**Elastic Collision**

Elastic collision definition: A collision is an event when two or more physical bodies come in direct contact with each other, exerting mutual forces in a short time. There are two main types of collisions— elastic collision and inelastic collision. There is no net loss of kinetic energy in elastic collisions, while in inelastic collisions, there is a net loss of kinetic energy. The kinetic energy lost converts into sound energy, thermal energy, or material deformation.

**Elastic Collision Definition**

An elastic collision definition: It is a type of collision characterized by no net loss of kinetic energy; rather, there is a conservation of both the kinetic energy and momentum; therefore, in this type of collision, the kinetic energy remains the same as before and after the collision. An elastic collision is either one or two-dimensional.

A perfectly elastic collision is not possible practically as some energy conversion is bound to occur in the process, even if small. Although the linear momentum of the whole system does not change, the individual momenta of the individual components change. As those moments are equal and opposite to each other, they cancel each other out, and the initial energy is conserved, and the total energy remains the same.

**Elastic Collision Examples**

Following are some elastic collision examples:

A ball at a billiard table hitting another ball is an example of elastic collision. When a ball bounces back to your hand after you throw it on the ground, the system’s net kinetic energy does not change. Therefore, this is an example of an elastic collision.

The Elastic Collision formula is:

m1u1 + m2u2 = m1v1 + m2v2

Here m1 is the first body’s mass

M2 is the second body’s mass

U1 is the starting velocity of the first body

U2 is the starting velocity of the second body

V1 is the final velocity of the 1st body

V2 is the final velocity of the 2nd body

The kinetic energy’s elastic collision formula is given by:

(1/2) m1u1^{2 }+ (1/2) m2u2^{2 }= (1/2) m1v1^{2 }+ (1/2) m2v2^{2}

**Problem on Elastic Collision**

On the collision of two billiard balls, one of the balls moves with a velocity of 6 m/s while the other ball is at rest. Ball 1 comes to a halt after the collision. What is the velocity of ball two post-collision? Also, is this collision inelastic or elastic when each ball’s mass is 0.20 kg?

**Solution:**

A momentum table can be used to determine the velocity of ball 2

Objects | Momentum before | Momentum after |

Ball 1 | 0.20 × 6 m/s = 1.2 | 0 |

Ball 2 | 0 | 0.20 kg × v2 |

Total | 1.2 kg × m/s | 0.20 kg × v2 |

1.2 kg × m/s = 0.20 kg × v2

v2 =1.2 / 0.20 = 6 m/s

To find out whether it is an elastic or inelastic collision, calculate the net kinetic energy of the system before the collision and after it.

Objects | KE Before (J) | KE After (J) |

Ball 1 | 0.50 × 0.20 × 62 = 3.6 | 0 |

Ball 2 | 0 | 0.50 × 0.20 × 62 = 3.6 |

Total | 3.6 | 3.6 |

We can see that the kinetic energy before the collision is the same as that after the collision. As the kinetic energy of the system is conserved during the process, it is an elastic collision.

## Difference between Elastic and Inelastic Collision

Elastic Collision | Inelastic Collision |

The system’s overall kinetic energy before and after the collision is conserved. | Here the overall kinetic energy of the system before and after the collision is different. |

The total momentum of the system is conserved. | The total momentum of the system is conserved. |

Energy is not converted into other forms such as heat or sound energy. | Here there is a change of kinetic energy into other forms. |

This type of collision is unlikely to occur in the real world as there is always a loss of energy in some form. | This form of collision is more practical and is likely to occur in real form. |

An elastic collision example is a spacecraft flying near a planet without getting affected by its gravity or swinging balls. | An example of an inelastic collision is the collision between two cars. |

**Applications of Elastic Collision**

The time of collision affects the extent of force experienced by an object during a collision. Higher the collision time, the lesser the force acting upon the object. Therefore, to make the force acting upon an object maximum, the time of collision must be reduced. Similarly, to maximize the force, the time of the collision must be increased.

Several real-world applications of these phenomena exist. Automobile airbags increase the time of collapse and decrease the effects of force on bodies at the time of a collision. This is accomplished by airbags by extending the time needed to halt the momentum of the driver and the passenger.

**Why would we ever approximate a collision as perfectly elastic?**

We know that we will hardly ever encounter a perfectly elastic collision. The concept is of no practical use, but how is it useful? This concept is helpful as the requirement that kinetic energy must be conserved causes additional constraints to our motion equations. This way, we can solve problems that have a lot of unknown values. The solution is often adequate as the collision is very close to being perfectly elastic.

For example, a head-on collision occurs between two trolleys on a track— trolley A and trolley B. We want to know the final velocities for the two trolleys when only the initial velocities vAi and vBi are given. If we apply the law of conservation of momentum, we find that we have one equation with two unknown values, vAf and vBf.

mAvAi + mBvBi = mAvAf + mBvBf

Because there is a conservation of kinetic energy also, we have another constraint:

½ mAv2Ai + ½ mBv2Bi = ½ mAv2Af + ½ mBv2Bf

As there are two equations with two unknown values, we can now solve the system completely using two equations simultaneously to find out the velocities. It is somewhat tedious to solve these equations. So, we simply state the result for now:

**Proof**

These solutions have an interesting thing about them; there are limiting cases applying for various configurations of head-on collisions. These can aid in gaining an intuitive understanding of the occurrences in situations like elastic vs inelastic collisions in Newton’s demonstration. There is a collision between object A with target B of equal mass, which is at rest:

vAf = 0 , vBf = vAi

The object that was impacting comes to a halt while the target object attains a speed exactly equal to that of the impacting object. This is the type of interaction seen in Newton’s demonstration. When we swing a ball on one side of the cradle, it always comes out from the other side. In principle, there could also be a conservation of momentum if the two balls came out, each with a speed half of the original value. However, most collisions are elastic. So, the only way to make sure that both the kinetic energy and momentum are conserved is by making only one ball come out.

**Conclusion**

Hope this discussion helped clear all your doubts about the elastic collision. Try solving more and more problems related to the topic of elastic vs inelastic collision to gain a fundamental understanding of it.

**Frequently Asked Questions**

**1. Does an elastic collision formula conserve momentum?**

Momentum is conserved in an elastic collision.

**2. Give an example of an elastic collision.**

When a ball is thrown on the ground, it bounces back in the hand. In this process, no net change in kinetic energy occurs, and therefore, it is an elastic collision.

**3. What is the difference between elastic and inelastic collision?**

In an elastic collision, there is a conservation of kinetic energy. In contrast, in an inelastic collision, there is different kinetic energy between the colliding bodies before and after the collision.

**4. What are the applications for elastic collision?**

The airbags in automobiles increase the collapse time and minimize the effect of force on objects during a collision.

**5. State the law of conservation of linear momentum.**

According to the law of conservation of linear momentum, if the net external force acting on a system of bodies is zero, then the system’s momentum remains constant.

**6. Write some applications of conservation of linear momentum.**

The launching of rockets is one of the laws of conservation of momentum. The exhaust gases are pushed downwards by the rocket fuel burns, and this pushes the rocket upwards. The same principle is used by motorboats. Water is pushed backwards by them, and they get propelled forwards in a reaction to conserve momentum.

**7. Define elastic potential energy.**

It is the energy stored due to the application of a force to deform an elastic object. Until the force is removed, the energy is stored, and the object springs back to its real shape performing work in the process.

**8. What is contact force?**

Any force that requires contact to occur can be classified as a contact force. If there is no contact between two surfaces, they can not apply a normal force on one another.

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