Collision in Two Dimensions (Non-Head-on/Oblique Collision)

When at least one of the initial velocities is NOT directed along the common line joining the two colliding bodies, the collision is OBLIQUE (non-head-on, or two-dimensional). It becomes convenient to resolve everything along two mutually perpendicular directions at the point of impact: the COMMON TANGENT (along which no force acts during the brief contact, so momentum along this direction is separately conserved for the whole interaction) and the LINE OF IMPACT, perpendicular to the tangent, along which the actual mutual (internal) contact forces act and change the bodies' momenta (Fig 4.4).

If $\alpha_1$, $\alpha_2$ are the angles the initial velocities $u_1$, $u_2$ make with the line of impact, and $\beta_1$, $\beta_2$ the angles the final velocities $v_1$, $v_2$ make with it, then: conservation of momentum ALONG the line of impact gives $m_1u_1\cos\alpha_1+m_2u_2\cos\alpha_2=m_1v_1\cos\beta_1+m_2v_2\cos\beta_2$; since there is no force along the common tangent, EACH body's momentum component along the tangent is separately conserved, giving $m_1u_1\sin\alpha_1=m_1v_1\sin\beta_1$ and $m_2u_2\sin\alpha_2=m_2v_2\sin\beta_2$; and the coefficient of restitution applies along the line of impact exactly as in the head-on case. Together these four equations can, in principle, be solved for the four unknowns $v_1$, $v_2$, $\beta_1$, $\beta_2$.

A short set of standing reminders for any collision problem: (1) colliding bodies exert internal, equal-and-opposite action-reaction forces on each other along the line of impact; (2) there is no force along the common tangent; (3) in reality some mechanical (kinetic) energy is always lost as sound/heat/light, though TOTAL energy of the system is always conserved; (4) the velocity of separation is in practice always less than the velocity of approach along the line of impact, so $e<1$ in every real collision; (5) only idealised atomic/molecular-scale elastic collisions have the velocity of separation exactly equal to the velocity of approach.

Worked illustration (exploding shell, Example 4.7): a 3 kg shell falling freely for 2 s (reaching $u=20$ m/s downward, using g=10 m/s^2) explodes into a 2 kg and a 1 kg fragment; momentum conservation gives $2v_1+v_2=60$, and the 300 J of kinetic energy supplied by the explosion gives a second (quadratic) relation between $v_1$ and $v_2$; solving simultaneously yields two valid solution pairs, $(v_1,v_2)=(40,-20)$ m/s or $(10,40)$ m/s, corresponding to the two different possible relative positions of the lighter fragment (above or below) after the explosion. …