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A collision between the molecules of a gas is such that there is no loss of kinetic energy. Also, the kinetic energy and the momentum remains conserved. Such type of collision is called the elastic collision.

However, when there is a loss of kinetic energy or it gets converted to other forms of energy, it is an inelastic collision.

So, the two types of collisions have their practical applications. Here, we are going to discuss the elastic and inelastic collisions.

Elastic Collisions

Molecules of an ideal gas possess an elastic collision. A gas comprises a huge number of molecules, which are perfectly elastic spheres. When these spherical molecules of gas come in a state of Brownian or random motion, they move in different directions with different speeds, ranging from zero to infinity, and obey Newton’s laws of motion.

Also, the size of these molecules is lesser than the distance between them, that’s why the volume occupied by each is considered negligible when compared with the gas volume.

The collision of these molecules with themselves and with the walls of the container is elastic such that their momentum and kinetic energy remains conserved.

Elastic Collision in One Dimension

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Before collision

m1 = the mass of the red ball

u1 = velocity of red ball

m2 = the mass of the green ball

u2 = velocity of the green ball

After collision

v1 = velocity of red ball

v2 = velocity of the green ball

For a collision to be elastic, the momentum is conserved, so the equation is:

(m1u1 + m2 u2) = (m1v1 + m2 v2)....(1)

⇒ m1 * (u1 - v1) = m2 * ( v2 - u2).....(2)

We know that in an elastic collision, the KE remains conserved before and after the collision, so our equation becomes:

\[\frac{1}{2}\] m1u1² + \[\frac{1}{2}\] m2u2² = \[\frac{1}{2}\] m1v1² + \[\frac{1}{2}\]m2v2²

⇒ m1u1² + m2u2² = m1v1² + m2v2²

⇒ m1 (u1² - v1²) = m2(v2² - u2²)

⇒ m1 (u1 + v1)(u1 - v1 ) = m2(v2 + u2)(v2 - u2) ....(3)

Now, dividing equation (2) by (1), we get:

(u1 + v1) = (v2 + u2) v1 = (v2 + u2 - u1)...(4)

On substituting the value of (4) in (1):

= m1 (u1 - (v2 + u2 - u1)) = m2 ( v2 - u2)

So, we finally get the equation as:

⇒ 2 m1 u1 + u2 ( m2 - m1) / (m1 + m2).....(5)

When you hit the basketball on the ground, it bounces back to your hand. Here, the kinetic energy remains conserved, which means the collision is elastic.

The collision between the billiards ball, between the atoms, is elastic.

We observed the collision at a microscopic level for the ideal gas. However, at the macroscopic level, we can observe inelastic collisions because the energy doesn’t remain conserved. Let’s take an example of inelastic collisions.

Let’s suppose that Rita is driving SUV at a high-speed, suddenly a car came on the way, and the car got crashed. Now what happens next is, some of the kinetic energy of each vehicle gets transformed to sound, heat, or deformation of the parts of the car.

So, when the car hits the SUV, the two vehicles collided with each other, and therefore, their kinetic energy didn’t remain conserved, it got changed to another form. Such type of collision in an inelastic collision.

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From the conservation of momentum:

m1v1 = (m1 + m2)v2

Where,

m1 = the mass of car SUV

v1 = velocity of SUV

m2 = the mass of the car

v2 = velocity of the car

⇒ v2 = (\[\frac{m1}{m1 + m2}\]) v1

The kinetic energy before and after hitting or collision will be:

KE2 = \[\frac{1}{2}\](m1 + m2) [(\[\frac{m_{1}}{m_{1} + m_{2}}\])v1]²

KE1 = \[\frac{1}{2}\]m1v1²

The ratio of kinetic energies is:

\[\frac{KE1}{KE2}\] = \[\frac{\frac{1}{2}m_{1}v_{1}^{2}}{\frac{1}{2}(m_{1}+m_{2}) [(\frac{m_{1}}{m_{1}+m_{2})v_{1}]^{2}}}\] = (\[\frac{m_{1}}{m_{1}+m_{2}}\])

The fraction of kinetic energy loss can be calculated by using the following formula:

= \[\frac{KE1 - KE2}{KE1}\] = \[\frac{[1 - (\frac{m_{1}}{m_{1}+m_{2})]KE1}}{KE1}\] = (\[\frac{m_{2}}{m_{1}+m_{2}}\])

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When two objects of mass m1 collide with a stationary object of mass m2, they both move in different directions. So, to represent them on the x-y plane, we have drawn figure.2. Also, the linear momentum remains conserved in the two-dimensional interaction of these two masses. Now, the equations for the components along the x and y-axis will be:

m1u1 CosΘ1= m1u2 CosΘ1 + m2v2 CosΘ2’

0 = m1u2 SinΘ1 - m2v2 SinΘ2’

This phenomenon of collision can be seen while playing carrom, billiards, etc.

A football

A mudball, when thrown against the wall, sticks to it.

Two cars hitting each other.

A dropped ball of clay or falling chalk piece.

FAQ (Frequently Asked Questions)

Question 1: Why Football is an Example of an Inelastic Collision?

Answer: A football game is considered an inelastic collision because of the following reason:

When two football players collide and move as one mass, the collision is perfectly inelastic. So when these two players tackle each other, they continue to fall together because of which there is a loss of energy.

Question 2: A Ball of Mass m, Travelling with a Speed u, Strikes a Surface Elastically at an Angle of Incidence.

If the speed of the ball before and after the collision is the same and the angle of incidence and ‘u’ towards the surface the angle of rebound equals the angle of incidence.

Now, if the velocity towards the surface is taken as negative then, the change in the momentum parallel to the surface will be?

Answer: Practically, we observed that the force acting on the body mass ‘m’ is the normal reaction force of the surface. Since no parallel force acts on the ball that’s why the change in momentum parallel to the surface will be zero.

Question 3: What Happens When Two Objects with Different Mass Collide?

Answer: When two bodies of equal mass collide, both objects experience equal and opposite forces on each other. However, if these bodies are of equal masses, they will have unequal accelerations because of the contact forces during the collision.

Question 4: List the Three Types of Collision.

Answer: The three types of collisions are:

Elastic

Inelastic

Completely inelastic