Question 5 of 6: Question 5 (paper Question V) — Pendulum Impact Between a Block and a Sphere (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2017-December. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in Part B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — figure reconstruction. Three figures needed a reconstructed reading, each corroborated by an independent numerical check rather than assumed: (1) Question 1's wall cable anchors C and D — the only geometry (C on the +x side, D on the −x side of the pipe) that returns positive (physically valid, tension-only) cable forces for both cables; any same-side reading returns a negative "tension" in one cable, which is impossible. (2) Question 2's overall frame height is not printed directly; it follows from the stated 30° leg angle and the 4 m offset of the lower joints, and the resulting member forces close exactly by symmetry (equal reactions, two exact zero-force members) — strong corroborating evidence the reconstruction is right. (3) Question 5's second suspension cable (block A) is read as the same 6 ft length as sphere B's cable, since the figure's one printed "6 ft" dimension spans a horizontal reference line common to both hanging positions.
Question 5 (paper Question V) — Pendulum Impact Between a Block and a Sphere (Part B · Dynamics, equal value)
Given. Block A ($W_A=20\ \text{lbf}$), suspended by a pair of parallel cables from C, is released from rest at $60^\circ$ from vertical; because both cables have the same length and pivot spacing, A translates (does not rotate) as it swings, so it behaves as a particle on a circular arc of radius equal to the cable length, $L=6\ \text{ft}$. Sphere B ($W_B=40\ \text{lbf}$), on a single 6 ft cable from E, hangs at rest directly below E, at the same depth as A's lowest swing position. Coefficient of restitution $e=0.80$.
Find. The velocities of A and B immediately after impact.
Figure 5 (reconstructed). Block A swings from 60° to the bottom of its arc and strikes sphere B, which is at rest.
Approach. Use energy conservation to find A's speed at the bottom of its swing (where its cables are vertical and its velocity is purely horizontal); sphere B is also momentarily at the bottom of its own arc (cable vertical) so its constrained velocity is horizontal too — the impact is therefore a standard direct central (1-D) impact, solved with momentum conservation and the restitution equation.
Speed of A just before impact (energy conservation). Height dropped from $60^\circ$ to the bottom: $h=L(1-\cos60^\circ)=6(1-0.5)=3.0\ \text{ft}$. $$v_A=\sqrt{2gh}=\sqrt{2(32.2)(3.0)}=\boxed{13.90\ \text{ft/s}}$$ (horizontal, toward B; sphere B is at rest, $v_B=0$.)
Momentum conservation (using weights in place of mass, since $g$ cancels). $$W_Av_A=W_Av_A'+W_Bv_B'\ \Rightarrow\ v_A=v_A'+2v_B'$$
