This is most important to know the line of action or impact during the collision.

The line of impact is the line along which the impulsive force acts on the bodies or the impact takes place.

How do you find the line of impact using this concept ?

Step 1: Let two bodies 1 and 2 collide. First of all find the point of contact just before the collision.

Step 2: Draw a straight line passing through the point of contact of the bodies so that it does not touch any part of their surfaces. That is known as tangent to the surfaces in contact or we can call it as common tangent.

Step 3: After drawing the tangent, draw a normal to the tangent at the point. This normal line is known as line of impact. That means impulsive force acts on the bodies along this line.

After finding the line of impact, calculate the components of their relative velocity just after and before the impact , along this line and set their ratio equal to e.

$ \displaystyle -\frac{\vec{v_{12}}_n}{\vec{u_{12}}_n} $ ; Where u_{12} & v_{12} – are the component velocity of 1 and 2 just before and after the impact along the normal (line of impact) respectively. (u_{21})_{n} is known as velocity of approach and (v_{21})_{n} is known as velocity of separation

Illustration : A triangular wedge of mass M is moving with uniform velocity v_{o} along a smooth horizontal surface in the leftward direction.

A particle of mass m falls from rest from h on to the inclined face, colliding elastically with it. Find velocity of the ball and wedge after the impact taking M = 2m.

Solution: Impact incline is a straight line perpendicular to the incline. Normal reaction force between the body and the wedge acts along the impact line.

This normal force becomes an internal force when we consider (wedge + body) as a total system – But , normal reaction and considerable in magnitude because the impact force during collision has contribution towards making of the normal force.

-> Momentum of the system is conserved along a line perpendicular to this normal force momentum of the system is conserved along horizontal. …(i)

-> Momentum of the body particle is conserved along the common tangent at the point of impact. ….. (ii)

-> As this is an elastic collision ,

relative velocity of separation along the impact line = relative velocity of approach. …(iii)

-> If the wedge were at rest then the body / particle would deflect in the horizontal direction after collision because it is an elastic collision …… (iv)

-> As, in this case, wedge is moving, the object would not be deflected horizontally ,

but at an angle ‘ α ‘ to the impact …. (v)

Velocity of body particle before Impact $ \displaystyle u = \sqrt{2 g h} $ along PO^{→}

Velocity of body particle after collision = v_{1} along OQ^{→}

Velocity of wedge before collision = v_{0} along RO^{→}

Velocity of wedge after collision = v_{2} along RO^{→}

v_{0} cos45° − u cos45° = − v_{1} cosα − v_{2}cos45°

v_{1} cosα = v^{2}/√2 = u/√2 − v_{0}/√2

√2v_{1} cosα + v_{2} = u − v_{0} … (A)

According to logic (i)

Mv_{0} = Mv_{2} − mv_{1} cos(45° – α)

⇒ 2v_{0} = 2v_{2} − v_{1}cos(45° – α)

⇒ 2v_{2} – v1cos(45° − α) = 2v_{0 } …. (B)

According to logic (ii)

mu sin 45° = mv_{1} sinα

⇒ v_{1} sinα = u …. (C)

Solving

$ \displaystyle v_1 = \sqrt{u^2 + [\frac{u(4-\sqrt2)-6v_0}{20}]^2} $

$ \displaystyle v_2 = \frac{v_0 + (\sqrt2 +1 )u}{5} $