# Work Energy Power

## Collisions

Let us consider two masses m_{1} and m_{2}. Particle m_{1} is moving with initial speed v_{i}, and particle m_{2} is at rest. Mass m_{1} collides with the stationary mass m_{2}. After collision, both the particles fly off in different directions.

### Elastic and Inelastic Collisions

The total linear momentum is conserved in all collisions. In other words, the initial momentum of the system is equal to the final momentum of the system. When two objects collide, the mutual impulsive forces acting over the collision time Δt cause a change in their respective momenta.

Δp_{1} = F_{12} Δt

Δp_{2} = F_{21} Δt

Here, F_{12} is the force exerted on the first particle by second particle. F_{21} is the force exerted on second particle by first particle. From Newton's third law, F_{12} = - F_{21}. So, the following implies:

Δp_{1} + Δp_{2} = 0

But the total kinetic energy of the system is not necessarily conserved. Part of the initial kinetic energy may get transformed into other forms of energy, like heat and sound. When the final kinetic energy is equal to the initial kinetic energy, the collision is called **elastic collision**. When deformation (due to collision) is not relieved and the two bodies move together after collision, it is called **completely inelastic collision**. When the deformation is partly relieved and some of the initial kinetic energy is lost, it is called **inelastic collision**.

#### Collision in One Dimension

Let us first consider a completely inelastic collision. In this case, (in given figure) θ_{1} = θ_{2} = 0

`m_1v_(1i)=(m_1+m_2)v_f`

Or, `v_f=(m_1)/(m_1+m_2)v_(1i)`

Loss in kinetic energy on collision

`ΔK=1/2m_1v_(1i)^2-1/2(m_1+m_2)v_f^2`

`=1/2m_1v_(1i)^2-1/2(m_1^2)/(m_1+m_2)v_(1i)^2`v

`=1/2m_1v_(1i)^2(1-(m_1)/(m_1+m_2))`

`=1/2(m_1m_2)/(m_1+m_2)v_(1i)^2`

Now, let us consider an **elastic** collision. In this case also θ_{1} =θ_{2} = 0. The equation for momentum and KE conservation are as follows:

`m_1v_(1i)=m_1v_(1f)+m_2v_(2f)` …………..(1)

`m_1v_(1i)^2=m_1v_(1f)^2+m_2v_(2f)^2`

From these two equation, we get

`m_1v_(1i)(v_(2f)-v_(1i))=v_(1i)^2-v_1f)^2`

Or, `v_(2f)(v_(1i)-v_(1f))=v_(1i)^2-v_(1f)^2`

`=(v_(1i)-v_(1f))(v_(1i)+v_(1f))`

Or, `v_(2f)=v_(1i)+v_(1f)`

Substituting this in equation (1), we get

`v_(1f)=(m_1-m_2)/(m_1+m_2)v_(1i)`

And, `v_(2f)=(2m_1v_(1i))/(m_1+m_2)`

**Case I:** If two masses are equal

`v_(1f)=0`

`v_(2f)=v_1i)`

The first mass comes to rest and pushes off the second mass with its initial speed on collision.

**Case II:** If one mass dominates, e.g. m_{2} > > m_{1}

`v_(1f)∼-v_(1i)`

`v_(2f)∼0`

The heavier mass is undisturbed while the lighter mass reverses its velocity.

#### Collision in Two Dimensions

Let us consider the plane determined by the final velocity directions of two particles and let us say it is the x-y plane. The conservation of the z-component of linear momentum implies that the entire collision is in the x-y plane. The x- and y-component equations are

`m_1v_(1i)=m_1v_(1f) co\s θ_1+m_2v_(2f) co\s θ_2`

`0=m_1v_(1f) si\n θ_1+m_2v_(2f) si\n θ_2`

**Example:** Consider the collision depicted between two billiard balls with equal masses m_{1} = m_{2}. The first ball is called the cue while the second balls is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ_{2} = 37°. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain θ_{1}

**Solution:** Since the masses are equal to, from momentum of conservation

`v_(1i)=v_(1f)+v_(2f)`

Or, `v_(1i)^2=(v_(1f)+v_(2f))(v_(1f)+v_(2f))`

`=v_(1f)^2+v+(2f)^2+2v_(1f).v_(2f)`

`=v_(1f)^2+v_(2f)^2+2v_(1f)v_(2f) co\s (θ_1+37°)` …………….(1)

Since the collision is elastic and m_{1} = m_{2} hence

`v_(1i)^2=v_(1f)^2+v_(2f)^2` ………..(2)

Comparing equations (1) and (2), we get

Cos (θ_{1} + 37°) = 0

Or, θ_{1} + 37° = 90°

Or, θ_{1} = 90° - 37° = 53°

This proves that when two equal masses undergo a glancing elastic collision with one of them at rest, they will move at right angles to each other after the collision.

In everyday life, collisions take place only when two bodies touch each other. But sometimes, we have to deal with forces involving action at a distance. Such an event is called scattering. In such a situation, the velocities and directions in which the two particles go away depend on their initial velocities as well as the type of interaction between them, their masses, shapes and sizes. A comet passing from near the sun is an example of scattering. Another example of scattering is of alpha particle going towards a nuclues and then going away in some direction.