Momentum and Kinetic Energy in Collisions — Question 1

Section 9-7 · Physics pages 1272–1273

Question 1 Audit both invariants before naming a collision Two 1 kg carts initially move at 4 and 0 m/s. Three proposed final velocity pairs are A: (0, 4), B: (1, 3) and C: (−1, 5) m/s. External impulse is negligible. Kf (J) 5 8 13 B A C Candidate energies are compared with the initial 8 J. (a) Check momentum and kinetic energy for each outcome. (b) Classify each and identify which requires release of internal energy.
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Question 1 - Solution Audit both invariants before naming a collision Kf (J) 5 8 13 B A C (a) Derivation and calculation For each candidate, momentum is Pi = 1(4) + 1(0) = 4, PA = 0 + 4 = 4, PB = 1 + 3 = 4, PC = −1 + 5 = 4. Thus all three satisfy the momentum constraint. (b) Result and interpretation Evaluate the energies separately: Ki = 1 2(42) = 8 J, (KA, KB, KC) = 1 2(16, 10, 26) = (8, 5, 13) J. A : elastic; B : loses 3 J; C : gains 5 J. Case C requires a source of stored internal energy. Physical check Equal total momentum does not imply equal kinetic energy.

Question 1

Audit both invariants before naming a collision

Two 1kg1\,\mathrm{kg} carts initially move at 44 and 0m/s0\,\mathrm{m/s}. Three proposed final velocity pairs are A: (0,4)(0,4), B: (1,3)(1,3) and C: (−1,5)m/s(-1,5)\,\mathrm{m/s}. External impulse is negligible.

Diagram shown in the original SVG worksheet.

  1. Check momentum and kinetic energy for each outcome.

  2. Classify each and identify which requires release of internal energy.

Question 1 - Solution

Audit both invariants before naming a collision

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For each candidate, momentum is Pi=1(4)+1(0)=4,PA=0+4=4,PB=1+3=4,PC=−1+5=4.P_i=1(4)+1(0)=4,\quad P_A=0+4=4,\quad P_B=1+3=4,\quad P_C=-1+5=4. Thus all three satisfy the momentum constraint.

(b) Result and interpretation

Evaluate the energies separately: Ki=12(42)=8J,(KA,KB,KC)=12(16,10,26)=(8,5,13)J.K_i=\tfrac12(4^2)=8\ \mathrm J,\quad (K_A,K_B,K_C)=\tfrac12(16,10,26)=(8,5,13)\ \mathrm J. A: elastic;B: loses 3J;C: gains 5J.\boxed{A:\text{ elastic};\quad B:\text{ loses }3\ \mathrm J;\quad C:\text{ gains }5\ \mathrm J.} Case C requires a source of stored internal energy.

Physical check

Equal total momentum does not imply equal kinetic energy.