Collisions and Momentum: Conceptual Doubts Cleared (4)

easy • 3 min read
Tags Collisions And Momentum

Question

Two balls of masses m1=2m_1 = 2 kg and m2=3m_2 = 3 kg collide head-on. Before collision, m1m_1 moves at 44 m/s and m2m_2 is at rest. A student claims kinetic energy is always conserved in a collision. Another says momentum is always conserved but KE may not be. Resolve, and find the final velocities for both an elastic and a perfectly inelastic case.

Solution — Step by Step

Momentum is always conserved in any collision (no external horizontal force during the brief collision time). Kinetic energy is conserved only in elastic collisions. In inelastic collisions, some KE converts to heat, sound, or deformation. Student 2 is right.

For 1D elastic collisions, use the standard formulas:

v1′=m1−m2m1+m2u1=2−35×4=−0.8 m/sv_1' = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{2 - 3}{5} \times 4 = -0.8 \text{ m/s}

v2′=2m1m1+m2u1=45×4=3.2 m/sv_2' = \frac{2 m_1}{m_1 + m_2} u_1 = \frac{4}{5} \times 4 = 3.2 \text{ m/s}

So m1m_1 bounces back at 0.8 m/s, m2m_2 moves forward at 3.2 m/s.

Momentum: before =2×4=8= 2 \times 4 = 8. After =2×(−0.8)+3×3.2=−1.6+9.6=8= 2 \times (-0.8) + 3 \times 3.2 = -1.6 + 9.6 = 8. ✓ KE: before =12(2)(16)=16= \tfrac{1}{2}(2)(16) = 16 J. After =12(2)(0.64)+12(3)(10.24)=0.64+15.36=16= \tfrac{1}{2}(2)(0.64) + \tfrac{1}{2}(3)(10.24) = 0.64 + 15.36 = 16 J. ✓

Common final velocity: v=m1u1m1+m2=85=1.6v = \dfrac{m_1 u_1}{m_1 + m_2} = \dfrac{8}{5} = 1.6 m/s.

KE before =16= 16 J. KE after =12(5)(1.6)2=6.4= \tfrac{1}{2}(5)(1.6)^2 = 6.4 J. Lost =9.6= 9.6 J — converted to heat and deformation.

Why This Works

Newton’s third law guarantees momentum conservation: the force m1m_1 exerts on m2m_2 equals the force m2m_2 exerts on m1m_1. Over the same collision time, these impulses are equal and opposite, so total momentum doesn’t change.

KE conservation requires that no energy is lost to non-conservative effects — which is rare in real collisions. Bouncing balls lose KE; hard steel balls lose less; only ideal elastic collisions (like billiard balls or atomic-scale collisions) conserve KE exactly.

For elastic collisions, memorize this trick: the relative velocity reverses. u1−u2=−(v1′−v2′)u_1 - u_2 = -(v_1' - v_2'). Combined with momentum conservation, this gives final velocities in two lines without the messy formulas.

Alternative Method

Use the relative velocity trick. For elastic: relative velocity before = 4−0=44 - 0 = 4 m/s. After: v2′−v1′=4v_2' - v_1' = 4. Momentum: 2v1′+3v2′=82v_1' + 3v_2' = 8. Solve: v2′=3.2v_2' = 3.2, v1′=−0.8v_1' = -0.8. Same answer, simpler algebra.

Students apply KE conservation in inelastic collisions and get nonsense answers. Never assume KE is conserved unless the problem says “elastic” or you can argue why no energy is lost.

Final answer: Elastic — v1′=−0.8v_1' = -0.8 m/s, v2′=3.2v_2' = 3.2 m/s. Inelastic — v=1.6v = 1.6 m/s.

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