force is acting perpendicular to the motion of body, it changes only the direction but not the magnitude of $\overrightarrow v $. Therefore, the path will be Circular.

Case 3:

If the force acts at an angle θ to the motion of a body, it changes both the magnitude and direction of $\overrightarrow v $. In this case the path of the body may be elliptical, non-uniform circular, parabolic or hyperbolic.

### Newtons Third Laws:

According to this laws, for every action, there is an equal and opposite reaction. E.g. when two bodies A and B exerts a force on each other i.e. ${F_A}$ & ${F_B}$. Then the force exerted by any of the bodies will be the same as the force exerted by another body but in opposite direction. \[{{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}}\]

The two forces involved in any interaction between two bodies are called action and reaction.

### Linear Momentum:

It is defined as the product of the mass of the body and its velocity i.e.

Linear momentum = \[mass\:\times\:velocity\]

If a body of mass $m$ is moving with a velocity $v$ , its linear momentum $\overrightarrow P $is given by:

\[\overrightarrow P = m\overrightarrow v \]

It is a vector quantity and its direction is the same as the direction of the velocity of the body.

SI unit of linear momentum is \[kg{\text{ }}m{s^{--1}}\]and the cgs unit of linear momentum is \[g{\text{ }}cm{\text{ }}{s^{--1}}\].

### Impulse:

The entire change in linear momentum is used to calculate the force's impulse, which is the product of the average force during impact and the duration of the impact created during the collision.

The force which acts on bodies for short time are called impulsive forces. E.g. hitting a ball with a bat, firing a bullet with a gun etc.

An impulsive force does not remain constant instead, it varies from zero to maximum and then back to zero. Therefore, it is not possible to measure easily the value of impulsive force because it changes with time.

\[{\vec I} = {\vec F_{av}} \times {\text{t}} = {\vec p_2} -{\vec p_1}\]

### Apparent Weight on a Body in a Lift

(a) When the lift is at rest, i.e. a=0:

\[mg - R = 0\]

\[\Rightarrow mg = R\]

\[R = mg(1 - \dfrac{a}{g})\]

\[{W_{app.}} = {W_o}(1 - \dfrac{a}{g})\]

\[R + mg - mg = 0\]

\[{F_{AB}}\times \Delta t = {f_{ms}} = {\mu _s}R\]

\[\angle AOC = \theta\]

\[{W_{app.}} = {W_o}\]

(b) When the lift moves upwards with an acceleration a:

\[R - mg - ma = 0\]

\[\Rightarrow R = m(g + a)\]

\[\Rightarrow R = mg(1 + \dfrac{a}{g})\]

\[{\therefore W_{app}} = {W_o}{\left ( 1 + \dfrac{a}{g} \right )}\]

(c) When the lift moves downwards with an acceleration a:

\[R + ma - mg = 0\]

\[\Rightarrow R = mg(1 - \dfrac{a}{g})\]

\[\therefore {W_{app}} = {W_o}(1 - \dfrac{a}{g})\]

(d) When the lift falls freely, i.e., a = g :

\[R + mg - mg = 0\]

\[\Rightarrow R\;=\;0\]

\[\therefore {W_{app}}\;=\;0\]

### Principle of Conservation of Linear Momentum-

According to this principle, in an isolated system, the vector sum of all the system's linear momenta is conserved and is unaffected by their interactions reciprocal action and response.

Mutual forces between pairs of particles in an isolated system (i.e., a system with no external force) can thus produce changes in the linear momentum of individual particles. The linear momentum changes cancel in pairs, and the overall linear momentum remains unaltered because the mutual forces for each pair are equal and opposing. As a result, an isolated system of interacting particles' total linear momentum is conserved. This principle is a direct result of Newton's second and third laws of motion.

Consider an isolated system that consists of two bodies A and B with initial linear momenta \[{P_A}\] and \[{P_B}\]. Allow them to collide for a short time t before separating with linear momenta \[{P_{A}}'\] and \[{P_{B}}'\], respectively.

If \[{F_AB}\]is force on A exerted by B, and \[{F_{BA}}\] is force on B exerted by A, then according to second law of newton:

\[{F_{AB}}\times \Delta t = \]Change in linear momentum of A = \[{\vec P_{A}}' - {\vec P_A}\]

\[{F_{BA}} \times \Delta t = \] Change in linear momentum of B= \[{\vec P_{B}}' - {\vec P_B}\]

Now, according to third law of newton:

\[\Rightarrow {{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}}\]

\[\therefore\:from\:equations\:we\:get\:\]

\[\Rightarrow {{\vec P_{A}}'-{\vec P_A}}\:=\:{{\vec P_{B}}'-{\vec P_B}}\]

or

\[\Rightarrow {{\vec P_{A}}'+{{\vec P_B}}'}={{\vec P_{A}}+{\vec P_{B}}}\]

which shows that the total final linear momentum of the isolated system is equal to its total initial linear momentum. This proves the principle of conservation of linear momentum.

### Friction:

Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

It is the force that comes into play when two surfaces come into contact with each other and oppose their relative motion.

I Frictional force is unaffected by the contact area. Because with an increase in the area of contact, the force of adhesion also increases.

When the surfaces in contact are extra smooth, the distance between the molecules of the surfaces in contact decreases, increasing the adhesive force between them. Therefore, the adhesive pressure increases, and so does the force of friction.

### Types of Friction:

There are 3 types of friction: Static, Limiting and Kinetic Friction.

Static Friction- The opposing force that comes into play when one body tends to move over the surface of another body, but the actual motion has yet not started is called Static friction.

Limiting Friction- Limiting friction is the maximum opposing force that comes into play when one body is just on the verge of moving over the surface of the other body.

Kinetic Friction - Kinetic friction or dynamic friction is the opposing force that comes into play when one body is actually moving over the surface of another body.

### Laws of Limiting Friction:

(i) Friction always works in the opposite direction of relative motion, making it a perverse force.

(ii) The maximum static friction force, fms (also known as limiting friction), is proportional to the normal reaction (R) between the two in contact surfaces i.e.

\[{f_{ms}} \propto R\]

(iii) The limiting friction force is tangential to the interface between the two surfaces and is determined by the kind and state of polish of the two surfaces in contact.

(iv) As long as the normal reaction stays constant, the limiting friction force is independent of the area of the surfaces in contact.

### Coefficient of Static Friction:

We know that,

\[{f_{ms}} \propto R\]

\[\Rightarrow {f_{ms}} = {\mu _s}R\]

Here, \[{\mu _s}\] is a constant of proportionality and is called the coefficient of static friction and its value depends upon the nature of the surfaces in contact and is usually less than unity i.e. 1 but never 0.

Since the force of static friction (fS) can have any value from zero to maximum (\[{f_{ms}}\] ), i.e. \[{f_s} \leqslant {f_{ms}}\],

Kinetic Friction:

The laws of kinetic friction are exactly the same as those for static friction. Accordingly, the force of kinetic friction is also directly proportional to the normal reaction:

\[{f_k} \propto R\]

\[\Rightarrow {f_k} = {\mu _k}R\]

Rolling Friction:

The opposing force that comes into play when a body rolls over the surface of another body is called rolling friction.

Friction caused by rolling. Consider a wheel that is rolling down a street. As the wheel travels down the road, it presses against the road's surface and is compressed slightly as shown in Fig. :

(Image will be uploaded soon)

### Angle of Friction:

The angle of friction between any two surfaces in contact is defined as the angle formed by the resultant of the limiting friction force F and the normal reaction direction R. It is represented by θ.

In the given fig. OA represents the normal reaction R that balances the weight mg of the body. OB represent F, the limiting force of sliding friction, when the body tends to move to the right. Complete the parallelogram OACB. Join OC. This represents the resultant of R and F. By definition, \[\angle AOC = \theta \] is the angle of friction between the two bodies in contact.

(Image will be uploaded soon)

The angle of friction is determined by the nature of the materials used on the surfaces in contact as well as the nature of the surfaces themselves.

Relation Between 𝛍 and 𝛉:

In $\Delta AOC$, $\tan \theta = \dfrac{{AC}}{{OA}} = \dfrac{{OB}}{{OA}} = \mu $$\tan \theta = \dfrac{{AC}}{{OA}} = \dfrac{{OB}}{{OA}} = \mu $

Hence, $\mu = \tan \theta $ ... (6)

(i.e.\[\mu\] is the coefficient of limiting friction)

### Angle of Repose or Angle of Sliding:

The minimum angle of inclination of a plane with the horizontal at which a body placed on the plane begins to slide down is known as the angle of repose or angle of sliding.

Represented by α. Its value depends on the material and nature of the surfaces in contact.

In fig., AB is an inclined plane such that a body placed on it just begins to slide down. \[\angle BAC\:\alpha\]= angle of repose.

(Image will be uploaded soon)

The various forces involved are :

weight, mg of the body,

normal reaction, R,

Force of friction F,

Now, mg can be resolved into two rectangular components: mg cosα

opposite to R and mg sinα opposite to F:

$F = mg\sin \alpha $ ... (7)

$R = mg\cos \alpha $ ... (8)

Diving these two eq. we get:

$\tan \alpha = \dfrac{F}{R}$

$\Rightarrow \tan \alpha = \mu $ ... (9)

Hence coefficient of limiting friction between any two surfaces in contact is equal to the tangent of the angle of repose between them.

From (6) and (9): $\mu = \tan \alpha = \tan \theta $

Therefore $\alpha = \theta $

i.e. (Angle of friction = Angle of repose)

### Methods of Changing Friction:

Some of the ways of reducing friction are:

By polishing.

By lubrication.

By proper selection of materials.

By Streamlining.

By using ball bearings.

### Dynamics of Uniform Circular Motion Concept of Centripetal Force:

The force required to move a body uniformly in a circle is known as centripetal force. This force acts along the circle's radius and towards the centre.

When a body moves in a circle, the direction of motion at any given time is along the tangent to the circle. According to Newton’s first law of motion, a body cannot change its direction of motion by itself an external force is needed. This external force is called the centripetal force.

An expression for centripetal force is:

\[F = m{v^2}/r = m{\omega ^2}r\]

\[R{\text{ }}--{\text{ }}mg{\text{ }} = {\text{ }}0{\text{ }}or{\text{ }}R{\text{ }} = {\text{ }}mg \]

\[F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R{\text{ }} = {\text{ }}{\mu _s}{\text{ }}mg{\text{ }} \]

\[R{\text{ }}cos\theta = {\text{ }}mg{\text{ }} + {\text{ }}F{\text{ }}sin\theta \]

\[From\:{\text{ }}\left( 3 \right),{\text{ }}R{\text{ }}\left( {cos\theta --{\text{ }}{\mu _s}{\text{ }}sin\theta } \right){\text{ }} = {\text{ }}mg \]

On account of a continuous change in the direction of motion of the body, there is a change in velocity of the body, and hence it undergoes an acceleration, called centripetal acceleration or radial acceleration.

### Centrifugal Force:

Centrifugal force is a force that arises when a body is moving actually along a circular path, by virtue of the tendency of the body to regain its natural straight-line path.

When a body is moving in a straight line centripetal force is applied on the body, it is forced to move along a circle. The body has a natural inclination to return to its normal straight-line course while moving in a circle. This tendency gives rise to a force called centrifugal force.

The magnitude of centrifugal force = $m{v^2}/r$, which is the same as that of centripetal force, but opposite in direction i.e. The centrifugal force acts along the circle's radius, away from the centre.

Note: Centripetal and Centrifugal forces, being the forces of action and reaction act always on different bodies. E.g. when a piece of stone tied to one end of a string is rotated in a circle, centripetal force F1 is applied on the stone by the hand. Due to the stone's desire to revert to its natural straight line route, centrifugal force F2 acts on it, pulling the hand outwards.. The centripetal and centrifugal forces are shown in Fig. :

(Image will be uploaded soon)

### Rounding A-Level Curved Road:

When a vehicle goes around a curved road, it requires some centripetal force. The vehicle's wheels have a tendency to depart the curved path and return to the straight-line path as it rounds the curve. Force of friction between the wheels and the road opposes this tendency of the wheels. This force (friction) therefore, acts, towards the centre of the circular track and provides the necessary centripetal force.

Three forces are acting on the car, fig.

(Image will be uploaded soon)

The weight of the car, mg, acting vertically downwards,

Normal reaction R of the road on the car, acting vertically upwards,

Frictional Force F, along the surface of the road, towards the centre of the turn.

As there is no acceleration in the vertical direction,

$R{\text{ }}--{\text{ }}mg{\text{ }} = {\text{ }}0{\text{ }}or{\text{ }}R{\text{ }} = {\text{ }}mg$ ...(1)

The centripetal force required for circular motion is applied along the road's surface, toward the turn's centre. As previously stated, static friction is what supplies the required centripetal force. Clearly $F \leqslant m{v^2}/r$ …(2)

where v is velocity of car and r is the radius of the curved path as,

$F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R{\text{ }} = {\text{ }}{\mu _s}{\text{ }}mg{\text{ }}$(${\mu _s}$ is coefficient of static friction between the tyres and the road)

Therefore from (2),

\[\dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}} \leqslant {{\mu_s}mg}\]

or

\[v\:\leqslant\:\sqrt{{\mu_s}rg}\]

\[\therefore {\vec v_{max}}= \sqrt{{\mu_s}rg}\] ……(3)

Therefore, the maximum velocity with which car can move without slipping is:

\[{\text{v}} = \sqrt {{\mu _{\text{s}}}{\text{rg}}} \]

### Banking of Roads:

The phenomenon of raising the outer edge of the curved road above the inner edge is called banking of roads.

The maximum permissible velocity with which a vehicle can go round a level curved road without skidding depends on μ (Coefficient of friction between the tyre and the road). As a result, the frictional force is not a reliable source of the required centripetal force for the vehicle.

In Fig., OX is a horizontal line. OA is the level of the banked curved road whose outer edge has been raised.

$\angle XOA{\text{ }} = \;\theta $= angle of banking.

(Image will be uploaded soon)

Forces are acting on the vehicle as shown in Fig:

Weight mg of the vehicle acting vertically downwards.

Normal reaction R of the banked road acting upwards in a direction perpendicular to OA.

Force of friction F between the banked road and the tyres, acting along with AO. R can be resolved into two rectangular components-

R cosθ, along a vertically upward direction

R sinθ, along the horizontal, towards the centre of the curved road. F can also be resolved into two rectangular components:

(i) F cosθ, along the horizontal, towards the centre of curved road

(ii) F sinθ, along the vertically downward direction.

As there is no acceleration along the vertical direction, the net force along this direction must be zero. Therefore,

$R{\text{ }}cos\theta = {\text{ }}mg{\text{ }} + {\text{ }}F{\text{ }}sin\theta $ ……(1)

If v is the vehicle's velocity on a banked circular road with radius r, then centripetal force is =$m{v^2}/r$. This is provided by the horizontal components of R and F as shown in Fig.

Therefore, \[R\sin \theta + F\cos \theta = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}\] ...(2)

But F < ${\mu _s}$ R, where ${\mu _s}$ is coefficient of static friction between the banked road and the tyres. To obtain ${v_{\max }}$, we put $F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R$ in (1) and (2)

\[{\text{R}}\cos \theta = {\text{mg}} + {\mu _{\text{s}}}{\text{R}}\sin \theta \] … (3)

And \[{\text{R}}\sin \theta + {\mu _{\text{s}}}{\text{R}}\cos \theta = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}\] …(4)

$From{\text{ }}\left( 3 \right),{\text{ }}R{\text{ }}\left( {cos\theta --{\text{ }}{\mu _s}{\text{ }}sin\theta } \right){\text{ }} = {\text{ }}mg$

\[{\text{R}} = \dfrac{{{\text{mg}}}}{{\cos \theta - {\mu _{\text{s}}}\sin \theta }}\] …(5)

From (4), $\quad {\text{R}}\left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right) = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$

$\operatorname{Using} (5),\dfrac{{\operatorname{mg} \left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right)}}{{\left( {\cos \theta - {\mu _{\text{s}}}\sin \theta } \right)}} = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$

$\therefore \quad {v^2} = \dfrac{{\operatorname{rg} \left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right)}}{{\left( {\cos \theta - {\mu _{\text{s}}}\sin \theta } \right)}} = \dfrac{{\operatorname{rg} \cos \theta \left( {\tan \theta + {\mu _{\text{s}}}} \right)}}{{\cos \theta \left( {1 - {\mu _{\text{s}}}\tan \theta } \right)}}$

\[{\text{v}} = {\left[ {\dfrac{{rg\left( {{\mu _{\text{s}}} + \tan \theta } \right)}}{{\left( {1 - {\mu _{\text{s}}}\tan \theta } \right)}}} \right]^{1/2}}\] …(6)

### Discussion:

If \[{\mu _s}\]= 0, i.e., if banked road is perfectly smooth, then from eqn. (51), \[{v_o} = {(rg\tan \theta )^{\dfrac{1}{2}}}\]

Even when there is no friction, this is the speed at which a banked road may be rounded. On a banked road, driving at this speed causes essentially little wear and tear tyres.

\[{v_o}^2 = {\text{ }}rg{\text{ }}\tan \theta \]

\[\Rightarrow {\overrightarrow F _{real}} + {\overrightarrow F _{pseudo}} = {m_p}{a_{P,O}}\]

\[{\overrightarrow F _{real}} - {m_p}{\overrightarrow a _o} = {m_p}{\overrightarrow a _{P,O}}\]

If speed of vehicle is less than \[{v_o}\] , frictional force will be up the slope. Therefore, the vehicle can be parked only if \[\tan \theta \leqslant {\mu _s}\].

The average speed of vehicles passing through a road is usually banked. However, if a vehicle's speed is slightly less or more than this, the self-adjusting static friction between the tyres and the road will operate, and the vehicle will not slide.

Note that curved railway tracks are also banked for the same reason. The level of the outer rail is raised a little above the level of the inner rail while laying a curved railway track.

### Bending of a Cyclist:

When a cyclist turns, he must also exert centripetal force. His weight is balanced by the usual reaction of the ground if he keeps himself vertical while spinning. In that case, he must rely on the friction between the tyres and the road to generate the required centripetal force. Because the force of friction is modest and unpredictable, relying on it is risky.

In order to avoid relying on friction to generate centripetal power, the cyclist must bend slightly inwards from his vertical position while turning. A component of normal reaction in the horizontal direction produces the required centripetal force in this way.

To calculate the angle of bending with vertical, suppose

m = mass of the cyclist,

v = velocity of the cyclist while turning,

r = radius of the circular path,

\[\theta\] = angle of bending with the vertical

In the fig. shown below the weight of the cyclist(mg) acts vertically downwards at the centre of gravity C. R is the normal reaction on the cyclist. It acts at an angle \[\theta\] with the vertical.

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R can be resolved into two rectangular components: \[R\:\cos\theta\], along the vertically upward direction,\[R\:\sin\theta\], along the horizontal, towards the centre of the circular track.

In equilibrium, \[R{\text{ }}cos\theta = {\text{ }}mg\]\[R{\text{ }}cos\theta = {\text{ }}mg\] ….(1)

\[R{\text{ }}\sin \theta = {\text{ }}m{v^2}/r\] ….(2)

Dividing (2) by (1), we get, \[\tan \theta = {\text{ }}\dfrac{{{v_{}}^2}}{{rg}}\]

For a safe turn, θ should be small, for which v should be small and r should be large i.e. Turning should be done slowly and on a greater radius track. This means, a safe turn should neither be fast nor sharp.

### Pseudo Force:

\[{\overrightarrow F _{pseudo}} = - {m_p}{\overrightarrow a _0}\]If observer O is non-inertial and still wants to apply Newton's Second Law on particle P, the observer must add a "Pseudo force" to the real forces on particle P.

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\[{\overrightarrow F _{real}} + {\overrightarrow F _{pseudo}} = {m_p}{a_{P,O}}\]

i.e \[{\overrightarrow F _{real}} - {m_p}{\overrightarrow a _o} = {m_p}{\overrightarrow a _{P,O}}\]

Where \[{\overrightarrow a _{P,O}}\] is acceleration of P with respect to observer O.

## Laws of Motion Class 11 Notes - Free PDF Download

Free Laws of Motion Class 11 NCERT notes PDF is available online to prepare well for their exams. Different teachers have prepared this PDF with years of experience and prepared different questions after learning the CBSE question paper. These notes are significantly helpful to the students as questions are prepared with appropriate answers to different levels of questions.

Physics Class 11 Chapter 4 notes are easy to read that covers all questions asked in the exams. Thus it helps students to prepare well before their final board class. All the answers are described in a precise manner such that students need not refer anywhere else for the same. According to marks assigned, the concepts are quickly clarified, and students become aware of answering a particular question.

### What Topics are Covered With Ch 4 Physics Class 11 Notes?

NCERT covers the entire syllabus as preferred by the CBSE board. Thus notes of Laws of Motion Class 11 are well prepared according to the question paper pattern designed by CBSE. These notes aid students while preparation and hence reduce their stress that comes during a hectic study schedule. These revision notes are best for self-study. Experienced teachers have prepared them according to the requirement. Below are the topics covered in physics Chapter 4 Class 11 notes:

The chapter begins with stating Dynamics, which is the study of forces and motions.

Then the first definition is covered: how to define forces?

The next concept is regarding Inertia along with its three different states.

Finally, it covers the definition of three different laws of motion along with the subtopics to prove these laws in detail. Numerical problems are also covered under this concept.

Now let us study each concept that is covered in Class 11 physics ch 4 notes.

### Dynamics

Dynamics is defined as the term in physics, which deals with the study and knowledge about motion forces and laws.

### Inertia

What is Inertia? Also, Describe Its Three Different States.

If an object moves in a straight line or is at rest or in a uniform motion, it cannot change its state to another until compelled. This property is known as inertia. It is measured according to the mass of a body.

There Are Three Different Types of Inertia

Inertia of motion

Inertia of rest

Inertia of direction

### Force

Class 11 physics laws of motion notes also cover a basic definition of force and its interaction with different objects. Force is the basic pull and pushes applied to an object, whether the object is at rest or moving. We can also say that it is an interaction of one object with another, due to which an object changes its state.

On general terms, there are two different types of forces:

Constant Force

Action Force

### Newton’s Laws of Motion

Newton has given three different laws of motion along with the proof to each law.

Newton’s First Law of Motion: Until you apply any external force to an object, the object will not change its state. It is basically if an object is at rest or moving uniformly in a straight line.

Also, this definition is called the Law of Inertia.

Newton’s Second Law of Motion: The net force is equivalent to the magnitude of the net force along with the direction of the applied net force.

Also, many define the second law of motion as the force directly proportional to acceleration produced while moving and the mass of the body.

\[\vec{F} = K \frac{d\bar{p}}{dt} = Km\vec{a}\]

Suppose F is the net force and m is the mass of the body, and a is the acceleration then Force is defined by the expression given above. Here, k is the constant of proportionality.

Newton’s Third Law of Motion: It is defined that if we apply a force to an object, we will experience an equal force in the opposite direction applied by the object on us.

### This Law Deals With Some Other Basic Subtopics Like

Linear Momentum

Impulse

Law of Conservation of Momentum

Concurrent forces and Equilibrium

Tension

Notes of Chapter 4 physics Class 11 will also cover the concept of the simple pulley, friction, actual and apparent weight, laws of friction, angle of friction, coefficient of friction, motion in a circle, and many other numerical problems associated with it.

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Students need a definitive tool to prepare well for exams. These physics ch 4 Class 11 notes are the best way to prepare well for the exams and avoid the last-minute rush. The most difficult chapter needs basic clarification for each topic and its subtopic if you want to score good marks.

Also, students get the least time to prepare their notes. Hence, these Vedantu notes will aid them at every step to clarify their concepts. According to CBSE format, the entire content of notes of ch 4 physics Class 11 is prepared by experienced teachers. Thus students need not scroll any other website for their help.

## Benefits for Taking Notes of Laws of Motions Class 11 CBSE Physics Chapter 4

The advantages of taking Laws of Motion Class 11 Notes CBSE Physics Chapter 4 in the form of a free PDF download aid students in effective learning, revision, and exam preparation, leading to improved performance in their academic journey.

Taking Laws of Motion Class 11 Notes CBSE Physics Chapter 4 in the form of a free PDF download offers several advantages to students:

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### Conclusion

The Laws of Motion Class 11 Notes CBSE Physics Chapter 4 in the form of a free PDF download offer a valuable resource to students pursuing physics. These comprehensive notes provide simplified explanations, examples, and practice questions, enhancing students' understanding and problem-solving skills related to motion. The convenience of accessing these notes as a PDF download allows for easy revision and quick reference during exam preparation. By utilizing these notes, students can efficiently cover the entire chapter, reinforcing their conceptual knowledge and preparing effectively for assessments. The free availability of these notes ensures equitable access to quality study material, promoting a holistic learning experience for all students.

## FAQs on Laws of Motion Class 11 Notes CBSE Physics Chapter 4 (Free PDF Download)

1. A 5 kg iron sphere feels an upsurge of velocity from 2 m/s to 4 m/s. The velocity rises within the 20 s because a constant force acts on it by unchanging the direction. Find the magnitude & direction of this force.

As per the question

Mass (m) = 5 kg

Time, t = 20 s

Initial velocity, u = 2 m/s

Final velocity, v = 4 m/s

Also, F = m * a …. [1]

As, v = u + at (equation of motion)

a = (v - u) / t

Or, m [(v - u) / t] = Force

Since, a = (v – u) / t

Force = 5 [(4 – 2) / 20]

= 0.5 N (along the direction of motion)

2. What are the chief roles played by Vedantu in CBSE board exams?

Vedantu is involved in so many educational activities. It is paying attention to the quality studies for Physics also. The awareness, along with thorough solved question and answer practice papers, is very helpful to our students for the CBSE board exams.

3. Explain Newton’s law of gravitation?

If two particles possessing different masses (M and m) are mutually attracted with identical and opposite forces such as F and - F.

The relation is F = G(Mm/r^{2})

Here, r = distance between the two particles

G = the universal constant of gravitation

4. How to prepare for Chapter 4 of Class 11 Physics?

The student should make a routine allotting equal hours to every subject and following this regularly to achieve better results. Apart from this, the student should study the given chapter line by line and highlight the important points and topics to retain so that they help during the examination. After this is done, the student should refer to the NCERT solutions and solve the given exercises to attain a better practice at the type of questions that can be asked in the question paper. Consistent practice and hard work will help the student in passing with flying colours.

5. Where can I get the NCERT Solutions for Chapter 4 of Class 11 Physics?

The NCERT solutions are planned and devised to meet the demands and needs of the students. These are planned by experts, containing various exercises considering the syllabus present in the curriculum. These exercises help the student to get a strong idea and understanding of the concept and topics so that the questions asked in the question papers don't perplex them. It is advised that the students have a copy of the NCERT solutions for every subject. These can be availed easily online and practised in the comforts of the home.

6. What are the important topics covered in Chapter 4 of Physics Class 11?

The important topics that are covered in Class 11 Chapter 4 Physics are:

Laws of motion

The law of inertia

Aristotle's fallacy

Newton’s first, second and third laws of motion

Conservation of momentum

Equilibrium of a particle

Common forces in mechanics

Circular motion

Solving problems in mechanics.

All these topics are important for the student to be well versed in. To achieve this they can refer to the NCERT Solutions offered by Vedantu and these solutions can be downloaded online. The exercises that these solutions offer will help the student to grasp a stronghold of the concepts present in Chapter 4 Physics.

7. Why is it important to refer to the NCERT Solutions for Chapter 4 of Class 11 Physics?

The NCERT solutions are devised by student matter experts in order to fulfil the needs and demands of the students. These solutions have numerous exercises that the student needs to practice to retain the important concepts better. Apart from this all these exercises have detailed and explained answers that will help the student in every step of the understanding process, clearing all their doubts and confusions. These solutions also have in the previous year’s question papers that the student can practice getting an idea of how the question might be asked. Thus, practising and acing the NCERT solutions will inevitably lead the student in grabbing good grades in the exam.

8. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Let m_{1} and m_{2} be the mass of the product respectively

Let v_{1} and v_{2} be the velocity respectively

Thus, the total linear momentum after disintegration becomes m_{1}v_{1}+m_{2}v_{2}

But it is to be noted that before disintegration the nucleus is at rest and therefore the linear momentum before disintegration becomes zero

Thus, following the principle of conservation of linear momentum;

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