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,
m1 u1 cosθ1 + m2 u2 cosθ2 = m1v1 cosβ1 + m2 v2 cos β2 …(1)
& m1 u1 sinθ1 + m2 u2 sinθ2 = m1 v1 sinβ1 + m2 v2 sin β2
Since no force is acting on m1 & m2 along the tangent, the individual momentum of m1 and m2 remains conserved.
⇒ m1 u1 sinθ1 = m2 v1 sinβ1 …(2)
& m2 u2 sinθ2 = m2 v2 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 v1, v2, β1 and β2. 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.