Evaluate Collision Design Solutions - Grade 11 Physics

1. Which crumple-zone design satisfies all three constraints?
   Answer: A deformation distance of $0.60\ \text{m}$ and an added mass of $34\ \text{kg}$ When a question gives you multiple design constraints, your job is to check every condition for each option — not just one. Here, you need to verify three things: deformation ≤ 0.60 m, average force ≤ 250 kN, and added mass ≤ 35 kg. Start by finding the required stopping force using the work-energy theorem. The car's kinetic energy is: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(15)^2 = 135{,}000\ \text{J}$ Since the crumple zone does work equal to $F \times d$, the average force for a given deformation distance is: $F = \frac{KE}{d} = \frac{135{,}000}{d}$ Now evaluate each option. Option B uses $d = 0.60\ \text{m}$, giving $F = \frac{135{,}000}{0.60} = 225{,}000\ \text{N} = 225\ \text{kN}$, which is under 250 kN. Its added mass of 34 kg is under 35 kg. All three constraints are satisfied — B is correct. Option A uses $d = 0.45\ \text{m}$: $F = \frac{135{,}000}{0.45} = 300\ \text{kN}$, which exceeds the 250 kN force limit, so it fails despite the acceptable mass. Option C uses $d = 0.75\ \text{m}$, which immediately violates the deformation limit of 0.60 m — no further calculation needed. Option D has an added mass of 41 kg, which exceeds the 35 kg limit, and it fails before you even check the force. The strategy here: when multiple constraints exist, eliminate options the moment any single constraint is violated. Work through each condition systematically rather than assuming one "good" value means the whole design passes.
2. Which airbag design satisfies both criteria?
   Answer: An average force of $6.5\ \text{kN}$ after a deployment delay of $0.010\ \text{s}$ When a question gives you multiple design constraints, your job is to check every condition for each option — one failure disqualifies the design entirely. Here's the physics: during the deployment delay $t_d$, the passenger travels $d_1 = v \cdot t_d$ at constant speed. Once the airbag engages, it decelerates the passenger from $12\ \text{m/s}$ to rest. Using the work-energy theorem, the stopping distance is $d_2 = \frac{mv^2}{2F}$. Total travel is $d = d_1 + d_2$, and you need $F \leq 7.0\ \text{kN}$ and $d \leq 0.90\ \text{m}$. For option C ($F = 6.5\ \text{kN}$, $t_d = 0.010\ \text{s}$): $d_1 = 12 \times 0.010 = 0.12\ \text{m}$, and $d_2 = \frac{70 \times 144}{2 \times 6500} \approx 0.775\ \text{m}$. Total: $0.12 + 0.775 = 0.895\ \text{m} \leq 0.90\ \text{m}$ ✓, and $6.5\ \text{kN} \leq 7.0\ \text{kN}$ ✓. Both criteria satisfied — C is correct. Option A ($6.2\ \text{kN}$, $0.010\ \text{s}$): force is fine, but $d_2 = \frac{70 \times 144}{2 \times 6200} \approx 0.813\ \text{m}$, giving total $0.933\ \text{m}$ — exceeds the distance limit. Option B ($7.4\ \text{kN}$): immediately disqualified — force exceeds $7.0\ \text{kN}$. Option D ($7.0\ \text{kN}$, $0.020\ \text{s}$): force is borderline acceptable, but $d_1 = 0.24\ \text{m}$ and $d_2 \approx 0.720\ \text{m}$, totaling $0.96\ \text{m}$ — over the limit. Your strategy: in multi-constraint problems, eliminate options the moment one condition fails. Check the force limit first (it's faster), then calculate travel distance only for surviving candidates.
3. Which barrier design satisfies the acceleration, tipping, and contact-time criteria?
   Answer: A contact time of $0.08\ \text{s}$ with the force acting at a height of $0.10\ \text{m}$ When a question asks you to evaluate multiple engineering constraints simultaneously, your best approach is to check each criterion independently and eliminate any option that fails even one. Three criteria must all be satisfied: (1) lateral acceleration $\leq 8g = 80\ \text{m/s}^2$, (2) collision torque $\leq$ restoring torque $mg(0.80\ \text{m}) = 2000(10)(0.80) = 16000\ \text{N}\cdot\text{m}$, and (3) contact time $\leq 0.10\ \text{s}$. For any design, the average force is $F = \frac{\Delta p}{\Delta t} = \frac{12000}{\Delta t}$, the acceleration is $a = \frac{F}{m} = \frac{12000}{2000\,\Delta t} = \frac{6}{\Delta t}$, and the collision torque is $\tau = F \times h$. Option C ($\Delta t = 0.08\ \text{s}$, $h = 0.10\ \text{m}$): $F = 150000\ \text{N}$, $a = 75\ \text{m/s}^2 = 7.5g\ ✓$, $\tau = 150000(0.10) = 15000\ \text{N}\cdot\text{m} < 16000\ ✓$, and $0.08 \leq 0.10\ ✓$. All three criteria pass — confirming C is correct. Option A ($\Delta t = 0.10\ \text{s}$, $h = 0.15\ \text{m}$): $F = 120000\ \text{N}$, $\tau = 18000\ \text{N}\cdot\text{m} > 16000$ — fails tipping criterion. Option B ($\Delta t = 0.06\ \text{s}$, $h = 0.07\ \text{m}$): $F = 200000\ \text{N}$, $a = 100\ \text{m/s}^2 = 10g > 8g$ — fails acceleration criterion. Option D ($\Delta t = 0.12\ \text{s}$): contact time exceeds 0.10 s — fails immediately without further calculation. Your strategy: always knock out options with the quickest check first. Contact time is trivially checked (eliminating D instantly), then acceleration, then torque. Organize multi-constraint problems as a checklist, and the moment one criterion fails, move on.
4. Which guardrail test satisfies all the design criteria?
   Answer: Contact for $0.050\ \text{s}$, deflection of $0.75\ \text{m}$, and outward speed of $3.0\ \text{m/s}$ When a question gives you multiple design constraints, your job is to check every criterion — a design fails if it violates even one. Here, you must verify three things for each option: average force ≤ 450 kN, deflection ≤ 0.80 m, and outward speed ≤ 4.0 m/s. Use the impulse-momentum theorem to find the average force. The car's momentum changes from $p_i = 1500 \times 12 = 18{,}000\ \text{N}\cdot\text{s}$ (toward rail) to $p_f = 1500 \times v_{out}$ (away from rail), so the total impulse magnitude is $F_{avg} \cdot \Delta t = 1500(12 + v_{out})$. Option D gives contact time 0.050 s, deflection 0.75 m, outward speed 3.0 m/s. Force: $F = \frac{1500(12+3.0)}{0.050} = \frac{22{,}500}{0.050} = 450{,}000\ \text{N} = 450\ \text{kN}$. Deflection 0.75 m ≤ 0.80 m ✓, outward speed 3.0 m/s ≤ 4.0 m/s ✓. All three criteria are satisfied — D is correct. Option A fails on deflection: 0.90 m exceeds the 0.80 m limit, regardless of the other values. Option B fails on outward speed: 5.0 m/s exceeds the 4.0 m/s limit. Also, force = $\frac{1500(17)}{0.050} = 510\ \text{kN}$, exceeding 450 kN. Option C has acceptable deflection and outward speed, but force = $\frac{1500(15)}{0.040} = 562.5\ \text{kN}$, well above the 450 kN limit. A useful strategy: always check the easiest criteria (deflection, outward speed) first by simple comparison — you can often eliminate choices before doing any calculation.
5. Which cushioning design satisfies all three constraints?
   Answer: Compression of $0.025\ \text{m}$ with a rebound speed equal to $0.20$ times the impact speed When a question gives you multiple constraints, your job is to check every option against every limit — one failure disqualifies a design. Here, you have three constraints: compression ≤ 0.040 m, average force ≤ 1,900 N, and rebound speed ≤ 1.5 m/s. Start with the given values: mass = 2.0 kg, impact speed = 6.0 m/s, so kinetic energy at impact is $K = \frac{1}{2}mv^2 = \frac{1}{2}(2.0)(6.0)^2 = 36\ \text{J}$. The average cushion force formula is $F = \frac{K}{d} + mg = \frac{36}{d} + 20$. For option C (compression $d = 0.025\ \text{m}$, rebound = $0.20 \times 6.0 = 1.2\ \text{m/s}$): Force = $\frac{36}{0.025} + 20 = 1440 + 20 = 1460\ \text{N}$. Compression: 0.025 m ≤ 0.040 m ✓. Force: 1,460 N ≤ 1,900 N ✓. Rebound: 1.2 m/s ≤ 1.5 m/s ✓. All three constraints satisfied — C is correct. Option A ($d = 0.020\ \text{m}$, rebound = $0.30 \times 6.0 = 1.8\ \text{m/s}$): Rebound speed 1.8 m/s exceeds the 1.5 m/s limit. Eliminated. Option B ($d = 0.018\ \text{m}$): Force = $\frac{36}{0.018} + 20 = 2000 + 20 = 2020\ \text{N}$, which exceeds 1,900 N. Eliminated. Option D ($d = 0.040\ \text{m}$, rebound = $0.40 \times 6.0 = 2.4\ \text{m/s}$): Rebound speed 2.4 m/s far exceeds 1.5 m/s. Eliminated. Your strategy: in multi-constraint problems, build a quick checklist and test each condition systematically. Smaller compression sounds safer but actually increases force — don't assume "less compression = better design."
6. Which echo-time threshold best satisfies both design requirements?
   Answer: Trigger when the echo time decreases to $0.195\ \text{s}$ When a system like this must stop a moving car before an obstacle, you need to account for every source of delay and distance: the processing lag, the braking distance, and the required safety buffer. The echo time itself tells you the distance to the obstacle at the moment of detection — but the car keeps moving during processing, and then needs additional distance to brake to a stop. Here's how to find the minimum trigger distance. During the 0.20 s processing delay, the car travels $d_1 = 18 \times 0.20 = 3.6\ \text{m}$. The braking distance from $v = 18\ \text{m/s}$ with $a = 6.0\ \text{m/s}^2$ is $d_2 = \frac{v^2}{2a} = \frac{324}{12} = 27\ \text{m}$. Adding the 2.0 m safety margin gives a minimum trigger distance of $3.6 + 27 + 2.0 = 32.6\ \text{m}$. The corresponding echo time is $t = \frac{2 \times 32.6}{340} \approx 0.192\ \text{s}$. Now apply both constraints: the system must trigger at a distance ≥ 32.6 m (safety) but ≤ 35 m (no false activation, echo time ≤ $\frac{2 \times 35}{340} \approx 0.206\ \text{s}$). Option B (0.195 s) corresponds to about 33.2 m — safely inside both bounds. Option A (0.180 s ≈ 30.6 m) triggers too late; the car won't stop in time. Option D (0.190 s ≈ 32.3 m) is just below the minimum safe distance, cutting it too close. Option C (0.210 s ≈ 35.7 m) exceeds the 35 m false-activation limit. Study tip: On multi-constraint problems, always solve for the valid range first, then check which answer falls inside it — eliminate answers that violate either boundary.
7. Which test result represents an acceptable helmet design?
   Answer: A contact time of $0.010\ \text{s}$ and a rebound speed of $0.50\ \text{m/s}$ When a question involves impact forces and rebound speeds, you need to apply the impulse-momentum theorem: $F_{avg} = \frac{\Delta p}{\Delta t}$. The momentum change must account for direction — if the headform rebounds, its velocity reverses sign. Here, the headform has mass $m = 5.0\ \text{kg}$, initial speed $v_i = 6.0\ \text{m/s}$, and a design is acceptable only if $F_{avg} \leq 3500\ \text{N}$ and rebound speed $\leq 1.0\ \text{m/s}$. For choice A, the rebound speed is $0.50\ \text{m/s}$ (within the 1.0 m/s limit), so $\Delta p = 5.0(6.0 + 0.50) = 32.5\ \text{N}\cdot\text{s}$. The average force is $F = \frac{32.5}{0.010} = 3250\ \text{N} = 3.25\ \text{kN}$, which is below 3.5 kN. Both conditions are satisfied — this is the correct answer. Choice B fails the rebound condition immediately: $1.50\ \text{m/s} > 1.0\ \text{m/s}$. No further calculation needed. Choice C has no rebound, so $\Delta p = 5.0 \times 6.0 = 30\ \text{N}\cdot\text{s}$. The force is $\frac{30}{0.008} = 3750\ \text{N} = 3.75\ \text{kN}$, which exceeds the 3.5 kN limit. Choice D fails the rebound condition: $2.0\ \text{m/s} > 1.0\ \text{m/s}$. A useful strategy: always check the simpler condition (rebound speed) first — it eliminates B and D instantly, saving time. Then apply the impulse formula carefully, remembering that rebound means the velocities add when calculating $\Delta p$.
8. Which design meets both approval criteria?
   Answer: Add $40\ \text{kg}$ and absorb $300\ \text{kJ}$ in the front structure When a question gives you multiple approval criteria, your job is to check both conditions for each design — a choice that fails even one criterion is eliminated. Start by calculating the car's initial kinetic energy at its original mass, then apply it to each scenario. The base kinetic energy is: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1600)(20)^2 = 320\ \text{kJ}$ Each design adds mass, so the new total kinetic energy increases. For a design adding mass $\Delta m$: $KE_{new} = \frac{1}{2}(1600 + \Delta m)(20)^2$ The energy remaining for the passenger compartment equals the new total KE minus what the front structure absorbs. This remainder must be $\leq 40\ \text{kJ}$, and the added mass must be $\leq 80\ \text{kg}$. Choice D adds $40\ \text{kg}$: $KE_{new} = \frac{1}{2}(1640)(400) = 328\ \text{kJ}$. The structure absorbs $300\ \text{kJ}$, leaving $328 - 300 = 28\ \text{kJ}$ — under the 40 kJ limit. Added mass of 40 kg is under 80 kg. Both criteria pass. Choice A adds only 20 kg ($KE = 324\ \text{kJ}$), but absorbs only 280 kJ, leaving $44\ \text{kJ}$ — exceeding the 40 kJ limit. Fails the energy criterion. Choice B adds 90 kg, which immediately exceeds the 80 kg mass limit. Eliminated without further calculation. Choice C adds 60 kg ($KE = 332\ \text{kJ}$) but absorbs only 285 kJ, leaving $47\ \text{kJ}$. Fails the energy criterion. Strategy tip: On multi-criteria problems, use the easiest condition to screen first — here, mass limit eliminates B instantly. Then calculate energy for the survivors. This saves time and reduces arithmetic errors.
9. Which docking-coupler design satisfies both criteria?
   Answer: A coefficient of restitution of $0.02$ and a contact time of $0.11\ \text{s}$ When a question gives you two safety criteria, your job is to check both for every option — a design fails if it violates either one. Start with the separation speed criterion. The relative separation speed after impact is $v' = e \times v = e \times 0.60\ \text{m/s}$. For safety, $v' \leq 0.050\ \text{m/s}$, which requires $e \leq 0.083$. This immediately eliminates B ($e = 0.10$), since $0.10 \times 0.60 = 0.060\ \text{m/s} > 0.050\ \text{m/s}$. Now check the average force criterion using $F_{avg} = J/t$, where $J = \mu(1+e)v = 667(1+e)(0.60)$. For A ($e = 0.02,\ t = 0.11\ \text{s}$): $J = 667(1.02)(0.60) \approx 408\ \text{N}\cdot\text{s}$, so $F = 408/0.11 \approx 3710\ \text{N} < 4000\ \text{N}$. ✓ Both criteria satisfied. For C ($e = 0.05,\ t = 0.10\ \text{s}$): $J = 667(1.05)(0.60) \approx 420\ \text{N}\cdot\text{s}$, so $F = 420/0.10 = 4200\ \text{N} > 4000\ \text{N}$. The separation speed is fine ($0.030\ \text{m/s}$), but the force is too large. ✗ For D ($e = 0,\ t = 0.09\ \text{s}$): $J = 667(1)(0.60) = 400\ \text{N}\cdot\text{s}$, so $F = 400/0.09 \approx 4440\ \text{N} > 4000\ \text{N}$. ✗ A is the only design passing both tests. The key strategy here: when multiple constraints are given, treat each one as a filter and eliminate options that fail any single criterion — don't stop after the first check.

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