Elastic and Inelastic Collisions in One and Two Dimensions

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 m₁ and m₂ collide with velocities u₁ and u₂, and their velocities after the collision are v₁ and v₂, respectively.

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

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂             (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, 

m₁(u₁ + v₁ ) (u₁ − v₁ ) = m₂ (v₂ + u₂ ) (v₂ − u₂ )    (6) 

Applying the conservation of momentum equation, 

m₁u₁ + m₂u₂ = m₁v₁+ m₂v₂   

Regroup the equation with the same masses, 

m₁u₁ − m₁v₁  = m₂v₂  − m₂u₂                                          (7) 

Therefore, 

m₁ (u₁ − v₁ ) = m₂(v₂ − u₂ )                      (8)

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

m₁(u₁ + v₁ ) (u₁ − v₁ ) =\[\frac{m_{2}(v_{2} + u_{2}) (v_{2}-u_{2})}{m_{1}(u_{1}-v_{1})}\] = \[m_{2}(v_{2}- u_{2})\]       (9)

u₁ + v₁ = v₂ + u₂          (10)

Now, 

v₁= v₂ + u₂  − u₁       (11)

Use the values of v₁  from equation (11) in law of conservation of momentum equation, we get, 

v₂ = [2 m₁u₁ + u₂(m₂ − m₁ ) ] / (m₁+ m₂ )          (12)

Now, use the value of v₂   in equation (11) v₁= v₂ + u₂  − u₁   

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

v₁= u₂  and v₂= u₁  

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, v₁= 0 and v₂= u₁  

And, If m₁ < m₂ 

Then, 

v₁= − u₁  and  v₂= 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,  m₁ > m₂ 

 Then,

v₁= u₁  and v₂= 2u₁    

Elastic Collision in Two Dimension

Assume that m₁ and  m₂ are two mass particles in a laboratory frame of reference and that m₁ collides with m₂ that is initially at rest. Let the velocity of mass m₁ before the collision be u₁, and after the collision, it moves with a velocity v₁ and is deflected by an angle θ₁ with its incident direction, while m₂ moves with a velocity v₂ and is deflected by an angle θ₂ with its incident direction.

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

m₁u₁=m₁v₁cosθ₁ +m₂v₂cosθ₂ ---(1)

For components along the y-axis,

0=m₁v₁sinθ₁ − m₂v₂sinθ₂ --(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 v₁, v₂, θ₁, θ₂ 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, K₂ < K₁; 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 m₁ and m₂ be the two mass particles, and m₁ collides with m₂, which is initially at rest. Let the velocity of mass m₁ before the collision be u₁, and after the collision, it moves with a velocity v₁ and is deflected by an angle θ₁ with its incident direction, while m₂ moves with a velocity v₂ and is deflected by an angle θ₂ with its incident direction.

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

m₁u₁=m₁v₁cosθ₁ +m₂v₂cosθ₂ ---(1)

And, for components along the y-axis,

0=m₁v₁sinθ₁ − m₂v₂sinθ₂ --(2)

Analyzing the above equations reveals that we must find the values of four unknown quantities v₁, v₂, θ₁, θ₂ 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.