Here the relative velocity of approach of the bodies doesn’t coincide with the line of impact. Therefore, this is a case of oblique impact.

Conserving of momentum of the system along & perpendicular to the line of impact,

we obtain,

m_{1} u_{1} cosθ_{1} + m_{2} u_{2} cosθ_{2} = m_{1}v_{1} cosβ_{1} + m_{2} v_{2} cos β_{2} …(1)

& m_{1} u_{1} sinθ_{1} + m_{2} u_{2} sinθ_{2} = m_{1} v_{1} sinβ_{1} + m_{2} v_{2} sin β_{2}

Since no force is acting on m1 & m2 along the tangent, the individual momentum of m_{1} and m_{2} remains conserved.

⇒ m_{1} u_{1} sinθ_{1} = m_{2} v_{1} sinβ_{1} …(2)

& m_{2} u_{2} sinθ_{2} = m_{2} v_{2} sinβ2 …(3)

Newton’s experimental Law : Along the line of impact

$ \displaystyle e = \frac{v_2 cos\beta_2 – v_1 cos\beta_1}{u_1 cos\theta_1 – u_2 cos\theta_2} $

Now we have four equations and four unknown’s v_{1}, v_{2}, β_{1} and β_{2}. Solving four equations for four unknown, we obtain,

Impulse = $ \displaystyle \frac{m_1 m_2 }{m_1 + m_2}(1+e)(u_1 cos\theta_1 – u_2 cos\theta_2) $

Energy Loss = $ \displaystyle \frac{m_1 m_2 }{2(m_1 + m_2)}(1-e^2)(u_1 cos\theta_1 – u_2 cos\theta_2)^2 $

##### Oblique Impact on A Fixed Plane :

Let a small ball collide with a smooth horizontal floor with a speed u at an angle θ to the vertical, as shown in the figure. Just after the collision, let the ball leave the floor with a speed v at an angle β to vertical.

It is quite clear that line of action is perpendicular to the floor. Therefore, the impact takes place along the (normal) vertical. Now we can use Newton’s experimental law as

$ \displaystyle e = \frac{velocity \; of \; separation}{velocity \; of \; approach} $

e (velocity of approach) = velocity of separation

$ \displaystyle e ( u cos \theta (-\hat{j})) = – v cos\beta (\hat{j}) $

$ \displaystyle v cos\beta = e u cos\theta $ …(i)

Since impulsive force N acts on the body along the normal we can not conserve its momentum. Since along horizontal, the component of N is zero, therefore the horizontal momentum of the body is conserved.