## What is Collision?

A collision is an event or process in which the velocity changes when approaching two bodies or particles, or when interacting, is called **Collision** of those two bodies or particles, or the interaction between two objects or particles is called a **Collision**.

It is not necessary to touch particles during collisions and in all types of collisions, the law of linear momentum conservation necessarily applies. While the law of conservation of mechanical energy is not necessarily applicable.

Examples - a cricket bat ball impact, rubber ball wall impact, billiard balls cross, etc.

### Types of Collision

In terms of mathematical convenience, the impact is divided into three parts.

#### 1. Elastic Collision

If two bodies come in their real shape and form after the impact ie no part of mechanical energy is stored in the bodies as the distorted potential energy, then the impact between the bodies is called **Elastic Collision**. Hence kinetic energy is also preserved with linear momentum. In this type of impact, the frictional force is absent, therefore, there is no loss of mechanical energy.

For example, the collisions between atoms, molecules, non-atomic particles, and parent particles are almost collapsed.

#### 2. Inelastic Collision

If the total kinetic energy of the body is not fixed at the time of impact, then this type of impact is called **Inelastic Collision**. Most of the occurrences occurring in a human's daily life are unpredictable.

Example - A shotgun shot hit the target.

#### 3. Perfectly Inelastic Collision

If two objects stick together after a collision, then such a collision is called a **Perfectly Inelastic Collision**. This type of impact results in a maximum loss of kinetic energy.

Example: A bullet fired from a gun collides with a piece of a stick and gets trapped inside it.

## Coefficient of Restitution or Resilience

The ratio of the relative velocity when moving away after the collision of any two bodies and the relative velocity when approaching objects before the collision is called the Coefficient of Restitution. Let us demonstrate with this e. If the velocity before the collision of any two bodies are u1 and u2 respectively and the velocity after the collision are v1 and v2 respectively, then,

Coefficient of Restitution (e) = Relative velocity after collision / Relative velocity before collision

e = Regression velocity / Access velocity

For a fully elastic collision, e=1

For a completely non-impact collision, e=1

For most of the collisions that occur in normal life, 0

## One-dimensional Elastic Collision

When two objects or objects collide with each other moving along a straight line, this type of impact is called One-dimensional Elastic Collision. Let there be two bodies M and N, whose mass is m1 and m2 respectively, and velocity u1 and u2 respectively. According to the picture given between the body M and N, the impact is

Relative velocity when approaching bodies before collision = (u_{1} - u_{2})

Total linear momentum of bodies before collision = (m_{1}u_{1} + m_{2}u_{2})

The relative velocity of objects moving away after the collision = (v_{2} – v_{1})

Total linear momentum of bodies after impact = (m_{1}v_{1} + m_{1}v_{2})

Linear momentum is conserved in a perfectly elastic collision. Therefore

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

m_{1}(u_{1}-v_{1}) = m_{2}(v_{2}-u_{2}) ...1

The total linear kinetic energy of the former bodies of collision = ½ m_{1}u_{1}^{2} + ½ m_{2}u_{2}^{2}

The total linear kinetic energy of the bodies after the collision = ½ m_{1}v_{1}^{2} + ½ m_{2}v_{2}^{2}

Linear kinetic energy is conserved in a perfectly elastic collision. Therefore

½m_{1}u_{1}^{2} + ½m_{2}u_{2}^{2} = ½m_{1}v_{1}^{2} + ½m_{2}v_{2}^{2}

m_{1}(u_{1}^{2} – v_{1}^{2}) = m_{2}(v_{2}^{2} – u_{2}^{2})

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

By dividing equation 2 from equation 1,

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

u_{1} – u_{2} = v_{2} – v_{1} ...3

Therefore, in a fully elastic collision, the relative velocity of the objects coming closer before the impact is equal to the relative velocity when moving away from the objects after impact.

From equation 3,

**Velocity of bodies after collision**

Substituting the value of v_{2} in Equation 3 in Equation 1

m_{1}(u_{1} – v_{1}) = m_{2} [(u_{1} – u_{2 }+ v_{1}) – u_{2}]

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

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

Putting the value of v_{1} in equation 3 in equation 4,

### Special Conditions

1. If the mass of the objects (M and N) is the same then

m_{1} = m_{2} = m ...6

Putting the value of equation 6 in equation 4,

v_{1} = u_{2} ...7

From equation 5,

v_{2} = u_{1} ...8

Therefore, we can say that if there are a fully elastic collision between two equal mass bodies, then their velocity changes after the collision.

2. If the object N is at rest i.e. u_{2} = 0, then

From equation 4,

From equation 5,

3. If m_{1} = m_{2}, then from equation 9 and 10, v_{1} = 0 and v_{2} = u_{1}

Hence, we can say that when one of the two objects of the same mass collides with the other stationary object, after the impact, the velocity of the first object becomes zero and the velocity of the second object becomes equal to the velocity of the first object.

4. If the first object is heavier than the second, ie m_{1}>>m_{2}, then from equation 9 and 10

v_{1} = u_{1}

v_{1} = 2u_{1}

Therefore, when a very heavy object collides with a stationary object, there is no change in the velocity of the heavy object after the impact, while the velocity of the light object doubles.

5. If the first object is lighter than the second, then (m_{1}<

v_{1} = -u_{1}

Therefore, when a light object collapses with a heavy object, after the impact, the light object returns to its former velocity, and the heavy object remains stationary.

## Two-dimensional Elastic Collision or Elastic Oblique Collision

Let two bodies of mass m_{1} and m_{2} be moving in the same direction with velocities u_{1} and u_{2} as shown in the figure. The perpendicular distance between their initial velocities is called the Impact Parameter b. 0**1 + r _{2}) for the oblique impact, where r_{1} and r_{2} are the radii of the impacting bodies. After the impact, both bodies move from the initial direction of their motion according to the picture at θ and φ angles respectively.**

By the law of momentum conservation,

(a) Along the X-axis, m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1}cosθ + m_{2}v_{2}cosφ ...1

(b) Along the Y-axis, 0 = m_{1}v_{1}sinθ – m_{2}v_{2}sinφ ...2

If the two bodies are of equal mass and the other body is at rest, then the scattering angle (θ + φ) of the directions for the two-dimensional elastic collision component is 90^{0}.

Let u_{2} = 0 and m_{1} = m_{2} Substituting in Eq. 1, 2, and 3,

u_{1} = v_{1}cosθ + v_{2}cosφ ...4

0 = v_{1}singθ – v_{2}sinφ ...5

u_{1}^{2} = v_{1}^{2} + v_{2}^{2} ...6

By adding equation 4 and equation 5 squared,

u_{1}^{2} = (v_{1}cosθ + v_{2}cosφ)^{2} + (v_{1}sinθ – v_{2}sinφ)^{2}

u_{1}^{2} = v_{1}^{2}cos^{2}θ + v_{2}^{2}cos^{2}φ + 2v_{1}v_{2}cosθcosφ +v_{1}^{2}sin^{2}θ + v_{2}^{2}sin^{2}φ – 2v_{1}v_{2}sinθsinφ

u_{1}^{2} = v_{1}^{2}(sin^{2}θ + cos^{2}θ) + v_{2}^{2}(sin^{2}φ + cos^{2}φ) + 2v_{1}v_{2}(cosθcosφ – sinθsinφ)

u_{1}^{2} = v_{1}^{2} + v_{2}^{2} + 2v_{1}v_{2}cos(θ+φ) ...7

From equation 6 and equation 7,

cos(θ+φ) = 0

θ+φ = π/2

## Perfectly Inelastic Collision

In this type of collision, the bodies remain dependent on each other before the collision and after collision move together as a body.

Let m_{1} and m_{2} be two masses of mass moving with u_{1} and u_{2} (u_{1} > u_{2}) velocity on a simple linear path, respectively. In the absence of external force, both bodies do a completely unpredictable impact and after mutual impact, they cling to each other. After the impact, the joint body moves in the same direction with v velocity on a simple linear path.

Total linear momentum of bodies before impact = m_{1}u_{1 }+m_{2}u_{2}

Linear momentum of the joint body after impact = (m_{1}+m_{2})v

### Case 1: when the bodies are moving in the same direction

By the law of momentum protection

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

Again the loss in kinetic energy is -∆K, so the loss (-∆K) = K_{f} - K_{i}

### Case 2: when the bodies are moving in the opposite direction

By the law of momentum conservation,

m_{1}u_{1} + m_{2}(-u_{2}) = (m_{1}+m_{2})v

Loss of kinetic energy = -∆K = K_{f} - K_{i}

If the second body is in resting position before the impact, then u_{2} = 0

Hence, according to Equation 1, the velocity of the composite body after the impact

And from equation 1, loss of kinetic energy after impact

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