You might have seen two billiard balls colliding with each other in the course of the game. This forceful coming together of two separate bodies is called collision. What happens after collisions? Can we determine the velocity or the trajectory of the colliding bodies? Let us find out!

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## What is a Collision?

Collision means two objects coming into contact with each other for a very short period. In other words, collision is a reciprocative interaction between two masses for a very short interval wherein the momentum and energy of the colliding masses changes. While playing carroms, you might have noticed the effect of a striker on coins when they both collide.

Collision involves two massesÂ Â *m _{1 }*and

*m*. The

_{2}*v*is the speed ofÂ particle

_{1i}Â*m*, where the subscript

_{1}*â€˜iâ€™*implies initial. The particle with mass

*m*is at rest. In this case, the object with mass

_{2}*m1*Â collides with the stationary object of mass

*m*

_{2}.As a result of this collision the massesÂ m1 and m2 move in different directions.

**Browse more Topics under Work Energy And Power**

- Concepts of Potential Energy
- Conservation of Mechanical Energy
- Potential Energy of a Spring
- Power
- The Scalar Product
- Work and Kinetic Energy
- Work-Energy Theorem
- Various Forms of Energy: The Law of Conservation of Energy

## Types Of Collision

Generally, the law of conservation of momentum holds true in the collision of two masses but there may be some collisions in which Kinetic Energy is not conserved. Depending on the energy conservation, conservation may be of two types:

**Elastic Collision:**In the elastic collision total momentum, the total energy and the total kinetic energy are conserved. However, the total mechanical energy is not converted into any other energy form as the forces involved in the short interaction are conserved in nature. Consider from the above graph two masses,*m*Â andÂ_{1}*m*moving with speed u_{2 }_{1}and u_{2}*.*The speed after the collision of these masses is*v*The law of conservation of momentum will give:_{1}and v_{2Â }.

Â Â Â Â *Â m _{1}u_{1Â }+ m_{2}u_{2Â }=Â m_{1}v_{1Â }+ m_{2}v_{2Â }*

Â Â Â Â Â Â The conservation of Kinetic Energy says:

*1/2Â m _{1}u^{2}_{1Â }+ 1/2 m_{2}u^{2}_{2Â }= 1/2 m_{1}v^{2}_{1Â }+1/2Â m_{2}v^{2}_{2Â }*

**Inelastic Collision:**In the inelastic collision, the objects stick to each other or move in the same direction. The total kinetic energy in this form of collision is notÂ conserved but the total momentum and energy are conserved. During this kind of collision, the energy is transformed into other energy forms like heat and light. Since during the phenomenon the two masses follow the law of conservation of momentum and move in the same direction with same the same speed*v*we have:

*m _{1}u_{1Â }+ m_{2}u_{2Â }= (m_{1}+ m_{2})v*

*v= (m _{1}u_{1Â }+ m_{2}u_{2})/(m_{1}+ m_{2})*

- The kinetic energy of the masses before the collision is : K.E
_{1Â }=Â*1/2Â m*_{1}u^{2}_{1Â }+ 1/2 m_{2}u^{2}_{2} - While kinetic energy after the collision is: K.E
_{2Â }= 1/2 (*m*) v_{1}+ m_{2}^{2} - But according to the law of conservation of energy:Â
*1/2Â m*=Â 1/2 (_{1}u^{2}_{1Â }+ 1/2 m_{2}u^{2}_{2}*m*) v_{1}+ m_{2}^{2Â }+ Q - ‘Q’ here is the change in energy that results in the production of heat or sound.

## The Coefficient of Restitution

The coefficient of restitution is the ratio between the relative velocity of colliding masses before interaction to the relative velocity of the masses after the collision. Represented by ‘e’, the coefficient of restitution depends on the material of the colliding masses. For elastic collisions, e = 1 while for inelastic collisions,e = 0. The value of e > 0 or e < 1 in all other kinds of forceful interactions.

## One Dimensional Collision

One dimensional sudden interaction of masses is that collision in which both the initial and final velocities of the masses lie in one line. All the variables of motion are contained in a single dimension.

### Elastic One Dimensional Collision

As already discussed in the elastic collisions the internal kinetic energy is conserved so is the momentum. Elastic collisions can be achieved only with particles like microscopic particles like electrons, protons or neutrons.

*m*_{1}u_{1}Â +Â *m*_{2}*u*_{2}Â =Â *m*_{1}*v*_{1}Â +Â *m*_{2}*v*_{2}

Since the kinetic energy is conserved in the elastic collision we have:

*1/2Â m _{1}u^{2}_{1Â }+ 1/2 m_{2}u^{2}_{2 =Â Â }1/2Â m_{1}v^{2}_{1Â }+ 1/2 m_{2}v^{2}_{2}*

This gives us :Â *m _{1}u^{2}_{1Â }+Â m_{2}u^{2}_{2 =Â }Â m_{1}v^{2}_{1Â }+Â m_{2}v^{2}_{2}Â Â Â Â Â Â Â *(Factoring out 1/2)

Rearranging we get: *m _{1}u^{2}_{1}–Â m_{1}v^{2}_{1}_{Â =Â }Â _{Â }Â m_{2}v^{2}_{2}Â –Â m_{2}u^{2}_{2}*

Therefore, *m _{1}(u^{2}_{1}–Â v^{2}_{1})=_{Â }Â _{Â }Â m_{2}(v^{2}_{2}Â –Â u^{2}_{2})*

Which if elaborated become *m _{1}(u_{1}+v_{1}Â ) (u_{1}–Â v_{1}Â )=_{Â }Â _{Â }Â m_{2Â }(v_{2}Â + u_{2})(v_{2}Â –Â u_{2})*

Using the conservation of momentum equation: *m _{1}u_{1Â }+ m_{2}u_{2Â }=Â m_{1}v_{1Â }+ m_{2}v_{2Â }*

We regroup it with same masses: *m _{1}u_{1}-m_{1}v_{1Â }_{Â }=Â m_{2}v_{2Â }-m_{2}u_{2}*

Hence, *m _{1}(u_{1}-v_{1})_{Â }=Â m_{2}(v_{2Â }-u_{2})*

Now dividing the two equations:

*m _{1}(u_{1}+v_{1}) (u_{1}–Â v_{1}) =_{Â }Â m_{2Â }(v_{2}Â + u_{2})(v_{2}Â –Â u_{2}) / m_{1}(u_{1}-v_{1Â })_{Â }=Â m_{2}(v_{2Â }-u_{2} )*

We get: *u _{1}+v_{1} =Â v_{2}Â + u_{2}*

Now, *v _{1Â }=Â v_{2}Â + u_{2}–Â u_{1}*

When we use this value ofÂ Â *v _{1Â }*

_{Â }in equation of conservation momentum we get :

*Â*

*v _{2 }= [2 m_{1Â }u_{1Â }+ u_{2Â }(m_{2}-m_{1})] / (m_{1Â }+ m_{2Â }*)

Now using the value of *v _{2} *in equationÂ

*v*

_{1Â }=Â v_{2}Â + u_{2}–Â u_{1Â } *v _{1Â }= [2 m_{1Â }u_{1Â }+ u_{2Â }(m_{2}-m_{1})] / (m_{1Â }+ m_{2}*

*)Â +Â u*

_{2}–Â u_{1}*v*_{1Â }= [*2 m _{1Â }u_{1Â }+ u_{2Â }(m_{2}-m_{1}) +Â u_{2}Â (m_{1Â }+ m_{2Â })_{Â }–Â u_{1}(m_{1Â }+ m_{2Â })]_{Â Â }/Â (m_{1Â }+ m_{2Â })_{Â }*

We finally get:

*v*_{1Â }*= [**2m _{2Â }u_{2Â }+ u_{1Â }( m_{1Â }– m_{2})] / (m_{1Â }+m_{2} )*

When masses of both the bodies are equal then generally after collision, these masses exchange their velocities.

*v _{1Â }= u_{2Â Â }and v_{2Â }= u_{1}*

This means that incourse of collision between objects of same masses, if the second mass is at rest and the firstmass collides with it then after collision the first mass comes to rest and the second mass moves with the speed equal to first mass. Therefore in such case, v_{1Â }= 0 and v_{2}Â _{=Â }u_{1}. In case if m_{1Â }< m_{2Â }then, v_{1Â }= -u_{1}Â and v_{2Â }= 0

This means that the lighter body will bombard back with its own velocity, while the heavier mass will remain static. However, if m_{1Â }> m_{2Â }then v_{1Â }= u_{1}Â and v_{2}Â = 2u_{1}

### Inelastic One Dimensional Collision

In inelastic one dimensional collision, the colliding masses stick together and move in the same direction at same speeds. The momentum is conserved and Kinetic energy is changed to different forms of energies. For inelastic collisions the equation for conservation of momentum is :

*m _{1}u_{1Â }+ m_{2}u_{2}_{Â }= (m_{1}_{Â }+ m_{2}) v*

Since both the objects stick, we take final velocity after the collision as *v.Â *Now *v *shall be:

*= m _{1}u_{1Â }+ m_{2}u_{2}/Â m_{1}_{Â }+ m_{2}*

The kinetic energy lost during the phenomenon shall be:

E = 1/2Â *m _{1}u^{2}_{2Â }– 1/2 (m_{1}_{Â }+ m_{2}) v^{2Â }*

### Collision in Two Dimensions

The above figure signifies collision in two dimensions, where the masses move in different directions after colliding. Here the moving mass m_{1}Â collides with stationary mass m_{2}. The linear momentum is conserved in the two-dimensional interaction of masses. In this case, we see the masses moving in x,y planes. The x and y component equations are:

m_{1}u_{1Â }= m_{1}u_{2}cosÎ¸_{1Â }+ m_{2}v_{2}cosÎ¸_{2}

0 = m_{1}u_{2}sinÎ¸_{1Â }-m_{2}v_{2}sinÎ¸_{2}

For spherical objects that have smooth surfaces, the collision takes place only when the objects touch with each other. This is what happens in the games of marbles, carom, and billiards.

## Solved Examples For You

Q: A sphere of mass m, moving with a speed v, strikes a wall elastically at an angle of incidence Î¸. If the speed of the sphere before and after the collision is the same and the angle of incidence and velocity normally towards the wall the angle of rebound is equal to the angle of incidence and velocity normally towards the wall is taken as negative then, the change in the momentum parallel to the wall is:

A) mv cosÎ¸Â Â Â B) 2mv cosÎ¸Â Â Â Â C) -2mv cosÎ¸Â Â Â Â D) zero

Solution: D) The mass of the object remains constant. To see the change in the momentum, we need to see the forces applied due to the collision. The only force applied to the sphere of mass ‘m’ is the normal reaction force due to the wall. This force has no parallel component along the wall. Therefore, the change in momentum in a direction parallel to the wall is zero.

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