Collisions in Two Dimensions — Question 6

Section 9-10 · Physics pages 1342–1343

Question 6 A measured component leaves two mirror-image scattering solutions Identical pucks collide elastically, with one initially at (8, 0) m/s and the other at rest. The first puck's final x component is 3 m/s. v1x v1y 3 8 Velocity-space circle; axes share the same scale, in m/s. (a) Find every possible final velocity pair. (b) Show that the first puck's velocity endpoint lies on a circle and explain the remaining ambiguity.
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Question 6 - Solution A measured component leaves two mirror-image scattering solutions v1x v1y 3 8 (a) Derivation and calculation Write⃗v1 = (3, b); momentum gives⃗v2 = (5, −b). Elastic energy requires 32 + b2 + 52 + b2 = 82 ⇒2b2 = 30.⃗ v1 = (3, ± √ 15),⃗ v2 = (5, ∓ √ 15) m/s . (b) Result and interpretation For a general endpoint (a, b), a2 + b2 + (8 −a)2 + b2 = 64 ⇒a2 + b2 = 8a. (a −4)2 + b2 = 16 . The line a = 3 intersects this circle twice. The two outcomes are reflections across the incident axis; a transverse-direction measurement selects one. Physical check The two signs cancel exactly in total transverse momentum.

Question 6

A measured component leaves two mirror-image scattering solutions

Identical pucks collide elastically, with one initially at (8,0)m/s(8,0)\,\mathrm{m/s} and the other at rest. The first puck's final xx component is 3m/s3\,\mathrm{m/s}.

Diagram shown in the original SVG worksheet.

  1. Find every possible final velocity pair.

  2. Show that the first puck's velocity endpoint lies on a circle and explain the remaining ambiguity.

Question 6 - Solution

A measured component leaves two mirror-image scattering solutions

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Write v→1=(3,b)\vec v_1=(3,b); momentum gives v→2=(5,−b)\vec v_2=(5,-b). Elastic energy requires 32+b2+52+b2=82⇒2b2=30.3^2+b^2+5^2+b^2=8^2\Rightarrow2b^2=30. v→1=(3,±15),v→2=(5,∓15)m/s.\boxed{\vec v_1=(3,\pm\sqrt{15}),\quad\vec v_2=(5,\mp\sqrt{15})\ \mathrm{m/s}}.

(b) Result and interpretation

For a general endpoint (a,b)(a,b), a2+b2+(8−a)2+b2=64⇒a2+b2=8a.a^2+b^2+(8-a)^2+b^2=64\Rightarrow a^2+b^2=8a. (a−4)2+b2=16.\boxed{(a-4)^2+b^2=16}. The line a=3a=3 intersects this circle twice. The two outcomes are reflections across the incident axis; a transverse-direction measurement selects one.

Physical check

The two signs cancel exactly in total transverse momentum.