Conservation of Linear Momentum — Question 10

Section 9-6 · Physics pages 1270–1271

Question 10 Conservation gives a family until another condition is supplied Two isolated carts of masses 2 and 3 kg have initial velocities 5 and 0 m/s. Only momentum conservation is assumed for their interaction. v1 v2 2v1 + 3v2 = 10 Schematic momentum constraint line; coordinates of examples are specified in the text. (a) Express all final velocity pairs allowed by momentum conservation alone. Give two examples. (b) Explain what extra physical information is needed to select an actual outcome.
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Question 10 - Solution Conservation gives a family until another condition is supplied v1 v2 2v1 + 3v2 = 10 (a) Derivation and calculation One momentum equation defines a line of possible final states: 2v1 + 3v2 = 2(5) + 3(0) = 10. v2 = 10 −2v1 3 , (v1, v2) = (2, 2) or (−1, 4) m/s. (b) Result and interpretation The example energies differ despite equal momentum: K(2,2) = 1 2(5)(2)2 = 10 J, K(−1,4) = 1 2(2)(1) + 1 2(3)(16) = 25 J. Sticking, elastic energy conservation, restitution, or another interaction law is needed to choose an outcome. Momentum alone also permits states requiring internal energy release. Physical check Both examples reproduce the same total momentum of 10 kg m/s.

Question 10

Conservation gives a family until another condition is supplied

Two isolated carts of masses 22 and 3kg3\,\mathrm{kg} have initial velocities 55 and 0m/s0\,\mathrm{m/s}. Only momentum conservation is assumed for their interaction.

Diagram shown in the original SVG worksheet.

  1. Express all final velocity pairs allowed by momentum conservation alone. Give two examples.

  2. Explain what extra physical information is needed to select an actual outcome.

Question 10 - Solution

Conservation gives a family until another condition is supplied

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

One momentum equation defines a line of possible final states: 2v1+3v2=2(5)+3(0)=10.2v_1+3v_2=2(5)+3(0)=10. v2=10−2v13,(v1,v2)=(2,2)or(−1,4)m/s.\boxed{v_2=\frac{10-2v_1}{3}},\qquad (v_1,v_2)=(2,2)\ \text{or}\ (-1,4)\ \mathrm{m/s}.

(b) Result and interpretation

The example energies differ despite equal momentum: K(2,2)=12(5)(2)2=10J,K(−1,4)=12(2)(1)+12(3)(16)=25J.K_{(2,2)}=\tfrac12(5)(2)^2=10\ \mathrm J,\qquad K_{(-1,4)}=\tfrac12(2)(1)+\tfrac12(3)(16)=25\ \mathrm J. Sticking, elastic energy conservation, restitution, or another interaction law is needed to choose an outcome. Momentum alone also permits states requiring internal energy release.

Physical check

Both examples reproduce the same total momentum of 10 kg m/s.