**This article is formulated according to the 5th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

**Motion of the object needs the concept of Velocity and acceleration. In earlier chapter we have studied the motion but not what causes the motion? In this chapter we study about what causes the motion.**

In early days it was known that some influence was needed to keep the body in motion and it was also known that rest is the natural state of an object.

## Aristotle’s Law and its Fallacy

**Law:** An external force is required to keep a body in motion.

**Fallacy in the law:** A moving object comes to rest because; the external force of friction on the object by the floor opposes its motion. If there is no friction no force is required to keep the object in motion.

## Law of Inertia

If the net external force is zero, a body at rest continues to be at rest and a body in motion continues to be in uniform motion.

**Note:** Aristotle’s viewpoint about the motion of the body was rejected by Galileo and gave the law of Inertia.

## Inertia

The property of a body to change its state of rest or uniform motion unless some external force acts on it. Mass of a body is measure of inertia.

## Types of inertia

**(i) Inertia of rest:** The property of a body to remain at rest.

**(ii) Inertia of motion:** The property of a body to oppose the change in its motion.

## Newton’s laws of motion

Based on the Galileo’s idea, the intimate relationship between force acting on a body and its motion executed by the body was first understood by Isaac Newton.

## Newton’s first law of motion

Everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

## Alternate statement of Newton’s first law

The first law can be stated in terms of acceleration as

“If the net external force on a body is zero, its acceleration is zero. Acceleration can non-zero only if there is a net external force on the body”.

Ex: A passenger in a bus is pushed back when the bus suddenly starts moving. A person in a moving vehicle tends to fall forward when the vehicle suddenly stops.

**Significance**

Newton’s first law of motion gives the definition for force and reveals Inertia, a fundamental property of all matter. Force is a vector quantity and dimensions are [𝑀𝐿𝑇^{−2}].

## Force

The external agency which changes or tends to change the state of rest or state of uniform motion of a body in a straight line.

## Momentum

Momentum of a body is defined as the product of its mass and velocity. **Momentum = mass × velocity**

𝒑⃗ = 𝒎𝒗⃗

Momentum is a vector quantity, and its SI unit is 𝑘𝑔𝑚𝑠^{−1}.

Dimensions of momentum are [𝑀𝐿𝑇^{−1}].

**Note: **The concept of momentum was introduced by Newton. It is a measure of the ability of a body to impart motion to another.

## Newton’s Second law of motion

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Mathematically, 𝑑𝑝/𝑑𝑡 ∝ 𝐹⃗

## Derivation of 𝑭 = 𝒎𝒂 using Newton’s second law of motion

Consider a body of mass 𝑚, moving with a velocity 𝑣⃗_{1} and having momentum 𝑝⃗_{1}. Let a force 𝐹 acts on it for time ∆𝑡. Then velocity changes to 𝑣⃗_{2} and momentum to 𝑝⃗_{2}. Then change momentum is, 𝑝⃗_{2} − 𝑝⃗_{1} = 𝑚𝑣⃗_{2} − 𝑚𝑣⃗_{1}

𝑑𝑝 = 𝑚(𝑣⃗_{2} − 𝑣⃗_{1})

𝑑𝑝 = 𝑚𝑑𝑣⃗

From second law of motion, 𝐹 ∝ 𝑑𝑝/𝑑𝑡

𝐹 = 𝑘(𝑑𝑝/𝑑𝑡)

𝐹 = 𝑘𝑚(𝑑𝑣/𝑑𝑡)

𝐹 = 𝑘𝑚(𝑑𝑣/𝑑𝑡)

𝐹 = 𝑘𝑚𝑎

For simplicity, we choose 𝑘 = 1.

Then, 𝑭⃗ = 𝒎𝒂⃗

Significance

Newton’s second law of motion signifies momentum and gives a formula to measure the force.

## Unit of Force

SI unit of force is 𝑛𝑒𝑤𝑡𝑜𝑛 or 𝑁 and 1𝑁 = 1𝑘𝑔𝑚𝑠^{−2}

## newton(N)

One newton is that force which causes an acceleration of 1𝑚𝑠^{−2} to a body of mass 1𝑘𝑔.

## Some applications of Newton’s second law

- A cricket player lowers his hands while catching a ball.
- A person falling on a cemented floor gets injured but a person falling on heap of sand is not.
- The vehicles are fitted with shockers (springs).
- Glass wares and China wares are wrapped with straw pieces before transportation.

## Components of force

We have, 𝐹 = 𝐹_{𝑥}𝑖̂+ 𝐹_{𝑦}𝑗̂+ 𝐹_{𝑧}𝑘̂ and 𝑎 = 𝑎_{𝑥}𝑖̂+ 𝑎_{𝑦}𝑗̂+ 𝑎_{𝑧}𝑘̂.

Then, 𝐹_{𝑥}𝑖̂+ 𝐹_{𝑦}𝑗̂+ 𝐹_{𝑧}𝑘̂ = 𝑚(𝑎_{𝑥}𝑖̂+ 𝑎_{𝑦}𝑗̂+ 𝑎_{𝑧}𝑘̂)

On comparing the co-efficient of 𝑖,̂ 𝑗̂𝑎𝑛𝑑 𝑘̂ We have, 𝐹_{𝑥} = 𝑚𝑎_{𝑥}, 𝐹_{𝑦} = 𝑚𝑎_{𝑦} and 𝐹_{𝑧} = 𝑚𝑎_{𝑧}

## Impulsive force

Large force acting on a body for a short time is called an impulsive force.

**Ex: **A ball hit by bat, kicking a football, hammering a nail etc.

## Impulse

It is the product of the force and time interval for which the force acts. It is denoted by J.

𝑖𝑚𝑝𝑢𝑙𝑠𝑒 (𝐽) = 𝐹𝑜𝑟𝑐𝑒 × 𝑡𝑖𝑚𝑒 = 𝐹𝑡

It is a vector. SI unit of Impulse is 𝑛𝑒𝑤𝑡𝑜𝑛–𝑠𝑒𝑐𝑜𝑛𝑑(𝑁𝑠). Dimensions are [𝑀𝐿𝑇^{−1}].

**Note: **Impulse and momentum have same dimensions.

## Impulse-Momentum theorem

Impulse is equal to change in momentum.

**Proof:** Impulse, **𝐽** = **𝐹**𝑡. But we have **𝐹** = 𝑚**𝑎**, **𝐽** = 𝑚**𝑎**𝑡, **𝐽** = 𝑚(**𝑣** − **𝑣**** _{0}**) (∵

**𝑣**−

**𝑣**

**=**

_{0}**𝑎**𝑡)

**𝐽** = 𝑚**𝑣** − 𝑚**𝑣**_{0}

**𝐽** = 𝑓𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚

𝑱 = 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎

## Newton’s third law of motion

“To every action there is always an equal and opposite reaction”.

𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝐴 𝑏𝑦 𝐵 = −(𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝐵 𝑏𝑦 𝐴)

𝑭⃗_{𝑨𝑩} = −𝑭⃗_{𝑩𝑨}

**Note:** The term action and reaction mean the force and the force on the object A by the object B and the force on object B by A act at the same instant. Action and reaction forces act on different bodies, not on the same body. So they do not cancel each other.

**Significance**

Newton’s third law signifies that forces never occur singly in nature, but they always occur in pairs. Launching of rocket is based on this law.

## Illustrations of Newton’s third law

- When a person jumps from a boat, he pushes the boat in the backward direction while the boat pushes him in the forward direction.
- A swimmer pushes the water in the backward direction and the water pushes the swimmer in the forward direction.
- A person walking on the floor.

## Law of conservation of momentum

“The total momentum of an isolated system of interacting particle is conserved”.

Ex: Recoil of gun, the motion of rocket is based on this principle.

**Proof:** Consider two bodies A and B, with initial momentum **𝑝 _{𝐴}** and

**𝑝**. Let the bodies collide and get apart with final momentum

_{𝐵}**𝑝**and

_{𝐴}’**𝑝**respectively. From Newton’s second law,

_{𝐵}’Force on 𝐴 by 𝐵 is, **𝐹 _{𝐴𝐵}**

**=**

**𝑝**

_{𝐴}**′**−

**𝑝**/𝑑𝑡 and Force on 𝐵 by 𝐴 is,

_{𝐴}**𝐹**=

_{𝐵𝐴}**𝑝**

_{𝐵}**′**

**−**

**𝑝**/𝑑𝑡 where 𝑑𝑡 is time for which the bodies are in contact. But from Newton’s third law,

_{𝐵}**𝐹 _{𝐴𝐵}** = −

**𝐹**

_{𝐵𝐴}(**𝑝 _{𝐴}**

**′**−

**𝑝**)/𝑑𝑡 = − (

_{𝐴}**𝑝**

_{𝐵}**′**−

**𝑝**)/𝑑𝑡

_{𝐵}**𝑝 _{𝐴}**

**′**−

**𝑝**= − (

_{𝐴}**𝑝**

_{𝐵}**′**−

**𝑝**)

_{𝐵}**𝑝 _{𝐴}**

**′**−

**𝑝**= −

_{𝐴}**𝑝**

_{𝐵}**′**+

**𝑝**

_{𝐵}**𝑝 _{𝐴}**

**′**+

**𝑝**

_{𝐵}**′**=

**𝑝**+

_{𝐴}**𝑝**

_{𝐵}𝑭𝒊𝒏𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 = 𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎

**Note:**

## Isolated system

It is a system with no external force acts on it.

## Resultant force

Resultant force is that single force which produces the same effect on the body as the net effect of all the forces together.

## Equilibrium

A set of forces are said to be in equilibrium if their resultant is zero.

## Equilibrant

The equilibrant is that force which when acts together with those forces keep the body in equilibrium.

## Equilibrium of a particle

The particle is said to be in equilibrium if the net external force acting on the particle is zero.

## Equilibrium under two forces

Let two forces, **𝐹**** _{1 }**and

**𝐹**

**act on a particle. The particle will be in equilibrium, if**

_{2}**𝐹**

**+**

_{1}**𝐹**

**= 0**

_{2}**𝐹**** _{1}** = −

**𝐹**

_{2}That is two forces on the particle must be equal and opposite.

## Equilibrium under three forces

Let three forces, 𝐹_{1}, 𝐹 and 𝐹 act on a particle. The particle will be in equilibrium, if 𝐹_{1} + 𝐹_{2} + 𝐹_{3} = 0

## Equilibrium under several forces

A particle is in equilibrium under the action of several forces, if the resultants of the resolved components of these forces in each of the X and Y-directions are independently zero. ∑𝐹_{𝑥} = 0 and ∑𝐹_{𝑦} = 0.

**Note:** For a particle to be in equilibrium, minimum number of forces acting on a particle must be two.

## Common forces in mechanics

Common forces in mechanics are

- Gravitational force.
- Spring force.
- Tension in the string.

## Gravitational force

It is the force of attraction between the two bodies due to their masses. Every object on the earth experiences the force of gravity due to earth and it can act at a distance without need of material medium.

## Weight

The force exerted by the earth on the object is called the weight of the object. It is given by, **𝑊** = 𝑚**𝑔**. Weight is vector quantity and its unit is 𝑛𝑒𝑤𝑡𝑜𝑛.

## Spring force (F)

When a spring is compressed or extended by an external force a restoring force is generated. This restoring force is called spring force. For small displacements, the spring force is proportional to the compression or elongation. The spring force is given by, **𝐹** = −𝑘**𝑥** where 𝑘 → spring constant and **𝑥** → Displacement. The −𝑣𝑒 sign denotes that the force is opposite to the displacement.

## Tension in a string (T)

The restoring force in a string is called tension. For an inextensible string the force constant is very high.

## Contact forces

When two bodies are in contact then they exert force on each other. These forces are called as contact forces.

Ex: Except the gravitational force all the above forces are contact forces. Buoyant force, Viscous force and air resistance are examples for contact forces. When bodies are in contact, there are mutual contact forces. They are in accordance with Newton’s third law. All the contact forces are electrical in nature. Although macroscopically bodies are unchanged, at microscopic level all the matter consists of charged particles namely, electrons and protons. The contact forces between objects in contact arising due to elasticity of bodies, molecular collisions and impacts etc.

## Normal reaction (N)

The component of contact force normal to the surface in contact is called Normal reaction.

## Friction(f)

The component of contact force parallel to the surface in contact is called Friction. Friction opposes impending or relative motion between the two surfaces.

There are two types of friction.

They are,

**Static friction****Kinetic friction**

## Static friction(𝒇_{𝒔})

Static friction is the force which balances the applied force when a body is in the state of rest. When there is no applied force, there is no static friction. It comes to play at that moment when there is an applied force. As the applied force increases, static friction also increases and remains equal and opposite to applied force up to a certain limit. Hence it is called a self-adjusting force. Limiting friction: The maximum static friction that a body can exert on the other body in contact with it is called limiting friction.

The limiting friction is directly proportional to the normal reaction between the two surfaces.

That is, **(****𝑓 _{𝑠}**

**)**

**∝**

_{𝑚𝑎𝑥}**𝑁**.

(𝒇_{𝒔})_{𝒎𝒂𝒙} = 𝝁_{𝒔}𝑵 where 𝜇_{𝑠} → co-efficient of static friction and it has no unit.

## Kinetic friction

Frictional force that opposes the relative motion between the surfaces in contact is called kinetic friction.

𝒇_{𝒌} = 𝝁_{𝒌}𝑵 where 𝜇_{𝑘} → co-efficient of kinetic friction and it has no unit. When the body begins, the force acting on the body is given by, **𝐹** − **𝑓 _{𝑘}** = 𝑚

**𝑎**.

If velocity is constant then = 0, **𝐹** − **𝑓 _{𝑘}** = 0

**𝐹** = **𝑓 _{𝑘}**

If the applied force is removed, **𝐹** = 0 then, −**𝑓 _{𝑘}** = 𝑚

**𝑎**

**𝑎** = −**𝑓 _{𝑘}**/𝑚, the body eventually comes to rest.

## Laws of friction

(1) The direction of static friction is opposite to the impending motion and the magnitude is given by, **𝑓 _{𝑠}** ≤ 𝜇

_{𝑠}

**𝑁**

(2) The direction of kinetic friction is opposite to relative motion of the body and the magnitude is given by, **𝑓 _{𝑘}** = 𝜇

_{𝑘}

**𝑁.**

(3) The values of 𝜇_{𝑠} and 𝜇_{𝑘} depend on the nature of the surfaces and 𝜇_{𝑘} is generally less than 𝜇_{𝑠}.

(4) The coefficients of friction are independent of area of contact, provided normal force is constant.

(5) Kinetic friction is nearly independent of velocity.

## Rolling friction

The force which opposes the rolling motion of a body is called rolling friction. In principle, a body like ring rolling without slipping over a horizontal plane will suffer no friction. But in practice some resistance to motion does occur. Rolling friction has a complex origin and somewhat different from that of static and kinetic friction. Rolling friction is much smaller than these.

## Advantages of friction

(1) Friction helps in walking on ground.

(2) Brakes of vehicle work on account of friction.

(3) Writing with chalk on the black board is possible because of friction.

(4) Nails and screws can be fixed an account of friction.

(5) A matchstick is lighted due to friction.

(6) Moving belt remains on the rim of wheel because of friction.

## Disadvantages of friction

(1) Friction causes wear and tear of machine parts.

(2) Efficiency of the machine is reduced on account of friction.

(3) Heat is generated because of friction that may damage the machinery.

(4) Friction restricts the speed of the vehicles.

## Methods of reducing friction

(1) By polishing the surfaces.

(2) Using lubricants like oil, grease etc., in machines.

(3) Using the materials of low co-efficient of friction.

(4) Using ball bearing in wheels, axels and shafts of automobiles.

(5) By providing the streamlined shape to moving vehicles (car, bus, aeroplanes etc.).

## Circular motion

Motion of a body in circular path is known as circular motion. In circular motion, the moving body possess an acceleration which is directed towards the centre of the circular path is given by, 𝑎 = 𝑣^{2}/𝑅 where 𝑅 is the radius of the circular path

## Centripetal force

The force which is directed towards the centre of the circular path is called centripetal force.

## Expression for centripetal force

Force acting on the body is given by, **𝑓 _{𝑐}** = 𝑚

**𝑎**where 𝑚 → mass of the body executing circular motion and

**𝑎**→ centripetal acceleration. But

**𝑎**=

**𝑣**

^{2}/𝑅

Then, **𝑓 _{𝑐}** = 𝑚

**𝑣**

^{2}/𝑅

𝒇_{𝒄} = 𝒎𝒗^{𝟐}/𝑹

## Examples of centripetal force

(1) When a stone is rotated in a circle by a string, the centripetal force is provided by the tension in the string.

(2) The centripetal force for the motion of planet around the sun is provided by gravitational force on the planet due to the sun.

(3) When a car takes turn on a horizontal road, the centripetal force is the force of friction.

## Motion of car on level road

Forces acting on the car are,

- The weight of the car, 𝑚
**𝑔**. - Normal reaction,
**𝑁**. - Friction,
**𝑓**.

## Expression for Maximum speed on the level road

There is no acceleration in the vertical direction; **𝑁** − 𝑚**𝑔** = 0, **𝑁** = 𝑚**𝑔**.

Since the car is moving in circular path, the centripetal force is provided by friction.

**𝑓 _{𝑠}** =

**𝑓**

_{𝑐}For maximum speed, **(****𝑓 _{𝑠}**

**)**

**= 𝑚**

_{𝑚𝑎𝑥}**𝑣**

_{𝑚𝑎𝑥}^{2}/𝑅

𝜇_{𝑠}**𝑁** = 𝑚**𝑣 _{𝑚𝑎𝑥}**

^{2}/𝑅

𝑣_{𝑚𝑎𝑥}^{2} = 𝜇_{𝑠}𝑅𝑁/𝑚

𝑣_{𝑚𝑎𝑥}^{2} = 𝜇_{𝑠}𝑅(𝑚𝑔)/𝑚 = 𝜇_{𝑠}𝑅𝑔

𝒗_{𝒎𝒂𝒙} = √(𝝁_{𝒔}𝑹𝒈)

## Motion of a car on a banked road

**Banking of road:** Rising of outer edge of the road as compared to the inner edge to provide centripetal for vehicles is called banking of roads. We can reduce the contribution of friction in circular motion of the car if the road is banked.

## Expression for Maximum speed on the banked road

As there is no acceleration in the vertical direction,

𝑁 cos 𝜃 = 𝑓 sin 𝜃 + 𝑚𝑔

𝑁 cos 𝜃 − 𝑓 sin 𝜃 = 𝑚𝑔 → (1)

The centripetal force is provided by horizontal components of 𝑁 and 𝑓

𝑁 sin 𝜃 + 𝑓 cos 𝜃 = 𝑚𝑣^{2}/𝑅 → (2)

Taking equation (2) ÷ (1),

(𝑁 sin 𝜃 + 𝑓 cos 𝜃)/(𝑁 cos 𝜃 − 𝑓 sin 𝜃) = (𝑚𝑣^{2}/𝑅) × (1/𝑚𝑔)

(𝑁 sin 𝜃 + 𝑓 cos 𝜃)/(𝑁 cos 𝜃 − 𝑓 sin 𝜃) = 𝑣^{2}/𝑅𝑔

Taking 𝑁 as common in LHS,

(𝑁 (sin 𝜃 + (𝑓/𝑁) cos 𝜃))/(𝑁 (cos 𝜃 – (𝑓/𝑁) sin 𝜃)) = 𝑣^{2}/𝑅𝑔

(sin 𝜃 + (𝑓/𝑁) cos 𝜃)/(cos 𝜃 – (𝑓/𝑁) sin 𝜃) = 𝑣^{2}/𝑅𝑔

To obtain maximum speed we put 𝑓/𝑁 = 𝜇_{𝑠}

(sin 𝜃 + 𝜇_{𝑠} cos 𝜃)/(cos 𝜃 − 𝜇_{𝑠} sin 𝜃) = 𝑣_{𝑚𝑎𝑥}^{2}/𝑅𝑔

𝑣_{𝑚𝑎𝑥}^{2}/𝑅𝑔 = (sin 𝜃 + 𝜇_{𝑠} cos 𝜃)/(cos 𝜃 − 𝜇_{𝑠} sin 𝜃)

Divide RHS of numerator and denominator by cos 𝜃

𝑣_{𝑚𝑎𝑥}^{2}/𝑅𝑔 = (tan 𝜃 + 𝜇_{𝑠})/(1 − 𝜇_{𝑠} tan 𝜃)

𝑣_{𝑚𝑎𝑥}^{2} = 𝑅𝑔((tan 𝜃 + 𝜇_{𝑠})(1 − 𝜇_{𝑠} tan 𝜃))

𝒗_{𝒎𝒂𝒙} = √(𝑹𝒈((𝐭𝐚𝐧𝜽 + 𝝁_{𝒔})(𝟏 − 𝝁_{𝒔}𝐭𝐚𝐧 𝜽))

This is the maximum speed, that a car can take turn without slipping in a circular path.

**Note:** The optimum speed to negotiate a curve can be obtained by putting

𝜇_{𝑠} = 0, 𝑣_{0} = √(𝑅𝑔 tan 𝜃) and banking angle, 𝜃 = tan^{−1}(𝑣_{0}^{2}/𝑅𝑔)

## Steps in solving problems in mechanics

(1) Draw the free body diagram showing a given body as a point.

(2) Consider a body of interest in a given problem and mark as a point. If the given problem has two bodies, then mark then as different points.

(3) Mark the various forces acting on each body.

(4) Write the equation of motion and solve the given unknown parameters.

## Differences between Mass and Weight

Mass | Weight |

It is the amount of matter contained in a body | It is the gravitational force attraction on a body |

It is a scalar | It is vector |

Mass of the body remains same at all places | Weight of the body varies from place to place |

SI unit is kilogram | SI unit is newton |

Problems

1) A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15ms^{-1}. How long does the body take to stop?

Given 𝐹 = −50𝑁, 𝑚 = 20 𝑘𝑔, 𝑣_{0} = 15𝑚𝑠^{−1}, 𝑣 = 0, 𝑡 = ?

𝐹 = 𝑚𝑎

−50 = 20 × 𝑎

𝑎 = −50/20 = −5/2𝑚𝑠^{−2}

𝑣 = 𝑣_{0} + 𝑎𝑡

0 = 15 + ((−5/2) × 𝑡)

(5/2)𝑡 = 15

𝑡 = (15 × 2)/5 = 6𝑠

2) A constant force acting on a body of mass 3kg changes its speed from 2ms^{-1} to 3.5ms^{-1} in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?

Given 𝑚 = 3𝑘𝑔, 𝑣_{0} = 2𝑚𝑠^{−1}, 𝑣 = 3.5𝑚𝑠^{−1}, 𝑡 = 25𝑠

𝑣 = 𝑣_{0} + 𝑎𝑡

3.5 = 2 + (𝑎 × 25)

25𝑎 = 3.5 – 2

𝑎 = (1.5/25)𝑚𝑠^{−2}

𝐹 = 𝑚𝑎

𝐹 = 3 × (1.5/25)

𝐹 = 4.5/25 = 0.18𝑁

The direction of the force is along the motion of the body.

3) A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.

Given 𝑚 = 5𝑘𝑔, 𝐹_{1} = 8𝑁, 𝐹_{2} = 6𝑁, 𝜃 = 90°

Magnitude of Resultant force, 𝐹 = √(𝐹_{1}^{2} + 𝐹_{2}^{2} + 2𝐹_{1}𝐹_{2} cos 𝜃)

𝐹 = √(8^{2} + 6^{2} + (2 × 8 × 6 × cos 90°)

𝐹 = √(8^{2} + 6^{2}) = 10𝑁

Direction of force, 𝛼 = tan^{−1}(6/8)

𝛼 = tan^{−1}(0.75)

𝛼 = 36°53′

Resultant force F makes an angle 36°53′ with 𝐹_{1}

Acceleration, 𝑎 = 𝐹/𝑚

𝑎 = 10/5 = 2𝑚𝑠^{−2}

Direction of acceleration is along resultant force F.

4) The Driver of a three-wheeler moving with a speed of 36kmph sees a child standing in the middle of the road and brings his vehicle to rest in 4s just in a time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400kg and the mass of the driver is 65kg.

Given 𝑣_{0} = 36 𝑘𝑚𝑝ℎ = 10𝑚𝑠^{−1}, 𝑡 = 4 𝑠, 𝑣 = 0

𝑚 = 𝑚_{1} + 𝑚_{2} = 400 + 65 = 465 𝑘g

𝑣 = 𝑣_{0} + 𝑎𝑡

0 = 10 + (𝑎 × 4)

4𝑎 = −10 𝑎 = −10/4 = −5/2𝑚𝑠^{−2}

𝐹 = 𝑚𝑎

𝐹 = 465 × (−5/2)

𝐹 = −1162.5 𝑁

Negative sign shows force is applied against the motion.

5) A shell of mass 0.02kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms^{-1}, what the recoil speed of the gun?

Given Mass of the shell, 𝑚_{1} = 0.020 𝑘𝑔

Initial speed of the shell, 𝑣_{1}_{𝑖} = 0𝑚𝑠^{−1}

Final Speed of the shell, 𝑣_{1}_{𝑓} = 80𝑚𝑠^{−1}

Mass of the gun, 𝑚_{2} = 100𝑘𝑔

Initial speed of the gun, 𝑣_{2}_{𝑖} = 0𝑚𝑠^{−1}

Final (Recoil) speed of the gun, 𝑣_{2}_{𝑓} = ?

From the law of conservation of momentum, we have 𝑝_{𝑖} = 𝑝_{𝑓}

Initially the gun and shell are at rest, therefore Initial momentum

𝑝_{𝑖} = 0

0 = 𝑚_{1}𝑣_{1} + 𝑚_{2}𝑣_{2}

0 = (0.020 × 80) + (100 × 𝑣_{2})

100𝑣_{2} = −(0.020 × 80)

𝑣_{2} = −(0.020 × 80)/100

𝑣_{2} = −0.016𝑚𝑠^{−1}

Negative sign shows that gun recoils in a direction opposite to direction of motion of shell.

6) A bullet of mass 0.04kg moving with a speed of 90ms^{-1} enters a heavy wooden block and is stopped after a distance of 60cm. What is the average resistive force exerted by the block on the bullet?

Given 𝑚 = 0.04𝑘𝑔, 𝑣_{0} = 90𝑚𝑠^{−1}, 𝑥 = 60𝑐𝑚 = 0.60𝑚, 𝑣 = 0

𝑣^{2} = 𝑣_{0}^{2} + 2𝑎𝑥

0 = (90 × 90) + (2 × 𝑎 × 0.60)

𝑎 = −(90 × 90)/(2 × 0.60)

𝑎 = −6750𝑚𝑠^{−2}

𝐹 = 𝑚𝑎

𝐹 = (0.04 × (−6750))

𝐹 = −270𝑁

Negative sign shows the force is resistive.

## Important Questions for exam.

One mark.

1. State Aristotle’s law of motion.

2. Define momentum or define linier momentum of a body.

3. Define SI unit of force.

4. On what principle does a rocket work?

5. Which law of motion is used to explain rocket propulsion?

6. What is friction?

7. Mention a way of reducing kinetic friction.

8. What is Normal reaction?

Two marks.

1. Name any two forces acting on a car when it is moving on the level road.

2. State Newton’s third law of motion. Give an illustration.

3. A net force of 5 𝑁 is acting on a body of mass 10 𝐾𝑔. What is the acceleration produced?

4. Calculate Impulse of force when a force of 10 𝑁 acting for 0.1 𝑠.

5. Define impulsive force. Give one example.

6. What are static friction and limiting friction?

7. Write any two differences between mass and weight.

Three marks.

1. State Newton’s first law of motion. Hence define Force and inertia.

2. Distinguish between impulse and impulsive force. Give an example for impulsive force.

3. Mention the common forces in mechanics.

4. Prove that the total momentum of an isolated system of interacting particle is conserved or Prove law of conservation of linear momentum from Newton’s law of motion. or Total momentum of an isolated system of interacting particle is conserved. Prove it in the case of collision of two bodies. or Prove law of conservation of linear momentum in the case of collision of two bodies.

5. Write any three methods to reduce friction.

6. Mention any three advantages of friction.

7. Give any three disadvantages of friction.

8. Show that Impulsive force is equal to change in momentum.

9. Show that 𝑣_{𝑚} = √(𝜇_{𝑠}𝑅𝑔) for the motion of a car on level road.

10. State laws of Friction.

11. Distinguish between mass and weight.

Five marks.

1. State Newton’s second law of motion and derive, 𝐹 = 𝑚𝑎.

2. State and prove law of conservation of linear momentum. or State and prove law of conservation of linear momentum from Newton’s third law of motion.

3. Derive an expression for maximum safe speed of a vehicle on a banked road in circular motion.

## Additional Problems

1. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?

2. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

3. A body of mass 6 𝑘𝑔 is moving with a velocity of 2 𝑚/𝑠. Calculate the force required to stop in 5 𝑠. How far it will travel before coming to rest.

4. A body of mass 0.2 𝑘𝑔 which is at rest is acted upon by a force of 8 𝑁 for 4 𝑠. Find (i) Change in momentum of the body and (ii) The velocity of the body.

5. Two masses 8 𝑘𝑔 and 12 𝑘𝑔 are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released (𝑔 = 10 𝑚/𝑠).