# Oblique Impact , Collision

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 = m1 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.