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 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.
FAQs on Elastic and Inelastic Collisions in One and Two Dimensions
1. What is the fundamental difference between an elastic and an inelastic collision?
The fundamental difference lies in the conservation of kinetic energy. In an elastic collision, both momentum and kinetic energy of the system are conserved. In an inelastic collision, momentum is still conserved, but the total kinetic energy of the system is not conserved as some of it is converted into other forms like heat, sound, or deformation of the objects.
2. What are some real-world examples of elastic and inelastic collisions?
Real-world examples help illustrate the difference between these collision types:
- Elastic Collision Examples: The collision between two billiard balls or the collision between atoms in an ideal gas are close approximations of elastic collisions. When a basketball is bounced properly, it returns to nearly the same height, showing a highly elastic collision with the ground.
- Inelastic Collision Examples: A car crash is a classic example where kinetic energy is lost to sound and metal deformation. A mudball thrown against a wall that sticks to it is a perfectly inelastic collision.
3. What happens to objects in a perfectly inelastic collision?
In a perfectly inelastic collision, the colliding objects stick together after the impact and move with a single, common velocity. This type of collision results in the maximum possible loss of kinetic energy from the system, while still conserving the total momentum.
4. Why is analyzing a collision in two dimensions more complex than in one dimension?
Analyzing a collision in two dimensions is more complex because momentum must be conserved independently along two perpendicular axes (usually the x and y axes). Instead of a single momentum equation, you need to apply the law of conservation of momentum separately for the x-components and y-components of velocity, leading to a system of equations to solve for the final state.
5. If momentum is conserved in all types of collisions, why do we need to distinguish between elastic and inelastic types?
While total momentum is always conserved in an isolated system, distinguishing between elastic and inelastic collisions is crucial for understanding the energy dynamics. The distinction tells us what happens to the kinetic energy. Knowing a collision is elastic allows us to use the conservation of kinetic energy as a second powerful tool to solve for final velocities. For inelastic collisions, it tells us that we must account for energy loss to accurately describe the outcome.
6. What does the coefficient of restitution tell us about a collision?
The coefficient of restitution (e) is a measure of the 'elasticity' of a collision. It is the ratio of the final relative velocity to the initial relative velocity between two objects after they collide. Its value tells us the type of collision:
- e = 1: Perfectly elastic collision (no kinetic energy is lost).
- e = 0: Perfectly inelastic collision (maximum kinetic energy is lost, objects stick together).
- 0 < e < 1: A real-world inelastic collision where some kinetic energy is lost.
7. What happens in a one-dimensional elastic collision if a light object hits a very heavy object at rest?
In a one-dimensional elastic collision where a very light object (m1) hits a very heavy object (m2) at rest, the light object will essentially bounce back with nearly the same speed it had initially, while the heavy object will remain almost stationary. This is because the massive object's inertia is too large for the light object's momentum to cause any significant change in its velocity.
