Explain with neat sketches (1) direct central impact and (2) oblique central impact. (asked in MU June 2025 KT)

Answer: In both cases the line of impact passes through the mass centres of the bodies. In direct central impact the velocities are along the line of impact; in oblique central impact one or both velocities are inclined to it.

Line of impact: the common normal to the contact surfaces at the point of contact. The impact is central when the mass centres of both bodies lie on this line.

(1) Direct central impact. Sketch: two spheres A and B on a horizontal line (the line of impact through both centres), with velocity arrows \(u_A\) and \(u_B\) drawn along that line. The velocities before impact are along the line of impact, so after impact the bodies also move along it. Equations: conservation of momentum \(m_Au_A+m_Bu_B=m_Av_A+m_Bv_B\) and coefficient of restitution \(e=\dfrac{v_B-v_A}{u_A-u_B}\). Example: two billiard balls colliding head-on.

(2) Oblique central impact. Sketch: the same two spheres with the line of impact (x-axis, through both centres) and the common tangent (y-axis) at the contact point; arrows \(u_A\) and \(u_B\) are drawn at angles \(\alpha\) and \(\beta\) to the line of impact. Resolve velocities along the line of impact (n) and the tangent (t):

• Along the line of impact: momentum is conserved and \(e=\dfrac{(v_B)_n-(v_A)_n}{(u_A)_n-(u_B)_n}\).
• Along the tangent (no impulsive force for smooth bodies): each body's tangential velocity is unchanged, \((v_A)_t=(u_A)_t\), \((v_B)_t=(u_B)_t\).

Example: a billiard ball striking another ball off-centre, or a ball striking a floor at an angle.