## Define Elastic and Inelastic Collision

A collision is a transfer of momentum or kinetic energy from one object to another. Collisions are classified into two types: elastic collisions and inelastic collisions.

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 remain conserved. This type of collision is called an 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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Assume that two particles with masses m_{1} and m_{2} collide with velocities u_{1} and u_{2}, and their velocities after the collision are v_{1} and v_{2}, respectively.

The collision of these two particles is a head-on elastic collision. As per the law of conservation of momentum equation,

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} (1)

We also have the law of kinetic energy conservation for elastic collisions

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

We get,

\[m_{1} u_{1}^{2}+ m_{2} u_{2}^{2} = m_{1} v_{1}^{2} + m_{2} v_{2}^{2} (Factoring \, out \frac{1}{2})\] (3)

On rearranging the equation, we get,

\[m_{1} u_{1}^{2} - m_{1} v_{1}^{2} = m_{2} v_{2}^{2} - m_{2} u_{2}^{2}\] (4)

Now,

\[m_{1} (u_{1}^{2}-v_{1}^{2}) = m_{2} (v_{2}^{2}-u_{2}^{2})\] (5)

If we elaborate the above equation, it will now become,

m_{1}(u_{1} + v_{1} ) (u_{1} − v_{1} ) = m_{2} (v_{2} + u_{2} ) (v_{2} − u_{2} ) (6)

Applying the conservation of momentum equation,

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1}+ m_{2}v_{2}

Regroup the equation with the same masses,

m_{1}u_{1} − m_{1}v_{1} = m_{2}v_{2}_{ } − m_{2}u_{2} (7)

Therefore,

m_{1} (u_{1} − v_{1} ) = m_{2}(v_{2} − u_{2} ) (8)

On dividing equation (6) by (8), we get,

m_{1}(u_{1} + v_{1} ) (u_{1} − v_{1} ) =\[\frac{m_{2}(v_{2} + u_{2}) (v_{2}-u_{2})}{m_{1}(u_{1}-v_{1})}\] = \[m_{2}(v_{2}- u_{2})\] (9)

u_{1} + v_{1} = v_{2} + u_{2} (10)

Now,

v_{1}= v_{2} + u_{2} − u_{1} (11)

Use the values of v_{1} from equation (11) in law of conservation of momentum equation, we get,

v_{2} = [2 m_{1}u_{1} + u_{2}(m_{2} − m_{1} ) ] / (m_{1}+ m_{2} ) (12)

Now, use the value of v_{2} in equation (11) v_{1}= v_{2} + u_{2} − u_{1}

\[ v_{1} = \frac {[2 m_{1}u_{1} + u_{2}(m_{2} − m_{1} )]}{(m_{1}+ m_{2} ) + u_{2} − u_{1}} \] (13)

\[v_{1} = \frac{ [2 m_{1}u_{1} + u_{2}(m_{2}-m_{1}) + u_{2}(m_{1}+m_{2} - u_{1}(m_{1}+m_{2})]}{(m_{1}+m_{2})}\] (14)

As a result, we get,

\[v_{1} = \frac{[2 m_{2}u_{2} + u_{1}(m_{1} - m_{2})]}{(m_{1} + m_{2})}\] (15)

When the masses of both bodies are equal, their velocities are generally exchanged after collision.

v_{1}= u_{2} and v_{2}= u_{1}

This means that if two objects of the same mass collide, if the second mass is at rest and the first mass collides with it, the first mass comes to rest and the second mass moves at the same speed as the first mass.

In this case, v_{1}= 0 and v_{2}= u_{1}

And, If m_{1} < m_{2}

Then,

v_{1}= − u_{1} and v_{2}= 0

This means that the lighter body will use its own velocity to bombard the heavier mass, while the heavier mass will remain static.

But if, m_{1} > m_{2}

Then,

v_{1}= u_{1} and v_{2}= 2u_{1}

### Elastic Collision in Two Dimension

Assume that m_{1} and m_{2} are two mass particles in a laboratory frame of reference and that m_{1} collides with m_{2} that is initially at rest. Let the velocity of mass m_{1} before the collision be u_{1}, and after the collision, it moves with a velocity v_{1} and is deflected by an angle θ_{1} with its incident direction, while m_{2} moves with a velocity v_{2} and is deflected by an angle θ_{2} with its incident direction.

According to the law of conservation of linear momentum, for components along the x-axis

m_{1}u_{1}=m_{1}v_{1}cosθ_{1} +m_{2}v_{2}cosθ_{2} ---(1)

For components along the y-axis,

0=m_{1}v_{1}sinθ_{1} − m_{2}v_{2}sinθ_{2} --(2)

As per the law of conservation of Kinetic energy,

\[\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}=\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}\] ---(3)

Analyzing the above equations reveals that finding values for four unknown quantities v_{1}, v_{2}, θ_{1}, θ_{2} using the above three equations is not possible. As a result, it cannot predict the variable because there are four of them. However, if we measure any one variable, we can uniquely determine the other variable using the above equation.

### Elastic Collision Examples

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.

### Inelastic Collision

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 an 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 hit 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 is an inelastic collision.

### Inelastic Collision in One Dimension

Consider the masses m1 and m2 of two particles. Let u1 and u2 represent the respective velocities before the collision.

Assume that both particles stick together after colliding and move at the same velocity v. As per the law of linear momentum conservation,

m1u1 + m2u2=(m1+ m2) v

or

\[V = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1}+m_{2}}\] (1)

On considering the second particle to be stationary or at rest, u2=0

then

m1u1 = v(m1+m2)

or

\[V =\frac{m_{1} u_{1}}{m_{1}+m_{2}}\] (2)

Hence, |v| < |u1|.

The Kinetic energy before the collision is as follows:

KE1= \[\frac{1}{2}\]m1\[u_{1}^{2}\]

The Kinetic Energy of the system after the collision is as follows:

KE2= \[\frac{1}{2}\](m1+m2) v2

\[KE_{2}=\frac{1}{2}(m_{1}+m_{2})(\frac{m_{1}u_{1}}{m_{1}+m_{2}})^{2} = \frac{1}{2} \frac{m_{1}^{2}u_{1}^{2}}{(m_{1} + m_{2})}\] (3)

Hence,

\[\frac{K.E._{2}}{K.E._{1}}\]=\[ \frac{\frac{1}{2} \frac{m_{1}^{2} u_{1}^{2}}{(m_{1}+m_{2})} }{\frac{1}{2} m_{1} u_{1}^{2}} \] = \[\frac{m_{1}}{m_{1}+m_{2}}\]< 1 (4)

As a result, K_{2} < K_{1}; hence, energy loss would be there after the particles collide.

**The Fraction of Kinetic Energy Loss can be Calculated by Using the Following Formula:**

\[\frac{K.E_{1} - K.E_{2}}{K.E_{1}} = \frac{[1 - (\frac{m_{1}}{m_{1} + m_{2}})]K.E_{1}}{K.E_{1}} = \frac{m_{2}}{m_{1} + m_{2}}\]

### Collision in Two Dimension

In a laboratory frame of reference, let m_{1} and m_{2} be the two mass particles, and m_{1} collides with m_{2}, which is initially at rest. Let the velocity of mass m_{1} before the collision be u_{1}, and after the collision, it moves with a velocity v_{1} and is deflected by an angle θ_{1} with its incident direction, while m_{2} moves with a velocity v_{2} and is deflected by an angle θ_{2} with its incident direction.

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According to the law of conservation of linear momentum, for components along the x-axis,

m_{1}u_{1}=m_{1}v_{1}cosθ_{1} +m_{2}v_{2}cosθ_{2} ---(1)

And, for components along the y-axis,

0=m_{1}v_{1}sinθ_{1} − m_{2}v_{2}sinθ_{2} --(2)

Analyzing the above equations reveals that we must find the values of four unknown quantities v_{1}, v_{2}, θ_{1}, θ_{2} with the help of the above three equations, which is impossible because we need at least four equations to find the values of four unknown quantities. As a result, we cannot predict the variables because there are four of them. However, if we measure any two variables, then the other variable can be uniquely determined using the above equation.

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

### Examples of Inelastic Collisions

A football

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

Two cars hit each other.

A dropped ball of clay or falling chalk piece.

## FAQs on Elastic and Inelastic Collisions in One and Two Dimensions

**1. What is a perfectly inelastic collision? Give an example.**

Perfectly inelastic collisions are those in which the colliding bodies stick together after impact. Some car accidents, a meteorite colliding with the Moon, and a collision between two balls of plasticine would be completely inelastic. Much, if not all, of the system's initial kinetic energy, is lost in these collisions.

**2. A ball of mass m, traveling 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?**

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.

**3. What happens when two objects with different masses collide?**

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

**4. List the three types of collision.**

The three types of collisions are as follows:

Elastic

Inelastic

Completely inelastic