Collisions in Two Dimensions — Question 3

Section 9-10 · Physics pages 1336–1337

Question 3 Inferring an unseen target mass from two components A 2 kg puck initially moves at (6, 0) m/s toward a stationary target of unknown mass M. After collision the incident puck has (2, 2) m/s and the target has (2, −1) m/s. External impulse is negligible.⃗ u1⃗ v1⃗ v2 Velocity directions are schematic; the text supplies magnitudes. (a) Infer M separately from each momentum component and test consistency. (b) Determine whether the collision is elastic.
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Question 3 - Solution Inferring an unseen target mass from two components⃗ u1⃗ v1⃗ v2 (a) Derivation and calculation The component equations independently determine the unknown mass: x : 2(6) = 2(2) + M(2) ⇒M = 4 kg, y : 0 = 2(2) + M(−1) ⇒M = 4 kg . The two measurements are consistent with zero external impulse. (b) Result and interpretation Compare the kinetic energies using squared vector magnitudes: Ki = 1 2(2)(6)2 = 36 J, Kf = 1 2(2)(22 + 22) + 1 2(4)(22 + (−1)2) = 18 J. L = 18 J ; the collision is inelastic. Physical check The final vertical momenta +4 and −4 kg m/s cancel.

Question 3

Inferring an unseen target mass from two components

A 2kg2\,\mathrm{kg} puck initially moves at (6,0)m/s(6,0)\,\mathrm{m/s} toward a stationary target of unknown mass MM. After collision the incident puck has (2,2)m/s(2,2)\,\mathrm{m/s} and the target has (2,−1)m/s(2,-1)\,\mathrm{m/s}. External impulse is negligible.

Diagram shown in the original SVG worksheet.

  1. Infer MM separately from each momentum component and test consistency.

  2. Determine whether the collision is elastic.

Question 3 - Solution

Inferring an unseen target mass from two components

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The component equations independently determine the unknown mass: x:2(6)=2(2)+M(2)⇒M=4kg,x:\quad2(6)=2(2)+M(2)\Rightarrow M=4\ \mathrm{kg}, y:0=2(2)+M(−1)⇒M=4kg.y:\quad0=2(2)+M(-1)\Rightarrow\boxed{M=4\ \mathrm{kg}}. The two measurements are consistent with zero external impulse.

(b) Result and interpretation

Compare the kinetic energies using squared vector magnitudes: Ki=12(2)(6)2=36J,K_i=\tfrac12(2)(6)^2=36\ \mathrm J, Kf=12(2)(22+22)+12(4)(22+(−1)2)=18J.K_f=\tfrac12(2)(2^2+2^2)+\tfrac12(4)(2^2+(-1)^2)=18\ \mathrm J. L=18J;the collision is inelastic.\boxed{L=18\ \mathrm J};\qquad\text{the collision is inelastic}.

Physical check

The final vertical momenta +4+4 and −4-4 kg m/s cancel.