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04-BS-3 · December 2016

Question 5 of 6: Question 5 (paper Question V) — Oblique Central Impact of Two Spheres (Part B · Dynamics, equal value)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination December 2016 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.

Reference texts: Hibbeler, Engineering Mechanics: Statics, 14th ed. (Questions 1–3); Hibbeler, Engineering Mechanics: Dynamics, 14th ed. (Questions 4–6).

Separately, Question 1's bracket figure (Figure 1) gives only one anchor point's coordinates (E) plus two angle pairs for the other two cables; every printed dimension (50 mm×2, 100 mm×2, 75 mm, the two 45° arcs at C, the 30°/60° pair at D) was used, but the exact 3-D position of points B and C required one reasonable reconstruction, disclosed at first use below. All other figures (2–6) reconstruct cleanly and unambiguously from the printed dimensions, confirmed by internal checks that return clean values (member lengths BD = 5 ft and DE = 10 ft in Question 6) — strong evidence the readings are correct.

Question 5 (paper Question V) — Oblique Central Impact of Two Spheres (Part B · Dynamics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Identical spheres A and B (equal mass $m$, cancels), line of impact horizontal (the line of centres). $V_A=30\ \text{m/s}$ at 30° to the line of impact (approaching B); $V_B=40\ \text{m/s}$ at 60° to the line of impact (approaching A); $e=0.90$; both surfaces frictionless (no tangential impulse).

Given data
QuantityValue
$V_A$30 m/s at 30°
$V_B$40 m/s at 60°
$e$0.90
MassesEqual, identical spheres

Find. $V_A'$ and $V_B'$ (magnitude and direction) immediately after impact.

A B V_A = 30 m/s V_B = 40 m/s 30° 60°
Figure 5 — oblique central impact; n (horizontal, dashed) is the line of impact, t (vertical) is tangent to both spheres at contact.

Approach. Resolve both velocities into normal (n, along the line of centres) and tangential (t) components. Frictionless spheres exchange no tangential impulse, so each sphere's tangential component is unchanged; the normal components are found from conservation of momentum along n plus the coefficient-of-restitution equation.

  1. Resolve into n–t components (n positive from A toward B). $$V_{An}=30\cos30^{\circ}=25.98\ \text{m/s},\quad V_{At}=30\sin30^{\circ}=15.00\ \text{m/s}$$ $$V_{Bn}=-40\cos60^{\circ}=-20.00\ \text{m/s},\quad V_{Bt}=40\sin60^{\circ}=34.64\ \text{m/s}$$
  2. Tangential components are unchanged (frictionless impact): $V_{At}'=15.00\ \text{m/s}$, $V_{Bt}'=34.64\ \text{m/s}$.
  3. Normal-direction momentum and restitution. Equal masses cancel from momentum conservation: $$V_{An}+V_{Bn}=V_{An}'+V_{Bn}'\ \Rightarrow\ V_{An}'+V_{Bn}'=5.98$$ $$e=\frac{V_{Bn}'-V_{An}'}{V_{An}-V_{Bn}}\ \Rightarrow\ V_{Bn}'-V_{An}'=0.9(45.98)=41.38$$ Solving: $V_{An}'=\boxed{-17.70\ \text{m/s}}$, $V_{Bn}'=\boxed{23.68\ \text{m/s}}$.
  4. Recombine into resultant velocities. $$V_A'=\sqrt{(-17.70)^2+15.00^2}=\boxed{23.20\ \text{m/s}},\quad \theta_A'=\arctan\!\frac{15.00}{17.70}=40.3^{\circ}\ \text{above n, away from B}$$ $$V_B'=\sqrt{23.68^2+34.64^2}=\boxed{41.96\ \text{m/s}},\quad \theta_B'=\arctan\!\frac{34.64}{23.68}=55.6^{\circ}\ \text{above n, away from A}$$
Post-impact velocities
SphereSpeedDirection
A23.20 m/s40.3° above the line of impact, rebounding away from B
B41.96 m/s55.6° above the line of impact, continuing away from A