# Laws of Motion

## Conservation of Momentum

**Law of conservation of momentum:** The total momentum of an isolated system of interacting particles is conserved.

Let us assume two bodies A and B have initial momenta p_{A} and p_{B}. The bodies collide, get apart and their final momenta become P'_{A} and P'_{B}. According to second law

`F_(AB)Δt=p’_A-p_A`

`F_(BA)Δt=p’_B-p_B`

Since as per the third law

`F_(AB)=-F_(BA)`

So, `p’_A-p_A=-(p’_B-p_B)`

Or, `p’_A+p’_B=p_A+p_B`

This shows that for an isolated system the total final momentum is equal to its initial momentum.

### Equilibrium of a Particle

If two forces F_{1} and F_{2} act on a particle, equilibrium requires

`F_1=-F_2`

If three concurrent forces F_{1}, F_{2} and F_{3} are acting on a particle then equilibrium requires the vector sum of the three forces is zero

`F_1+F_2+F_3=0`

In other words, the resultant of any two forces F_{1} and F_{2} (obtained by parallelogram law of forces) must be equal and opposite to the third force F_{3}

The above equation also implies following:

`F_(1x)+F_(2x)+F_(3x)=0`

`F_(1y)+F_(2y)+F_(3y)=0`

`F_(1z)+F_(2z)+F_(3z)=0`

A particle is in equilibrium under the action of forces F_{2}, F_{2} ………..F_{n} if the forces can bre represented by the sides of a closed n-sided polygong with arrows directed in the same sense.

## Common Forces in Mechanics

Gravitational force can act at a distance without the need of any intevening medium. So, it is a non-contact force. All other forces common in mechanics are contact forces. A contact force on an object arises due to contact with some other object. The component of contact force normal to the surfaces in contact is called normal reaction. The componet of contract force parallel to the surfaces in contact is called friction.

**Force due to Spring:** When a spring is compressed or extended by an external force, a restoring force is generated. This is called spring force.

Spring force `F = - kx`

Negative sign shows that the force is opposite the displacement from unstretched state.

**Tension in String:** The force constant k is very high for an inextinsible string. The restoring force in a string is called tension. It is customary to use a constant tension T throughout the string. This assumption is true for a string of neglibile mass.

### Friction:

**Static Friction:** The frictional force which opposes the impending motion of a body is called static friction. The limiting vlue of static friction (f_{s})_{max} is independent of the area of contact and varies with the normal force (N):

`(f_x)_(ma\x)=μ_s\N`

Where μ_{s} is a constant of proportionality and it depends only on the nature of surfaces in contact. This is callse the coefficient of static friction. The law of static friction may be written as

`f_x=≤μ_s\N`

**Kinetic Friction:** The frictional force which comes into play when two surfaces are in relative motion to each other is called kinetic friction. Kinetic friction is independent of the area of contact and is nearly independent of velocity.

`f_k=μ_k\N`

Where μ_{k} is coefficient of kinetic friction, which depends only on the surfaces in contact. Experiment show that μ_{k} is less than μ_{s}.

**Rolling Friction:** When an object is rolling over a horizontal plane, the friction that opposes the rolling is called rolling friction. Rolling friction is much smaller than static or sliding friction.

In principle, when a body; like a ring or a sphere; is rolling without slipping over a horizontal plane, it will suffer no friction. At every instance there is just one point of contact betweenthe body and theplane and this point has no motion relative to the plane. This is an ideal situation in which the body should continue to roll with constant velocity because of absence of friction.

But during rolling the surfaces in contact get momentarily deformed a little, and this results in a finite area (not a point) of the body coming in contact with surface. The net effect is that the component of the contact force parallel to the surface opposes motion. This somewhat explains the origin of rolling friction.

### Circular Motion

Let us assume that a body is moving in a circle of radius R with uniform speed v. Then acceleration of the body

`a=(v^2)/R`

According the the second law, the force f_{c} providing this acceleration is

`f_c=(mv^2)/R`

This force is directed towards the centre and is caleld centripetal force.

#### Motion of a car on a level road

When a car is moving on a level road, three forces act on the car

- Weight of car = mg
- Normal reaction N
- Frictional force f

As there is no acceleration in vertical direction

`N-mg=0`

Or, `N=mg`

The centripetal force required for circular motion is along the surface of the road, and is provided by the component of contact force between road and the car tyres along the surface. By definition, this is the frictional force. It is important to note that it is the static friction which provides centripetal acceleration.

So, `f=(mv^2)/R≤μ_s\N`

Or, `v^2≤(μ_s\RN)/m`

Or, `v^2=μ_s\Rg`

(Since N = mg)

This shows that for a given value of μ_{s} and R there is a maximum speed of circular motion of the car possible which is given as follows:

`v_(ma\x)=sqrt(μ_s\Rg)`

#### Motion of a car on a banked road

If the road is banked, we can reduce the contribution of friction to the circular motion of the car. Since there is no acceleration in vertical direction, the net force along this direction must be zero. So,

N cos θ = mg + f sin θ

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

`N sin θ + f cos θ = (mv^2)/R`

But `f≤μ_s\N`

Then prvious equations can be written as

N cosθ = mg + μ_{s}N sin θ

N sin θ + μ_{s}N cos θ `=(mv^2)/R`

Or, `N = (mg)/(co\s\θ-μ_s\si\n\θ)`

Substituting the value of N in previous equation

`(mg(si\n\θ-μ_s\co\s\θ))/(co\s\θ-μ_s\si\n\θ)=(mv^2_(ma\x))/R`

Or, `v_(ma\x)=(Rg(μs-ta\n\θ)/(1-μs\ta\n\θ))^(1/2)`

This shows that maximum possible speed of a car on a banked road is greater than that on a flat road.

If μ_{s} = 0 in previous equation

Then v_{0} = (Rg tan θ)^{1/2}

At this speed, frictional force is not needed at all to provide the necessary centripetal force. So, driving at this speed on a banked road will cause little wear and tear of the tyres. For v < v_{0}, frictional force will be up the slop, and a car can only be parked if tan θ ≤ μ_{s}