Question 5 of 6: Question 5 (paper Question V) — Impact and Spring Compression (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 — 2018-May. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from 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 — source reconstruction notes. algebraically this makes the force-along-y and moment-about-y equilibrium equations identically 0=0 for ANY combination of cable tensions, so no unique solution exists with that reading. Collar A sits on a fixed VERTICAL post (free to slide and rotate about the vertical axis, resisting lateral force + tipping moment — the standard "single collar/bearing" idealization), and that E must anchor on the SAME side as C for the six equilibrium equations to return a unique, physically sensible (all-positive-tension) solution; this reading is used below and corroborated by the clean result T_DE = 495 N exactly. (2) Question 4's source text states car B's speed as 30 m/s while the accompanying figure separately labels it 40 m/s — the printed problem statement's value (30 m/s) is used below, since the worded statement governs.
Question 5 (paper Question V) — Impact and Spring Compression (Part B · Dynamics, equal value)
Given. Block A: mass 10 kg, released from rest, falls 2.5 m to plate P. Plate P: mass 5 kg, rests on a spring of stiffness $k=1600$ N/m, slides freely (no friction) on vertical guides. Coefficient of restitution $e=0.75$.
Given data
Quantity
Value
Mass of block A, $m_A$
10 kg
Mass of plate P, $m_P$
5 kg
Drop height
2.5 m
Coefficient of restitution, $e$
0.75
Spring stiffness, $k$
1600 N/m
Find. (a) the velocities of block A and plate P just after impact; (b) the maximum additional spring compression caused by the impact.
Figure 5 (reconstructed). Block A falling onto plate P, which rests on the spring between guides BC and DE.
Approach. Use free-fall kinematics for A's impact speed, then conservation of momentum plus the restitution equation for the impact itself. After separation, apply the work-energy principle to plate P alone (spring + gravity) to find the additional compression beyond its initial static equilibrium position.
Speed of A just before impact. $v_{A0}=\sqrt{2g h}=\sqrt{2(9.81)(2.5)}$, so $\boxed{v_{A0}=7.004\text{ m/s}}$ (downward).
Momentum and restitution. Taking down as positive, with P initially at rest: $m_A v_{A0}=m_A v_A'+m_P v_P'$ and $e\,v_{A0}=v_P'-v_A'$. Substituting $m_A=10$, $m_P=5$, $v_{A0}=7.004$, $e=0.75$: $70.04=10v_A'+5v_P'$ and $5.253=v_P'-v_A'$.
Solve the two equations. Eliminating $v_P'=v_A'+5.253$: $70.04=10v_A'+5(v_A'+5.253)=15v_A'+26.26$, so $\boxed{v_A'=2.918\text{ m/s}}$ (down) and $\boxed{v_P'=8.171\text{ m/s}}$ (down). Since $v_P'\gt v_A'$, plate P pulls away from block A after impact — they separate, confirming A does not ride on P.
Static (pre-impact) spring compression. Before impact, P alone is in equilibrium on the spring: $x_0=m_P g/k=5(9.81)/1600$, so $x_0=30.7$ mm.
Additional compression from work-energy (P alone, after separation). Measuring displacement $d$ from the static-equilibrium position, gravity's work on $d$ exactly cancels the linear term in the spring's PE change (since $m_P g=k x_0$ at equilibrium), leaving simply $\tfrac12 m_P (v_P')^2=\tfrac12 k d^2$. Solving, $d=v_P'\sqrt{m_P/k}=8.171\sqrt{5/1600}$, so $\boxed{d=0.457\text{ m}}$.
