Collisions and Momentum: Common Mistakes and Fixes (5)

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Tags Collisions And Momentum

Question

A 2 kg2 \text{ kg} ball moving at 5 m/s5 \text{ m/s} collides head-on with a 3 kg3 \text{ kg} ball at rest. Find the final velocities if the collision is (a) perfectly elastic, (b) perfectly inelastic. Then check the energy lost in case (b).

Solution — Step by Step

Both cases conserve momentum:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

(2)(5)+(3)(0)=2v1+3v2  ⟹  10=2v1+3v2(⋆)(2)(5) + (3)(0) = 2v_1 + 3v_2 \implies 10 = 2v_1 + 3v_2 \quad (\star)

For 1D elastic collision, the standard result:

v1=m1−m2m1+m2u1+2m2m1+m2u2v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2

v2=2m1m1+m2u1+m2−m1m1+m2u2v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{m_2 - m_1}{m_1 + m_2}u_2

Plug in: v1=−15×5=−1 m/sv_1 = \dfrac{-1}{5}\times 5 = -1 \text{ m/s}, v2=45×5=4 m/sv_2 = \dfrac{4}{5}\times 5 = 4 \text{ m/s}.

The first ball bounces back at 1 m/s1 \text{ m/s}; the second moves forward at 4 m/s4 \text{ m/s}.

Initial KE: 12(2)(5)2=25 J\tfrac{1}{2}(2)(5)^2 = 25 \text{ J}.

Final KE: 12(2)(1)2+12(3)(4)2=1+24=25 J\tfrac{1}{2}(2)(1)^2 + \tfrac{1}{2}(3)(4)^2 = 1 + 24 = 25 \text{ J}. ✓

Both balls move together after collision: v1=v2=vv_1 = v_2 = v.

10=(2+3)v  ⟹  v=2 m/s10 = (2 + 3)v \implies v = 2 \text{ m/s}

KE after: 12(5)(2)2=10 J\tfrac{1}{2}(5)(2)^2 = 10 \text{ J}. Energy lost: 25−10=15 J25 - 10 = 15 \text{ J} (turned into heat, sound, deformation).

Why This Works

Momentum is conserved in all collisions because there are no external horizontal forces during the brief impact. KE is conserved only in elastic collisions. So momentum gives you one equation always; the second equation is either KE conservation (elastic) or “they stick” (inelastic).

The elastic-collision formula falls out of solving the two equations simultaneously — most of JEE Main expects you to know the result, not derive it. NEET sometimes asks for the derivation.

Alternative Method

Use the coefficient of restitution approach:

e=relative velocity of separationrelative velocity of approache = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}

For elastic, e=1e = 1: v2−v1=u1−u2=5v_2 - v_1 = u_1 - u_2 = 5. Combine with momentum equation (⋆)(\star):

10=2v1+3(v1+5)=5v1+15  ⟹  v1=−1, v2=410 = 2v_1 + 3(v_1 + 5) = 5v_1 + 15 \implies v_1 = -1, \, v_2 = 4

For perfectly inelastic, e=0e = 0: v2=v1v_2 = v_1. Then (⋆)(\star) gives v1=v2=2v_1 = v_2 = 2. Same answers.

Memorise the special case “equal masses elastic head-on collision”: the moving ball stops and the stationary one takes off at the original speed. This is the principle behind Newton’s cradle and shows up in PYQs surprisingly often.

Common Mistake

The two big mistakes in collision problems:

  1. Sign of the rebounding velocity. Many students miss that v1v_1 can be negative when m1<m2m_1 < m_2. Always set up positive direction first and let the algebra produce signs — don’t intuit them.

  2. Assuming KE is conserved in inelastic collisions. It isn’t. Momentum stays, but KE drops. The energy goes into deformation, sound, heat. Writing 12m1u12=12(m1+m2)v2\tfrac{1}{2}m_1u_1^2 = \tfrac{1}{2}(m_1+m_2)v^2 for an inelastic collision gives a wrong final velocity by a factor of m1/(m1+m2)\sqrt{m_1/(m_1+m_2)}.

Final answer: Elastic: v1=−1 m/sv_1 = -1 \text{ m/s}, v2=4 m/sv_2 = 4 \text{ m/s}. Inelastic: v=2 m/sv = 2 \text{ m/s}, energy lost =15 J= 15 \text{ J}.

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