m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Elastic vs inelastic collision example.

Conservation of Momentum — Two Balls Collide

Question

A ball of mass 2 kg moving at 4 m/s collides head-on with a ball of mass 1 kg at rest. After an elastic collision, find the velocities of both balls.

Solution — Step by Step

In any collision (elastic or not), total momentum is conserved. We write:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Substituting: $2(4) + 1(0) = 2v_1 + v_2$

8 = 2v_1 + v_2 \quad \text{...(i)}

This is the step many students miss. For an elastic collision, kinetic energy is also conserved — but there’s a smarter way to use this. For elastic collisions, the relative velocity of approach equals the relative velocity of separation:

u_1 - u_2 = v_2 - v_1

Substituting: $4 - 0 = v_2 - v_1$

v_2 - v_1 = 4 \quad \text{...(ii)}

From equation (ii): $v_2 = v_1 + 4$

Substitute into (i):

8 = 2v_1 + (v_1 + 4) = 3v_1 + 4

3v_1 = 4 \implies v_1 = \frac{4}{3} \approx 1.33 \text{ m/s}

v_2 = v_1 + 4 = \frac{4}{3} + 4 = \frac{16}{3} \approx 5.33 \text{ m/s}

Initial KE $= \frac{1}{2}(2)(4^2) = 16$ J

Final KE $= \frac{1}{2}(2)\left(\frac{4}{3}\right)^2 + \frac{1}{2}(1)\left(\frac{16}{3}\right)^2 = \frac{16}{9} + \frac{128}{9} = \frac{144}{9} = 16$ J ✓

Energy is conserved — confirms elastic collision.

Final Answer: $v_1 = \frac{4}{3}$ m/s (forward), $v_2 = \frac{16}{3}$ m/s (forward)

Why This Works

The two conservation laws — momentum and energy — give us two independent equations for two unknowns. That’s why elastic collisions are fully solvable. Inelastic collisions only give us one equation (momentum), so unless we’re told what happens (e.g., “they stick together”), we can’t find both final velocities.

The relative velocity shortcut ( $u_1 - u_2 = v_2 - v_1$ ) is derived from combining both conservation laws algebraically. It’s not a new law — it’s a consequence. Using it saves you from solving a quadratic in the KE equation, which is a real time-saver in JEE Main where every minute counts.

Notice both balls move forward after collision. This happens when the moving ball is heavier than the stationary one. If the masses were swapped, the lighter ball would bounce back — a result worth memorising for MCQs.

Alternative Method — Direct Formula

For elastic collisions, we can directly apply the standard result (derived once, used forever):

v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1

v_2 = \frac{2m_1}{m_1 + m_2} u_1

(valid when $u_2 = 0$ , i.e., target is at rest)

Substituting $m_1 = 2$ kg, $m_2 = 1$ kg, $u_1 = 4$ m/s:

v_1 = \frac{2-1}{2+1} \times 4 = \frac{1}{3} \times 4 = \frac{4}{3} \text{ m/s}

v_2 = \frac{2 \times 2}{2+1} \times 4 = \frac{4}{3} \times 4 = \frac{16}{3} \text{ m/s}

Same answer, much faster. Memorise this formula — it appears directly in JEE Main numericals.

Special case worth memorising: when $m_1 = m_2$ in an elastic collision, the first ball stops completely and the second moves off with the first ball’s original velocity. This is the Newton’s cradle result. MCQs love this.

Common Mistake

Students apply the relative velocity condition as $u_1 - u_2 = v_1 - v_2$ (wrong sign). The correct form is $u_1 - u_2 = -(v_1 - v_2) = v_2 - v_1$ . The sign flip comes from the definition: velocity of approach = velocity of separation, meaning the gap was closing before and opening after. Getting this backwards gives $v_1 = \frac{20}{3}$ m/s, which violates energy conservation — always do the sanity check.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Formula

Common Mistake

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