Evaluate Collision Design Solutions
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A helmet prototype spreads impact force over different contact areas to reduce pressure on the skull. Two designs produce the same average impact force $F_{\text{avg}}=3000,\text{N}$, but have different contact areas.
Design A: contact area $A=30,\text{cm}^2$
Design B: contact area $A=60,\text{cm}^2$
Design C: contact area $A=90,\text{cm}^2$
Safety criterion: keep average pressure below $P_{\max}=50,\text{kPa}$. Use $P=F/A$ and convert areas: $1,\text{cm}^2=1\times 10^{-4},\text{m}^2$.
Which design(s) meet the pressure limit?
Designs B and C, because doubling or tripling area cuts pressure and both fall below 50 kPa.
Design C only, because $P=3000/(90\times 10^{-4})\approx 3.3\times 10^5,\text{Pa}=330,\text{kPa}$ is below 50 kPa.
Design A only, because smaller area concentrates force and therefore lowers pressure.
None, because even Design C gives $P\approx 3000/0.009=3.3\times 10^5,\text{Pa}=330,\text{kPa}>50,\text{kPa}$.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = F d shows that kinetic energy KE = ½mv² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greater contact area of 90 cm² compared to Design A's 30 cm², which reduces the pressure by a factor of 3. Using P = F/A, the pressure in Design C is P = 3000/0.009 ≈ 333000 Pa = 333 kPa, which is above the safety threshold of 50 kPa, while Design A produces P = 3000/0.003 = 1000000 Pa = 1000 kPa which exceeds the limit. This demonstrates that greater area is critical for keeping pressure within safe ranges, but here even the largest area fails. Choice C is correct because it properly calculates pressures showing no design meets the safety threshold. Choice B selects designs with larger areas but ignores that both still exceed 50 kPa (500 kPa and 333 kPa >50). When evaluating collision protection designs: (1) identify momentum change Δp = mv or energy to absorb KE = ½mv², (2) for each design, determine collision time Δt or deformation distance d, (3) calculate F = Δp/Δt or F = KE/d for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
An automotive engineer compares restraint designs using impulse. A 60 kg passenger moves forward at 15 m/s relative to the car just before the restraint engages and is brought to rest. The maximum allowable average force on the passenger is 5000 N.
Design A: standard seatbelt, $\Delta t=0.080,\text{s}$
Design B: seatbelt with pretensioner + load limiter, $\Delta t=0.120,\text{s}$
Design C: seatbelt + airbag, $\Delta t=0.200,\text{s}$
Which design(s) keep $F_{\text{avg}}=\Delta p/\Delta t$ below 5000 N?
Designs B and C, because both have longer stopping times than A and therefore both must be below 5000 N.
Design A only, because a shorter stopping time reduces the impulse and therefore reduces force.
Design C only, because $F=(60\cdot 15)/0.200=4500,\text{N}<5000,\text{N}$, while A and B are above 5000 N.
All designs, because the passenger mass is moderate and the speed is only 15 m/s.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The impulse-momentum theorem F_avg = Δp/Δt shows that extending collision time Δt reduces average force F_avg for a given momentum change, which is the fundamental principle for comparing collision protection designs—solutions that provide longer collision times (through progressive deformation, controlled crumpling, or gradual compression) result in lower forces on occupants or contents. Design C provides longer collision time of 0.200 s compared to Design B's 0.120 s, which reduces the average force by a factor of about 1.67. Using F = Δp/Δt, the force in Design C is F = (60·15)/0.200 = 4500 N, which is below the safety threshold of 5000 N, while Design B produces F = (60·15)/0.120 = 7500 N which exceeds the limit. This demonstrates that longer time is critical for keeping forces within safe ranges. Choice A is correct because it correctly identifies the design with longest Δt, which produces forces below the threshold, while others exceed. Choice B selects designs with longer stopping times but ignores the calculation showing Design B exceeds 5000 N. When evaluating collision protection designs: (1) identify momentum change Δp = mv or energy to absorb KE = ½mv², (2) for each design, determine collision time Δt or deformation distance d, (3) calculate F = Δp/Δt or F = KE/d for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A 1200 kg car traveling at $v=25,\text{m/s}$ hits a rigid barrier and must absorb all its kinetic energy through front-end deformation. Three crumple-zone designs are possible, but the available front-end space limits deformation to at most $0.70,\text{m}$. Assume the average stopping force can be estimated by $F_{\text{avg}}\approx \text{KE}/d$.
- Design A: deformation distance $d=0.30,\text{m}$
- Design B: deformation distance $d=0.60,\text{m}$
- Design C: deformation distance $d=1.00,\text{m}$
Which design should be selected to minimize average stopping force while still meeting the space constraint?
Design A, because shorter deformation reduces the time and therefore reduces force.
Design B, because it uses the largest allowed deformation distance ($0.60,\text{m}$) under the $0.70,\text{m}$ limit.
Design C, because the largest deformation always minimizes force and space limits do not matter.
All designs are equivalent because they absorb the same kinetic energy.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = Fd shows that kinetic energy KE = ½mv² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation absorb the same energy with smaller forces. Design B provides deformation distance of 0.60 m which is within the 0.70 m space constraint, while Design C's 1.00 m exceeds the limit and Design A's 0.30 m underutilizes available space. Using F = KE/d where KE = 0.5(1200)(25²) = 375,000 J, the force in Design A is F = 375,000/0.30 = 1,250,000 N, Design B gives F = 375,000/0.60 = 625,000 N, and Design C would give F = 375,000/1.00 = 375,000 N but violates the space constraint. This demonstrates that using maximum allowable deformation distance minimizes force. Choice B is correct because it uses the largest allowed deformation distance (0.60 m) under the 0.70 m limit, thereby minimizing force while meeting the constraint. Choice C ignores the stated space constraint, selecting based on minimum force alone when the design must fit within 0.70 m, making Design C infeasible despite its lower force. When evaluating collision protection designs: (1) identify energy to absorb KE = ½mv², (2) determine maximum allowable deformation from constraints, (3) calculate F = KE/d for feasible designs, (4) select design with lowest force that meets all constraints. Remember: maximize deformation distance within constraints to minimize force.
A $m=2.0\ \text{kg}$ laptop in a shipping box is dropped and hits the ground at $v=4.0\ \text{m/s}$. The packaging must keep the average stopping force below $F_{\max}=300\ \text{N}$.
Design A: styrofoam corners that compress $d=0.020\ \text{m}$
Design B: suspension straps that allow $d=0.060\ \text{m}$
Design C: thin cardboard inserts, $d=0.010\ \text{m}$
Assume constant average force during stopping, so $F_{\text{avg}}\approx \frac{\tfrac12 mv^2}{d}$. Which design(s) meet the force limit?
Designs A and B, because both have $d\ge 0.02\ \text{m}$ so both give $F<300\ \text{N}$.
Only Design A, because it is stiffer and therefore reduces the force on the laptop.
Only Design B, because $\text{KE}=\tfrac12(2)(4^2)=16\ \text{J}$ and $F_B\approx 16/0.06\approx 267\ \text{N}<300\ \text{N}$, while A and C exceed 300 N.
All three, because the force depends only on mass and speed, not on stopping distance.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = F d shows that kinetic energy KE = ½ m v² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design B provides greater deformation distance of 0.060 m compared to Design A's 0.020 m, which reduces the peak force by a factor of 3; using F = KE / d, the force in Design B is F = 16 / 0.060 ≈ 267 N, which is below the safety threshold of 300 N, while Design A produces F = 16 / 0.020 = 800 N which exceeds the limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice A is correct because it correctly identifies the design with greatest d that produces lowest forces below the threshold and notes that A and C exceed. Choice D ignores the stated force criterion, selecting based on irrelevant factor that force depends only on mass and speed, when the primary goal is to keep forces below threshold via d. When evaluating collision protection designs: (1) identify energy to absorb KE = ½ m v², (2) for each design, determine deformation distance d, (3) calculate F = KE / d for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
Two car interior designs use different combinations of stopping time and contact area to reduce chest injury risk in a crash. An occupant of mass $m=70\ \text{kg}$ goes from $v_i=10\ \text{m/s}$ to rest.
Design A: airbag deploys early giving $\Delta t=0.14\ \text{s}$ and contact area $A=0.18\ \text{m}^2$, cost $\$900
Design B: airbag deploys later giving $\Delta t=0.08\ \text{s}$ and $A=0.25\ \text{m}^2$, cost $\$600
Design C: no airbag, padded steering wheel gives $\Delta t=0.05\ \text{s}$ and $A=0.10\ \text{m}^2$, cost $\$200
Safety limits: $F_{\text{avg}}<5000\ \text{N}$ and $P=F/A<30\ \text{kPa}$. Budget cap: $\$700. Using $F_{\text{avg}}=\Delta p/\Delta t$ with $\Delta p = m(0-v_i)$, which design best satisfies all requirements?
Design B, because $F\approx \frac{70\cdot 10}{0.08}=8750\ \text{N}$ which is below 5000 N when spread over $0.25\ \text{m}^2$, and it meets the budget.
Design A, because its longer stopping time gives the lowest force and it is under the $\$700 budget.
Design C, because it is cheapest and pressure is what matters most, not force or time.
None, because A fails the budget ($\$900>$\$700), B fails the force limit ($8750\ \text{N}>5000\ \text{N}$), and C fails both force and pressure limits.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The impulse-momentum theorem F_avg = Δp/Δt shows that extending collision time Δt reduces average force F_avg for a given momentum change, which is the fundamental principle for comparing collision protection designs—solutions that provide longer collision times (through progressive deformation, controlled crumpling, or gradual compression) result in lower forces on occupants or contents. Design A provides longer collision time of 0.14 s compared to Design B's 0.08 s, but F_A = 700 / 0.14 = 5,000 N = limit but cost $900 > $700 budget, F_B = 700 / 0.08 = 8,750 N > 5,000 N and P = 8,750 / 0.25 = 35,000 Pa > 30 kPa, while Design C gives F = 700 / 0.05 = 14,000 N > 5,000 N and P > 30 kPa. This demonstrates that longer time is critical for keeping forces within safe ranges, but none meet all here. Choice D is correct because it correctly identifies that no design meets all requirements, noting specific failures in budget, force, and pressure. Choice B makes a calculation error by claiming F ≈ 8,750 N is below 5,000 N when spread over area, but force exceeds regardless of area (pressure is separate). When evaluating collision protection designs: (1) identify momentum change Δp = m v, (2) for each design, determine collision time Δt, (3) calculate F = Δp/Δt for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A $m=10\ \text{kg}$ package is dropped from $h=1.2\ \text{m}$ (ignore air resistance). It hits the ground at speed $v=\sqrt{2gh}$ and must be cushioned so the average impact force on the contents stays below $F_{\max}=400\ \text{N}$. The package must fit in a shipping box that allows at most $d=0.08\ \text{m}$ of compression distance.
Three cushioning designs are proposed:
Design A: bubble wrap, compression distance $d=0.03\ \text{m}$, cost $\$2
Design B: foam insert, $d=0.06\ \text{m}$, cost $\$5
Design C: air-pillows, $d=0.08\ \text{m}$, cost $\$9
Assume the cushion provides an approximately constant average force while stopping the package, so $F_{\text{avg}}\approx \frac{\text{KE}}{d}$ with $\text{KE}=\tfrac12 mv^2 = mgh$. Budget limit: $\$6. Which design best satisfies both the force limit and the budget?
Design A, because smaller compression distance lowers the stopping force ($F=\text{KE}/d$).
Design B, because $\text{KE}=mgh\approx 10\cdot 9.8\cdot 1.2\approx 118\ \text{J}$ so $F\approx 118/0.06\approx 1960\ \text{N}$, which is below 400 N and within budget.
Design C, because $F\approx 118/0.08\approx 1475\ \text{N}$ and it is the only design under $400\ \text{N}$.
None of the designs, because even Design C gives $F\approx 118/0.08\approx 1.5\times 10^3\ \text{N}>400\ \text{N}$ (and C also exceeds the $\$6 budget).
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = F d shows that kinetic energy KE = ½ m v² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greater deformation distance of 0.08 m compared to Design B's 0.06 m, but calculations show F_C ≈ 118 / 0.08 ≈ 1,475 N > 400 N threshold, F_B ≈ 118 / 0.06 ≈ 1,967 N > 400 N, and F_A ≈ 118 / 0.03 ≈ 3,933 N > 400 N, with Design C also exceeding the $6 budget at $9. This demonstrates that greater distance is critical for keeping forces within safe ranges, but none achieve it here while meeting budget. Choice D is correct because it correctly identifies that no design meets both criteria, with accurate force calculations showing all exceed the limit and noting the budget violation for C. Choice C makes a calculation error by claiming F ≈ 118 / 0.08 ≈ 1,475 N is under 400 N, leading to incorrect force estimate that doesn't reflect actual performance. When evaluating collision protection designs: (1) identify energy to absorb KE = m g h, (2) for each design, determine deformation distance d, (3) calculate F = KE / d for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A car plus occupant (total mass $m = 1500\ \text{kg}$) crashes head-on into a rigid barrier at $v_i = 20\ \text{m/s}$ and comes to rest. Engineers are comparing three front-end safety designs that change how long the car takes to stop. Safety requirement: keep the average stopping force on the occupants below $F_{\max}=500{,}000\ \text{N}$ (use $F_{\text{avg}} = \Delta p/\Delta t$ with $\Delta p = m(0-v_i)$). Space constraint: maximum allowable crumple length is $0.7\ \text{m}$.
Design A: rigid front end, stopping time $\Delta t = 0.05\ \text{s}$, crumple distance $d=0.10\ \text{m}$
Design B: moderate crumple zone, $\Delta t = 0.12\ \text{s}$, $d=0.50\ \text{m}$
Design C: long crumple zone, $\Delta t = 0.20\ \text{s}$, $d=0.90\ \text{m}$
Which design should be selected to meet both the force limit and the space constraint?
Design A, because the shortest stopping time reduces the force on the occupants.
Design C, because it gives the lowest force and also fits the 0.7 m space limit.
Design A, because $F_{\text{avg}}=\frac{1500\cdot 20}{0.05}=6.0\times 10^4\ \text{N}$ which is below the limit.
Design B, because $F_{\text{avg}}=\frac{1500\cdot 20}{0.12}\approx 2.5\times 10^5\ \text{N}$ and $d=0.50\ \text{m}$ is within the 0.7 m space limit.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The impulse-momentum theorem F_avg = Δp/Δt shows that extending collision time Δt reduces average force F_avg for a given momentum change, which is the fundamental principle for comparing collision protection designs—solutions that provide longer collision times (through progressive deformation, controlled crumpling, or gradual compression) result in lower forces on occupants or contents. Design B provides longer collision time of 0.12 s compared to Design A's 0.05 s, which reduces the average force by a factor of about 2.4; using F = Δp/Δt, the force in Design B is F = 30,000 / 0.12 ≈ 250,000 N, which is below the safety threshold of 500,000 N, while Design C produces F = 30,000 / 0.20 = 150,000 N but exceeds the 0.7 m space limit with d=0.90 m. This demonstrates that longer time is critical for keeping forces within safe ranges, but constraints like space must also be met. Choice B is correct because it correctly identifies the design with an appropriate Δt that produces forces below the limit while fitting the space constraint and provides the proper calculation. Choice A selects the design with shorter collision time, which actually produces higher forces (F = Δp/Δt means smaller Δt gives larger F), making it less effective for protection despite the claimed advantage of shortest time. When evaluating collision protection designs: (1) identify momentum change Δp = m v, (2) for each design, determine collision time Δt, (3) calculate F = Δp/Δt for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A vehicle designer can either increase crumple distance or increase contact area with an airbag. In a $m=75\ \text{kg}$ occupant test, the occupant’s forward speed relative to the car is reduced from $v_i=12\ \text{m/s}$ to $0$.
Two proposed restraint designs:
Design A (short stop, large area): stopping time $\Delta t=0.06\ \text{s}$, contact area $A=0.20\ \text{m}^2$
Design B (longer stop, smaller area): stopping time $\Delta t=0.12\ \text{s}$, contact area $A=0.10\ \text{m}^2$
Assume average force $F_{\text{avg}}=\Delta p/\Delta t$ and average pressure on the chest $P=F/A$. Chest safety limits: $F_{\text{avg}}<5000\ \text{N}$ and $P<40\ \text{kPa}$. Which design meets both limits?
Design A, because larger area always reduces force and pressure.
Neither design, because Design A has $F=\frac{75\cdot 12}{0.06}=15000\ \text{N}$ and Design B has $F=\frac{75\cdot 12}{0.12}=7500\ \text{N}$, so both exceed 5000 N (even though pressures differ).
Design B, because longer time reduces force enough and pressure stays below 40 kPa.
Both designs, because they have the same impulse $\Delta p$ so both have the same force.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The impulse-momentum theorem F_avg = Δp/Δt shows that extending collision time Δt reduces average force F_avg for a given momentum change, which is the fundamental principle for comparing collision protection designs—solutions that provide longer collision times (through progressive deformation, controlled crumpling, or gradual compression) result in lower forces on occupants or contents. Design B provides longer collision time of 0.12 s compared to Design A's 0.06 s, which reduces the average force by a factor of 2; using F = Δp/Δt, the force in Design B is F = 900 / 0.12 = 7,500 N, which is above the safety threshold of 5,000 N and P = 7,500 / 0.10 = 75,000 Pa > 40 kPa, while Design A produces F = 900 / 0.06 = 15,000 N > 5,000 N and P = 15,000 / 0.20 = 75,000 Pa > 40 kPa. This demonstrates that longer time is critical for keeping forces within safe ranges, but neither meets both limits here. Choice D is correct because it correctly identifies that neither design meets both force and pressure limits, with proper calculations. Choice C ignores the stated force and pressure criteria, claiming both have the same impulse so same force, when actually different Δt lead to different forces. When evaluating collision protection designs: (1) identify momentum change Δp = m v, (2) for each design, determine collision time Δt, (3) calculate F = Δp/Δt for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A new package design must protect a fragile instrument (mass $m=5\ \text{kg}$) from a drop that produces an impact speed of $v=5\ \text{m/s}$. The instrument can tolerate at most $F_{\max}=600\ \text{N}$ average stopping force. The shipping department also requires the total added packaging mass to be $\le 1.5\ \text{kg}$.
Design A: dense foam (compression $d=0.04\ \text{m}$), packaging mass $1.2\ \text{kg}$
Design B: lighter foam (compression $d=0.08\ \text{m}$), packaging mass $1.8\ \text{kg}$
Design C: mixed foam + air pockets (compression $d=0.10\ \text{m}$), packaging mass $1.4\ \text{kg}$
Assume constant average stopping force so $F\approx \frac{\tfrac12 mv^2}{d}$. Which design meets both the force and mass constraints?
Design B, because it has a larger stopping distance and thus lower force, and its mass is within the limit.
Design C, because $\text{KE}=\tfrac12(5)(5^2)=62.5\ \text{J}$ so $F\approx 62.5/0.10=625\ \text{N}<600\ \text{N}$ and mass is 1.4 kg.
Design A, because it is under the 1.5 kg mass limit and stiffer foam reduces the force.
None, because A gives $F\approx 62.5/0.04=1563\ \text{N}$, C gives $F\approx 625\ \text{N}$, and B violates the 1.5 kg packaging-mass limit.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = F d shows that kinetic energy KE = ½ m v² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greater deformation distance of 0.10 m compared to Design A's 0.04 m, but calculations show F_C = 62.5 / 0.10 = 625 N > 600 N threshold despite mass 1.4 kg <=1.5 kg, F_B = 62.5 / 0.08 = 781 N > 600 N with mass 1.8 kg >1.5 kg, and F_A = 62.5 / 0.04 = 1,563 N > 600 N. This demonstrates that greater distance is critical for keeping forces within safe ranges, but none achieve it here while meeting mass limit. Choice D is correct because it correctly identifies that no design meets both criteria, with accurate force calculations and noting mass violation for B. Choice C makes a calculation error by claiming F ≈ 62.5 / 0.10 = 625 N < 600 N, leading to incorrect force estimate that doesn't reflect actual performance. When evaluating collision protection designs: (1) identify energy to absorb KE = ½ m v², (2) for each design, determine deformation distance d, (3) calculate F = KE / d for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).
A cyclist’s helmet must limit head deceleration to below $100g$ during a crash. A test headform of mass $m=5.0\ \text{kg}$ hits a rigid surface at $v_i=6.0\ \text{m/s}$ and is brought to rest by helmet padding. Assume approximately constant deceleration while the padding compresses a distance $d$, so $v^2 = 2ad$ and $a=\frac{v^2}{2d}$.
Padding options:
Design A: soft foam, $d=0.010\ \text{m}$
Design B: multi-density foam, $d=0.030\ \text{m}$
Design C: air-pocket liner, $d=0.050\ \text{m}$
Which design(s) meet the deceleration requirement ($a<100g \approx 980\ \text{m/s}^2$)?
Only Design C meets the requirement, since $a=\frac{6^2}{2(0.05)}=360\ \text{m/s}^2<980\ \text{m/s}^2$ while A and B exceed the limit.
Only Design A meets the requirement because it stops the head fastest.
Designs B and C meet the requirement, since $a_B=\frac{36}{2(0.03)}=600\ \text{m/s}^2$ and $a_C=360\ \text{m/s}^2$, both below $980\ \text{m/s}^2$.
All three meet the requirement because $a$ depends on mass, and the mass is only 5 kg.
Explanation
This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship W = F d shows that kinetic energy KE = ½ m v² must be absorbed through work during collision, so for fixed energy to absorb, increasing deformation distance d reduces peak force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greater deformation distance of 0.050 m compared to Design A's 0.010 m, which reduces the acceleration by a factor of 5; using a = v² / (2 d), the acceleration in Design C is a = 36 / 0.1 = 360 m/s², which is below the safety threshold of 980 m/s², while Design A produces a = 36 / 0.02 = 1,800 m/s² which exceeds the limit, and Design B gives a = 36 / 0.06 = 600 m/s² below the limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice C is correct because it correctly identifies the designs with greatest d that produce lowest accelerations below the threshold. Choice B makes a calculation error by claiming Design B exceeds the limit (600 m/s² is actually below 980 m/s²), leading to incorrect assessment that only C meets the requirement. When evaluating collision protection designs: (1) identify energy to absorb KE = ½ m v², (2) for each design, determine deformation distance d, (3) calculate F = KE / d or a = v² / (2 d) for each design, (4) compare to safety threshold, and (5) consider constraints (cost, weight, space). Designs that extend time and increase deformation distance provide the best protection. Remember: stiffer = shorter time = higher forces (worse), deformable = longer time = lower forces (better).