CENTRE OF MASS

COLLISIONS

&

CONSERVATION LAWS

BSc – I (UNIT IV)

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Contents

1) Centre of Mass

Velocity of centre of mass 3

Acceleration of centre of mass 3

Linear Momentum of centre of mass 4

Reduced mass 4

2) Collision Perfectly Elastic Collision

Laboratory frame 5

Centre of mass frame 6 Perfectly Inelastic Collision

Laboratory frame 8

Centre of mass frame 8

3) Conservation Laws

Linear momentum 9

Angular momentum 10

Angular impulse 11

Conservation of energy 11

4) Rocket 12

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Suppose particles constitute the mass of a rigid body. Let the masses of the particles be

whose position vectors from the origin are . Then the

position vector of the centre of mass, is defined as,

If the centre of mass coincides with the origin of the system then

The centre of mass is defined as a point in space such that the vector sum of the moments of the mass

points around it is zero.

Velocity of Centre of Mass

We can express the above expression as,

Differentiating with respect to time , we get

is called the velocity of the centre of mass. The momentum of the centre of mass is given as

. If no external force is applied the momentum is conserved and hence the velocity

remains constant.

Acceleration of Centre of Mass

We have the above expression as,

Differentiating with respect to time , we get

is called the acceleration of the centre of mass. Hence, we can write the vector sum of all external

forces on the individual particles to be,

If no external force is applied, the acceleration of the centre of mass is zero.

R

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Linear Momentum of the Particles

Let us consider a mass having the position vector . Let the centre of mass have the position vector

. The position vector between the centre of mass and the mass be . Hence, we have

We also have,

Hence, the total linear momentum about the centre of mass is zero.

Reduced Mass

Let us consider two masses and separated by a distance . Let their position vectors be

respectively and . Let the centre of mass have the position vector . It is given as,

If no external force is applied then the velocity of the centre of mass, , is constant.

The force on due to and on due to are both directed towards the centre of mass. Hence the

forces are central forces.

Let the force on be and that on be . Hence, we get

On subtracting second equation from the first, we get

But, and let us define the reduced mass as

. Hence we get

If is the gravitational force, then

. Hence, we get

Let, . Hence

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In the centre of mass frame, total linear momentum is conserved. Hence, the velocity of centre of mass

remains constant. In the centre of mass frame, . Therefore, the two bodies move around the

centre of mass in such a way that remains constant. Hence to an observer in the centre of mass frame,

heavier mass seems to describe a smaller circle and the lighter a larger circle. Hence,

When two bodies are approaching each other, a force comes into play between them for a finite time and

brings about a measurable change in their velocities, momenta and energy according to the respective

laws of conservation, a collision is said to have taken place.

Scattering: If the nature of particles does not change after collision, it is termed as scattering.

Elastic Scattering: If in a scattering, sum of kinetic energies is conserved then it is termed as elastic

scattering.

Inelastic Scattering: If in a scattering, sum of kinetic energies is not conserved then it is termed as

inelastic scattering.

Reaction: If the final products are different from the initial reactants and the sum of kinetic energies is

not conserved, then the collision is said to be a reaction.

Exoergic Reaction: If loss of energy occurs in a reaction then it is called as exoergic reaction.

Endoergic Reaction: If energy adds up in a reaction then it is called as endoergic reaction.

Laboratory Frame: If the origin of the reference system is a point rigidly fixed to the laboratory, it is

known as the laboratory frame.

Centre of Mass Frame: If the origin of the reference system is a point rigidly fixed to the centre of mass

of a system of particles on which no external force is acting, it is known as the centre of mass

frame of reference.

Perfectly Elastic Collision

Laboratory Frame (One – Dimension)

Let us consider two particles having mass and . Let their initial velocities before the collision be

and and their final velocities after collision be and .

Hence, by principal of conservation of linear momentum, we get

By principal of conservation of energy, we get

On dividing by , we get,

Before Collision After Collision

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Hence, we see that the relative velocity is equal. Hence, the velocity after collision is given as,

Hence,

Cases:

i) Let one particle be initially at rest. ( . Hence, the final velocities are given as,

ii) Let the particles have same mass, then

Here, the final velocities just exchange their values with the initial velocities.

iii) Let the particle at rest be massive. ( . Hence,

Here, the massive particle remains at rest and the lighter particle rebounds with the same velocity

with which it collided.

iv) Let the particle at rest be light. ( . Hence,

Here, the massive particle moves with it initial velocity and the lighter particle moves with twice

the initial velocity of the massive particle.

Centre of Mass Frame (One – Dimension)

The centre of mass for a two particle system is given as,

According to Galilean transformation, velocity before collision relative to the centre of mass frame is

given as,

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After collision, the velocities are given as

Hence, we see that in the centre of mass frame, the initial and final velocities are same in magnitude with

just change in direction.

Let one of the mass be initially at rest. Say . Hence, the initial velocities in centre of mass frame

are given as,

In the centre of mass frame, the centre of mass is at rest. Hence, the linear momentum is conserved and is

zero. Thus,

Since, the sum of kinetic energies is conserved, we get

Also, we get

Hence, the velocities remain same in magnitude but there is a change in the direction with respect to the

centre of mass.

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Perfectly Inelastic Collision

Laboratory Frame (One – Dimension)

In a perfectly inelastic collision, the particles stick together after the collision. Hence by principle of

conservation of linear momentum, we get

Let one of the particles (say ) be at rest. .

Let and

be the initial and final kinetic energies respectively. They are given as,

We find that,

Thus,

as .

Hence, the loss in kinetic energy is given as,

Centre of Mass Frame (One – Dimension)

The centre of mass for a two particle system is given as,

As , we get

The initial velocities in the centre of mass frame are given as,

In the centre of mass frame, the centre of mass is at rest. Hence, the linear momentum is conserved and is

zero. Thus,

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The final velocity in the centre of mass frame is given as,

Let and

be the initial and final kinetic energies respectively. They are given as,

Hence, the loss in kinetic energy is given as,

Hence, we see that the loss of energy in both laboratory and centre of mass frame is same.

Linear Momentum

The linear momentum is defined as the product of its mass and linear velocity. i.e.,

The total linear momentum of a system of particles free from the action of external forces and subjected

only to their mutual interaction remains constant, no matter how complicated the forces are.

In an isolated system, . Hence the linear momentum remains conserved.

Let us consider a system of two particles which are acted upon by the gravitational force. Let be the

force of particle on particle and Let be the force of particle on particle . Hence,

Now,

But,

Thus, we see that linear momentum is conserved.

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Angular Momentum

The angular momentum of a particle about a fixed point is defined as the moment of its linear momentum

about that point.

Thus, angular momentum is perpendicular to the plane containing the position vector and the velocity

. If is the angular velocity then taking magnitudes, we find .

Let us consider a rigid body rotating with angular velocity about an axis. Let us consider two particles

having mass and at a distance of and from the axis. The sum of the moments of the linear

momentum of all the particles of a rotating rigid body about the axis of rotation is called the angular

momentum. Hence, we get

The linear momentum is given by

The angular momentum is given by

The total angular momentum for a system of particles is given by

where, is defined as the moment of inertia.

If a force acts on a particle at a point whose position with respect to the origin of the inertial frame is

given by the displacement vector , then torque is defined as

We know that angular momentum is given by,

Differentiating with respect to time , we get

But,

Hence, torque is defined as the rate of change of angular momentum.

For a rigid body, . Hence,

where, is termed as the angular acceleration.

If no torque acts, then

Hence when no torque acts on a rigid body, angular velocity is conserved.

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Angular Impulse

If an unbalanced external torque is applied to a body rotating with an angular velocity for a time

so that angular velocity changes to , then

The torque is given by

This quantity is called as the angular impulse.

Conservation of Energy

Let the force be defined as a function of potential energy and position vector . Hence,

Where, potential energy is a function of force and position vector . Hence,

and

The kinetic energy

also does not explicitly depend on time . Hence, we get

The total energy is also defined as,

Taking differential, we get

But,

and

But as is not a function of ,

.

Hence,

Therefore, we get

But , hence,

Thus, energy is conserved.

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Rocket propulsion is based on the principle of conservation of momentum. When the rocket is fixed, the

exhaust gases rush downward at a high speed and pushes the rocket forward.

Let us suppose, at any instant

Rocket’s mass along with its fuel = and velocity of rocket =

After a time , the mass ejected be with an exhaust velocity with respect to moving rocket.

Velocity is constant and an intrinsic negative quantity. Therefore, velocity of ejected gases with

respect to earth is given as,

As mass decreases to , velocity increases to .

Let,

Initial momentum,

Final momentum,

Hence,

But as is a very small quantity, we can neglect the product . Therefore, we get

Taking limits, we get

is an intrinsic negative quantity. Taking magnitudes, we get

Now, rate of change of momentum is defined as the external force applied on the rocket. Hence,

Hence, the net force is given by,

If the rocket is outside the gravitational pull of the earth, . Hence, we get

is the reaction force and is the thrust which pushes the rocket forward.

Since, &

are both negative quantities,

is positive and is directed upward.

Near the earth, the upward thrust on the rocket is opposed by the force due to gravity. Hence, .

Hence for producing a large thrust, exhaust velocity should be large as should the rate of change of

mass

.

To get the equation of motion of a rocket far away from earth, we solve the equation

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If is the initial mass and is the initial velocity, then

Taking magnitudes, we get

This is the equation of motion of a rocket at any instant.

If is the mass of the burnt out rocket and be the final velocity, then

Let us take, , then

Velocity Achieved by a Rocket at Height

Let the initial velocity of the rocket be , be the radius of the earth, be the mass of the rocket and

be the mass of earth.

At the earth’s surface,

Kinetic energy of the rocket,

Potential energy of the rocket,

When the rocket reaches height,

Kinetic energy of the rocket,

Potential energy of the rocket,

According to law of conservation of energy,

But, acceleration due gravity is given as,

. Hence, we get

This is the expression for velocity of rocket at a height .

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