Q.A ball of mass $2\ \text{kg}$, moving at $5\ \text{m/s}$ along the $x$-axis, strikes an identical, stationary ball of mass $2\ \text{kg}$ in an elastic, oblique collision. After the collision, the first ball moves off at $30^\circ$ above the $x$-axis and the second at $60^\circ$ below the $x$-axis. Find the speed of each ball after the collision, and verify that kinetic energy is conserved.

Elastic collision: v2 = 2 m1 u/(m1 + m2) = 8 m/s. Answer (D).

For a one-dimensional elastic collision where mass m2 is initially at rest, the velocity of the struck body is:

v2 = 2 m1 u /(m1 + m2).

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A perfectly elastic collision has coefficient of restitution e = 1.

The coefficient of restitution is defined as:

$e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$

For a perfectly elastic collision, both momentum and kinetic energy are conserved, which mathematically requires the relative speed of separation to exactly equal the relative speed of approach, giving $e = 1$. For an inelastic collision …

Ivory is a hard, elastic material, so its collision with the floor is close to perfectly elastic and it rebounds to nearly the same height it was dropped from; wet clay is soft and deforms permanently on impact (a highly inelastic collision), losing almost all of its kinetic energy, so it barely rises after striking the floor.

When a ball is dropped and strikes the floor, the height h' to which it rebounds depends on how much of its kinetic energy just before impact is retained just after impact (rather than being lost to permanent deformation, heat, and sound during the collision) — this is measured by the coefficient of restitution, e, of the collision (e close to 1 = nearly elastic, e close to 0 = highly inelastic).

Ivory is a hard, relatively rigid and elastic material. When an ivory ball strikes the floor, it undergoes very little permanent deformation — it compresses briefly on impact and then springs back to its original shape, converting most of the kinetic energy back into the ball's motion. So the collision is close to perfectly elastic (e close to 1), and the ball rebounds to nearly the height it was dropped from.

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Elastic collision + equal masses + target initially at rest ⇒ velocities are exchanged, so the striking bob (A) stops completely and rises to zero height.

For a head-on elastic collision between two bodies of equal mass $m$, where the second body ($B$) is initially at rest, momentum and kinetic-energy conservation together give the standard result that the velocities are exchanged: body A (the one that was moving) ends up at rest, and body B moves off with A's original velocity.

Let A be released from $30°$ to the vertical. By energy conservation on the way down (before the collision), A arrives at the bottom (table level) with some speed $v$, having converted its initial gravitational PE fully into KE.

At the collision: since $m_A = m_B = m$ and it is elastic, the exchange-of-velocity result gives:

• A's velocity after collision $= 0$
• B's velocity after collision $= v$ (A's velocity just before collision) …

Elastic collision = momentum conserved AND kinetic energy conserved; this is what distinguishes it from an inelastic collision.

In any collision (elastic or inelastic), as long as no external force acts on the system, linear momentum is always conserved — this follows directly from Newton's third law (the colliding bodies exert equal and opposite forces on each other for equal times, so their net impulse on the system is zero).

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