Collision

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

COLLISIONS

When two particles approach each other, their motion changes or their momentum changes due to their mutual interactions. This phenomenon is called collision. During collision (i) an impulse (a large force for a relatively short time) acts on each colliding particle (ii) the total momentum of the particles remain conserved. The collision is infact a redistribution of total momentum of the particles. Physical interaction is not necessary for collision. Generally, the collisions are of two types:

(1) Elastic collision (2) Inelastic collision

1. Elastic collision: A collision is said to be elastic if kinetic energy is also conserved along with the linear momentum. There is no loss or transformation of kinetic energy in this collision

Consider two particles of masses m₁ and m₂ moving with velocities $u_{1}$ and $u_{2}$ respectively. They collide and their velocities after the collision become $v_{1}$ and $v_{2}$ respectively.

By conservation of the linear momentum,

Diagram being restored — will be back shortly

Total linear momentum before collision = Total linear momentum after collision , $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

or $m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2})$ … (1)

By conservation of kinetic energy,

Total kinetic energy before collision = Total kinetic energy after collision

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}$ = $\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

or $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2})$ … (2)

And only for one dimension collision or head on collision

On solving the above equation (1) and (2), $u_{1} - u_{2} = - (v_{1} - v_{2})$ … (3)

i. e. Relative velocity before collision = Relative velocity after collision.

Thus in elastic collision, the relative velocity of approach of particles before collision is equal to the relative velocity with which the particles recedes after collision. i.e. the×magnitude of relative velocity remains the same, but the direction is reversed.

On solving the equation (1) and (3) we get

$v_{1} = (\frac{m _{1} - m _{2}}{m _{1} + m _{2}}) u_{1} + (\frac{2 m _{2}}{m _{1} + m _{2}}) u_{2}$ and $v_{2} = (\frac{2 m _{1}}{m _{1} + m _{2}}) u_{1} + (\frac{m _{2} - m _{1}}{m _{1} + m _{2}}) u_{2}$

Illustration 1: A ball of 0.4kg mass and a speed of 3 m/s has a head-on, completely elastic collision with a 0.6-kg mass initially at rest. Find the speeds of both bodies after the collision:

Solution: By Conservation of momentum:

(0.4 x 3) + 0 = 0.4v + 0.6V or v + 1.5V = 3

We know that:

velocity of separation = –velocity of approach or –v + V = 3

We solve by adding the two equations to yield

2.5V = 6 V = 2.4 m/s v = –0.6 m/s

2. Inelastic collision: A collision in which the linear momentum is conserved, but a part of kinetic energy change into the other forms (such as heat, vibration, excitation energy etc) is called the inelastic collision. In other words, the kinetic energy is not conserved.(Actually there is no×violation of the law of conservation of energy, but a part of the kinetic energy, changes into a useless form.) In this, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. However, in the absence of external forces, law of conservation of linear momentum still holds good.

Let two particles of masses m₁ and m₂ moving with initial velocities $u_{1}$ and $u_{2}$

Collide and travel with velocities $v_{1}$ and $v_{2}$ respectively after the collision. By conservation of linear momentum

\left. \begin{align} {{m}_{1}}\overrightarrow{{{u}_{1}}}+{{m}_{2}}\overrightarrow{{{u}_{2}}}={{m}_{1}}\overrightarrow{{{v}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}} \\ or{{m}_{1}}\left( \overrightarrow{{{u}_{1}}}-\overrightarrow{{{v}_{1}}} \right)={{m}_{2}}\left( \overrightarrow{{{v}_{2}}}-\overrightarrow{{{u}_{2}}} \right) \\\end{align} \right\}

and by conservation of kinetic energy, $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}$ = $\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ + E

Where E is the part of energy which changes into the useless form due to inelastic collision;

For example, if two particles coalesce after collision and the combined system travel with a velocity $v$ after the collision in same direction, then

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v or v = \frac{m _{1} u _{1} + m _{2} u _{2}}{( m _{1} + m _{2} )}$

And loss in kinetic energy, $E = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2}$

or $E = \frac{1}{2} (m_{1} u_{1}^{2} + m_{2} u_{2}^{2}) - \frac{1}{2} (\frac{m _{1} u _{1} + m _{2} u _{2}}{( m _{1} + m _{2} )})^{2}$ $= \frac{1}{2} \frac{m _{1} m _{2}}{( m _{1} + m _{2} )} (u_{1} - u_{2})^{2}$

CONCEPT OF COEFFICIENT OF RESTITUTION

When two bodies collide head–on, the ratio of their relative velocities after collision and their relative velocities before collision is called the coefficient of restitution e.

Thus $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}$ … (i)

Or $e = \frac{se p er a t i o n s p ee d}{a pp r o a c h s p ee d}$ (along line of impact) … (ii)

Where v₁ = The speed of first body after collision.

v₂ = The speed of second body after collision.

u₁ = the speed of first body before collision.

u₂ = the speed of second body before collision.

The ratio e is called the coefficient of restitution and is constant for two particular objects.

In general,

$0 \leq e \leq 1$

e = 0, for completely inelastic collision, as both the objects stick together. So, their separation speed is zero or e = 0.

e = 1, for an elastic collision, as we can show that from equation (i)

v₂ – v₁ = u₁ – u₂ … (iii)

Or separation speed = approach speed

Or e = 1

Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e.

Diagram being restored — will be back shortly

Applying conservation of linear momentum, $m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{1}^{^{'}} + m_{2} v_{2}^{^{'}}$ … (iv)

Further, separation speed = e (approach speed) or u₁ – u₂ = e (v₂ – v₁) … (v)

Solving equations (iv) and (v) we get , $v_{1}^{^{'}} = (\frac{m _{1} - e m _{2}}{m _{1} - m _{2}}) v_{1} + (\frac{m _{1} + e m _{2}}{m _{1} + m _{2}}) v_{2}$ … (vi)

and $v_{2}^{^{'}} = (\frac{m _{2} - e m _{1}}{m _{1} + m _{2}}) v_{2} + (\frac{m _{1} + e m _{1}}{m _{1} + m _{2}}) v_{1}$ … (vii)

Illustration 2: A moving particle of mass m, makes a head–on collision with a particle of mass 2m, which is initially at rest. Show that the colliding particle loses (8/9) of its energy after collision.

Solution: Let u be the initial velocity of particle of mass m and $ν$ its velocity after the collision. Let V be the velocity of particle of mass 2m after the collision.

From the principle of conservation of linear momentum, we have

mu = mv + (2m) V Or u – v = 2 V … (i)

The conservation of kinetic energy gives

$\frac{1}{2}$ mu² = $\frac{1}{2}$ mv² + $\frac{1}{2}$ (2m)V² Or u² – v² = 2V² … (ii)

Or (u – v) (u + v) = 2V²

Using Eq. (1) in Eq. (2) we have

2V (u + v) = 2V² or u + v = V Or 2(u + v) = 2V … (iii)

Comparing (1) and (3) we get, u – v = 2 (u + v)× Or v = $- \frac{u}{3}$ … (iv)

Now, initial kinetic energy of the colliding mass is, K_i = $\frac{1}{2}$ mu²

Final kinetic energy, K_f = $\frac{1}{2}$ mv²

Loss in kinetic energy is, $Δ$ K = K_i – K_f = $\frac{1}{2}$ mu²– $\frac{1}{2}$ mv²

$∴$ Fractional loss = $\frac{Δ K}{K _{i}}$ = $\frac{\frac{1}{2} m u ^{2} - \frac{1}{2} m v ^{2}}{\frac{1}{2} m u ^{2}}$ = $\frac{u ^{2} - v ^{2}}{u ^{2}}$ = 1 – $(\frac{v}{u})^{2}$

= 1 – $(- \frac{1}{3})^{2}$ = $\frac{8}{9}$ ( $∵$ v = – u/3)

OBLIQUE COLLISION

When two bodies collide such that there velocities are not in the line of action of contact force then the collision is known as oblique collision.

We can divide also oblique collision in elastic, and inelastic collision. It is important to notice that momentum will conserve in each direction curing oblique collision also.

When two bodies collide obliquely, their relative velocity resolved along their common normal after impact is in a constant ratio to their relative velocity before impact (resolved along common normal) and is in the opposite direction.

$\frac{v _{1} cos θ - v _{2} cos ϕ}{u _{1} cos α - u _{2} cos β} = - e$ $\Rightarrow v_{1} cos θ - v_{2} cos ϕ = - e (u_{1} cos α - u_{2} cos β)$

Diagram being restored — will be back shortly

Illustration 3: A ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. The energy of the balls after the collision is 3/4^th of the original. Find the coefficient of restitution.

Diagram being restored — will be back shortly

Solution: As we have seen in the above discussion, that under the given conditions :

$v_{1}^{^{'}} = (\frac{1 + e}{2}) v an d v_{2}^{^{'}} = (\frac{1 - e}{2}) v$

Given that $K_{f} = \frac{3}{4} K_{i}$ Or $\frac{1}{2} m v_{1}^{^{'}^{2}} = \frac{3}{4} (\frac{1}{2} m v^{2})$

Substituting the value, we get, $(\frac{1 + e}{2})^{2} + (\frac{1 - e}{2})^{2} = \frac{3}{4}$

Or (1 + e)² + (1 – e)² = 3 Or 2 + 2e² = 3 Or $e^{2} = \frac{1}{2}$ 22²

Or $e = \frac{1}{2}$

Ready to master System Of Particles And Rotational Motion?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test