Q.Two identical ball bearings, labelled 2 and 3, are in contact with each other and rest on a smooth (frictionless) table. A third identical ball bearing, labelled 1, of the same mass moves along the line of centres and strikes them head-on with speed $V$. Assuming the collision is elastic, which of the listed outcomes is a physically possible result after the collision?

Let each ball have mass $m$. Before the collision only ball 1 moves, so the initial momentum is $p_i=mV$ and the initial kinetic energy is $K_i=\tfrac12 mV^2$.

Option (A): ball 1 stops, balls 2 and 3 (combined mass $2m$) move off together at $V/2$.

Momentum: $2m\times(V/2)=mV$ — matches $p_i$.

Kinetic energy: $\tfrac12(2m)(V/2)^2=\tfrac14 mV^2$ — only half of $K_i$. Kinetic energy is NOT conserved, so this outcome is not possible for an elastic collision.

Option (B): balls 1 and 2 stop, ball 3 (mass $m$) moves off at $V$.

Momentum: $m\times V=mV$ — matches $p_i$.

Kinetic energy: $\tfrac12 mV^2$ — exactly equal to $K_i$. Both conservation laws hold, so this is the physically valid outcome (it is exactly the familiar "Newton's cradle" behaviour: the impact transmits through the touching balls 2 and 3, leaving 1 and 2 at rest and only the far ball moving away). …

Concept and conditions

Let each ball have mass $m$. Before impact only ball 1 moves, so

An elastic collision requires both

Testing each outcome

Option A -- 1 stops, 2 and 3 move with $V/2$: momentum $p_f=2m\cdot\tfrac{V}{2}=mV$ (ok), but $K_f=\tfrac12(2m)\left(\tfrac{V}{2}\right)^2=\tfrac{mV^2}{4}=\tfrac12 K_i$ -- kinetic energy is halved, so this is not elastic.

Option B -- 1 and 2 stop, 3 moves with $V$: momentum $p_f=m\cdot V=mV$ (ok), and $K_f=\tfrac12 mV^2=K_i$ (ok) -- both conserved, so this is possible.

Option C -- all three move with $V/3$: momentum $p_f=3m\cdot\tfrac{V}{3}=mV$ (ok), but $K_f=\tfrac12(3m)\left(\tfrac{V}{3}\right)^2=\tfrac{mV^2}{6}=\tfrac13 K_i$ -- energy is lost, so not elastic. …

Concept: Testing Each Outcome Against BOTH Conservation Laws

An elastic collision must conserve momentum and kinetic energy simultaneously — momentum alone does not certify a result as physically possible.

Step 1: Write the "before" values (mass $m$ each, only ball 1 moving)

Step 2: Test option (A) — 1 stops, 2 & 3 move together at $V/2$

Step 3: Test option (B) — 1 and 2 stop, 3 moves at $V$

$p_f = mV \ \checkmark \qquad K_f = \tfrac12 mV^2 = K_i\ \checkmark$ …