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Physics Quiz

Physics Quiz: Evaluate Collision Design Solutions

Practice Evaluate Collision Design Solutions in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

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?

What this quiz covers

This quiz focuses on Evaluate Collision Design Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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?

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.
Designs B and C, because doubling or tripling area cuts pressure and both fall below 50 kPa.
None, because even Design C gives $P\approx 3000/0.009=3.3\times 10^5\,\text{Pa}=330\,\text{kPa}>50\,\text{kPa}$ . (correct answer)
Design A only, because smaller area concentrates force and therefore lowers pressure.

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).

Question 2

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?

Design C only, because $F=(60\cdot 15)/0.200=4500\,\text{N}<5000\,\text{N}$ , while A and B are above 5000 N. (correct answer)
Designs B and C, because both have longer stopping times than A and therefore both must be below 5000 N.
All designs, because the passenger mass is moderate and the speed is only 15 m/s.
Design A only, because a shorter stopping time reduces the impulse and therefore reduces 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 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).

Question 3

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?

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. (correct answer)
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.
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).

Question 4

A car’s steering wheel airbag is intended to reduce the average force on a driver’s chest. During a crash, the driver’s upper body (effective mass 40 kg) moves forward at 10 m/s relative to the car and must be brought to rest by the restraint system. Two restraint designs are compared:

Design A: no airbag (seat belt only), stopping time $\Delta t=0.040\,\text{s}$ , contact area on chest $A=0.020\,\text{m}^2$
Design B: airbag, stopping time $\Delta t=0.120\,\text{s}$ , contact area on chest $A=0.080\,\text{m}^2$

Use $F_{\text{avg}}=\frac{m\Delta v}{\Delta t}$ and pressure $P=\frac{F}{A}$ . Criteria: $F_{\text{avg}}<5000\,\text{N}$ and average pressure on the chest $P<100\,\text{kPa}$ . Which statement is correct?

Design A meets both limits because it has the smaller contact area, which lowers pressure.
Design B meets both limits because it increases $\Delta t$ (reducing force) and increases $A$ (reducing pressure). (correct answer)
Design B fails the force limit because a longer $\Delta t$ increases impulse and therefore increases force.
Design A meets the force limit since $F_{\text{avg}}=\frac{(40)(10)}{0.040}=1000\,\text{N}<5000\,\text{N}$ .

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 0.120 s compared to Design A's 0.040 s, which reduces the average force by a factor of 3. Using F = Δp/Δt, the force in Design B is F = (40×10)/0.120 ≈3333 N, which is below the safety threshold of 5000 N with P ≈41.7 kPa below 100 kPa, while Design A produces F = (40×10)/0.040 = 10000 N which exceeds the force limit and P=500 kPa exceeds pressure limit. This demonstrates that longer time is critical for keeping forces within safe ranges. Choice B is correct because it accurately explains the physics principle: extending time or distance reduces force and properly evaluates both force and pressure meeting thresholds. Choice D makes a calculation error by stating F=1000 N for Design A, when it's actually 10000 N, leading to incorrect force estimate that doesn't reflect actual performance. When evaluating collision protection designs: (1) identify momentum change Δp = mΔv, (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 (pressure). 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).

Question 5

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.
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.
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). (correct answer)

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).

Question 6

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 A, because its longer stopping time gives the lowest force and it is under the $\$ 700 budget.
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 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. (correct answer)

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).

Question 7

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). (correct answer)

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).

Question 8

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 A, because it is under the 1.5 kg mass limit and stiffer foam reduces the force.
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.
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. (correct answer)

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).

Question 9

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 A meets the requirement because it stops the head fastest.
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.
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 $980\ \text{m/s}^2$ . (correct answer)
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).

Question 10

An automotive team must redesign a bumper system for a low-speed collision test. A $m=1200\ \text{kg}$ car hits a barrier at $v=8.0\ \text{m/s}$ and stops. The bumper system must keep the average collision force below $F_{\max}=150{,}000\ \text{N}$ , and it must fit within a maximum crush distance of $d_{\max}=0.40\ \text{m}$ .

Proposed designs:

Design A: crush distance $d=0.20\ \text{m}$ , measured stopping time $\Delta t=0.05\ \text{s}$ Design B: $d=0.35\ \text{m}$ , $\Delta t=0.10\ \text{s}$ Design C: $d=0.45\ \text{m}$ , $\Delta t=0.14\ \text{s}$

Using impulse-momentum ( $F_{\text{avg}}=\Delta p/\Delta t$ ) and the distance constraint, which design meets the requirements?

Design A, because $d$ is smallest so it fits best and therefore reduces force.
Design B, because $F\approx \frac{1200\cdot 8}{0.10}=96{,}000\ \text{N}<150{,}000\ \text{N}$ and $d=0.35\ \text{m}\le 0.40\ \text{m}$ . (correct answer)
Design C, because it has the longest stopping time and thus the smallest force, and it also fits within $0.40\ \text{m}$ .
None, because $F_{\text{avg}}$ depends on crush distance, not on stopping time.

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 0.10 s compared to Design A's 0.05 s, which reduces the average force by a factor of 2; using F = Δp/Δt, the force in Design B is F = 9,600 / 0.10 = 96,000 N, which is below the safety threshold of 150,000 N, while Design C produces F = 9,600 / 0.14 ≈ 68,571 N but exceeds the 0.40 m space limit with d=0.45 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 appropriate Δt that produces forces below the limit while fitting the space constraint. Choice C selects the design with longer collision time but ignores the stated space criterion, selecting based on irrelevant factor of lowest force without checking d <= 0.40 m. 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).

Question 11

A bicycle helmet liner is tested with the same headform mass (5.0 kg) and the same initial speed (7.0 m/s) in each trial. The only major difference is the liner compression distance $d$ before the headform stops. Assume constant average force so $F_{\text{avg}}=\frac{\tfrac12 mv^2}{d}$ .

Design A: $d=0.010\,\text{m}$ (1 cm)
Design B: $d=0.030\,\text{m}$ (3 cm)
Design C: $d=0.050\,\text{m}$ (5 cm)

Safety target: $F_{\text{avg}}<2500\,\text{N}$ . Which design(s) meet the target?

Only Design A.
Only Design B.
Only Design C. (correct answer)
Designs B and C.

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 average force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greatest deformation distance of 0.050 m compared to Design B's 0.030 m, which reduces the average force by a factor of about 1.67. Using W = Fd, the force in Design C is F = (0.5×5.0×7.0²)/0.050 = 2450 N, which is below the safety threshold of 2500 N, while Design B produces F = (0.5×5.0×7.0²)/0.030 ≈4083 N which exceeds the limit; Design A has F ≈12250 N exceeding the limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice C is correct because it correctly identifies the design with greatest d that produces lowest forces and meets the safety target. Choice A selects the design with smaller deformation distance, which actually produces higher forces (F = KE/d means smaller d gives larger F), making it less effective for protection. When evaluating collision protection designs: (1) identify energy to absorb KE = ½mv², (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. 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).

Question 12

A bicycle helmet must be lightweight while keeping average impact force low. In a standardized test, the effective head mass is 4.0 kg and impact speed is 5.0 m/s. Approximate $F_{\text{avg}}\approx \text{KE}/d$ where $\text{KE}=\tfrac12 mv^2$ and $d$ is the liner compression distance.

Design A: $d=0.015\,\text{m}$ , mass of helmet 0.30 kg Design B: $d=0.030\,\text{m}$ , mass of helmet 0.45 kg Design C: $d=0.050\,\text{m}$ , mass of helmet 0.70 kg

Requirements: $F_{\text{avg}}<1000\,\text{N}$ and helmet mass $\le 0.50\,\text{kg}$ .

Which design meets BOTH requirements?

Design A, because it is lightest and therefore reduces impact force the most.
Design B, because $\text{KE}=\tfrac12(4)(25)=50\,\text{J}$ and $F\approx 50/0.030\approx 1.67\times 10^3\,\text{N}<1000\,\text{N}$ with mass under 0.50 kg.
Design C, because $F\approx 50/0.050=1000\,\text{N}$ but it fails the mass limit.
None, because A and B exceed 1000 N and C exceeds the mass limit (and is not below 1000 N). (correct answer)

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 deformation distance of 0.050 m compared to Design B's 0.030 m, which reduces the peak force by a factor of about 1.67. Using F = KE/d, the force in Design C is F = 50/0.050 = 1000 N, which is not below the safety threshold of 1000 N, while Design B produces F = 50/0.030 ≈ 1667 N which exceeds the limit, and C exceeds mass limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice D is correct because it properly calculates forces showing no design meets both the force and mass thresholds. Choice C ignores the stated mass criterion, selecting based on force, when the primary goal is to minimize weight while meeting force limit, and C does not strictly meet <1000 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).

Question 13

A playground must be designed so that a child’s head deceleration stays below $100\,g$ in a fall. Model the child as falling from height $h=2.0\,\text{m}$ and coming to rest over a stopping distance equal to the surface compression depth $d$ . Ignore air resistance. Candidate surfaces:

Design A: rubber mat, $d=0.15\,\text{m}$
Design B: wood chips, $d=0.30\,\text{m}$
Design C: sand, $d=0.45\,\text{m}$

Use $v=\sqrt{2gh}$ to find impact speed and then $a\approx \dfrac{v^2}{2d}$ to estimate average deceleration. Which surfaces meet the $100\,g$ requirement?

Only Design A.
Designs B and C only.
Only Design C.
All three designs. (correct answer)

Explanation: This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. For a fall from height h, impact velocity is v = √(2gh), and the work-energy relationship shows that deceleration a = v²/(2d), so increasing deformation distance d reduces deceleration—surfaces allowing greater compression absorb the same energy with smaller accelerations. Design C provides greatest compression of 0.45 m compared to Design B's 0.30 m and Design A's 0.15 m. Using v = √(2×9.8×2.0) = 6.26 m/s, then a = v²/(2d), the deceleration in Design A is a = 39.2/(2×0.15) = 131 m/s² = 13.4g which is below 100g, Design B gives a = 39.2/(2×0.30) = 65 m/s² = 6.6g, and Design C produces a = 39.2/(2×0.45) = 44 m/s² = 4.5g. This demonstrates that all three surfaces provide adequate protection for this fall height. Choice D is correct because it correctly identifies that all three designs meet the 100g requirement for a 2.0 m fall. Choice C incorrectly selects only Design C, failing to recognize that even the thinnest padding (Design A) provides sufficient deceleration distance to keep forces below the safety threshold for this specific fall height. When evaluating collision protection designs: (1) calculate impact velocity from fall height v = √(2gh), (2) determine deceleration a = v²/(2d) for each surface, (3) convert to g's by dividing by 9.8 m/s², (4) compare to safety threshold. Even modest padding can be sufficient for lower fall heights.

Question 14

A 75 kg occupant in a crash goes from $20\,\text{m/s}$ to $0\,\text{m/s}$ . Two airbag designs produce the same stopping time $\Delta t=0.15\,\text{s}$ but different contact areas on the chest:

Design A: contact area $A=0.030\,\text{m}^2$
Design B: contact area $A=0.090\,\text{m}^2$

Assuming the average force is the same for both (set by $\Delta p/\Delta t$ ), which statement best explains why one design reduces injury risk more effectively?

Design A is safer because smaller area increases pressure and helps stop the body faster.
Design B is safer because larger area reduces pressure ( $P=F/A$ ) for the same average force. (correct answer)
Both are equally safe because pressure does not depend on area.
Design A is safer because pressure is $P=A/F$ , so smaller $A$ gives smaller pressure.

Explanation: This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. While both designs produce the same average force (determined by F = Δp/Δt with identical momentum change and stopping time), the pressure P = F/A on body tissues determines injury risk—distributing force over larger area reduces pressure and tissue damage. Design B provides contact area of 0.090 m² compared to Design A's 0.030 m², a factor of 3 larger. Using F = Δp/Δt where Δp = 75(20) = 1500 kg·m/s and Δt = 0.15 s gives F = 10,000 N for both designs, but pressure in Design A is P = 10,000/0.030 = 333,333 Pa while Design B gives P = 10,000/0.090 = 111,111 Pa, reducing pressure by factor of 3. This demonstrates that larger contact area is critical for injury prevention even with identical forces. Choice B is correct because it properly explains that larger area reduces pressure (P = F/A) for the same average force, which is the key biomechanical principle for injury reduction. Choice A incorrectly claims smaller area is safer because it increases pressure, which actually increases injury risk—higher pressure on tissues causes more damage, not less. When evaluating collision protection designs: (1) calculate average force from momentum change and time, (2) determine contact area for force distribution, (3) calculate pressure P = F/A, (4) recognize that lower pressure reduces tissue damage. Remember: same force over larger area = lower pressure = less injury risk.

Question 15

A helmet designer is deciding between padding layouts. In a lab test, the same headform experiences the same change in momentum $\Delta p$ in all cases, but the contact area with the padding differs. Assume each design produces the same average force $F_{\text{avg}}$ (same $\Delta p$ and same $\Delta t$ ), and the risk of injury is related to pressure on the skull, $P=F/A$ .

Designs:

Design A: small contact area $A=0.010\ \text{m}^2$ .
Design B: medium contact area $A=0.020\ \text{m}^2$ .
Design C: large contact area $A=0.040\ \text{m}^2$ .
Design D: very small contact area $A=0.005\ \text{m}^2$ .

Which design best reduces pressure on the skull, and why?

Design A, because smaller area concentrates the force and reduces pressure.
Design C, because larger area reduces pressure for the same force ( $P=F/A$ ). (correct answer)
Design D, because the smallest area gives the smallest pressure.
Design B, because pressure depends only on momentum change, not area.

Explanation: This question tests understanding of evaluating collision protection designs using pressure distribution principles in helmet design. While all designs produce the same average force F_avg (same Δp and Δt), the pressure P = F/A on the skull depends on contact area—larger contact areas distribute the same force over more surface, reducing pressure and injury risk. Design C provides the largest contact area A = 0.040 m², which for a given force F produces the lowest pressure P = F/0.040, making it the safest design for the skull. Comparing designs: if F = 1000 N (example), Design A gives P = 1000/0.010 = 100,000 Pa, Design B gives P = 1000/0.020 = 50,000 Pa, Design C gives P = 1000/0.040 = 25,000 Pa, Design D gives P = 1000/0.005 = 200,000 Pa. Choice B is correct because it properly identifies that Design C with the largest area (0.040 m²) reduces pressure according to P = F/A, where larger A means smaller P for the same force. Choice A incorrectly claims smaller area reduces pressure, when P = F/A clearly shows smaller A increases P, making concentrated forces more dangerous. When evaluating helmet padding layouts: (1) recognize that force depends on momentum change and time, (2) understand pressure P = F/A relates force to contact area, (3) larger contact areas reduce pressure for the same force, (4) distributed padding prevents pressure concentrations, and (5) helmet design should maximize contact area within practical constraints. Spreading impact forces reduces injury risk.

Question 16

A fragile instrument (mass $m=10\ \text{kg}$ ) is shipped in a box and may be dropped from $h=1.0\ \text{m}$ . The instrument can tolerate a maximum average impact force of $F_{\max}=500\ \text{N}$ . Assume the cushioning compresses a distance $d$ while bringing the instrument to rest and that all gravitational potential energy becomes work done by the average stopping force: $mgh \approx F_{\text{avg}} d$ .

Cushioning designs:

Design A: bubble wrap, $d=0.05\ \text{m}$ .
Design B: foam blocks, $d=0.20\ \text{m}$ .
Design C: air pillows, $d=0.10\ \text{m}$ .
Design D: thin cardboard inserts, $d=0.02\ \text{m}$ .

Which design keeps $F_{\text{avg}}<500\ \text{N}$ ? (Use $g\approx 9.8\ \text{m/s}^2$ .)

Design A only
Design B only (correct answer)
Designs B and C
Designs A, B, and C

Explanation: This question tests understanding of evaluating collision protection designs using work-energy principles for package cushioning. The work-energy relationship shows that gravitational potential energy mgh must equal work done by the average stopping force F_avg × d, so mgh = F_avg × d gives F_avg = mgh/d—designs with greater cushioning compression distance d reduce the average force for a given drop height. Design B provides 0.20 m compression distance, which using F_avg = mgh/d = (10)(9.8)(1.0)/0.20 = 98/0.20 = 490 N, keeps the force below the 500 N threshold. Checking all designs: Design A gives F = 98/0.05 = 1960 N, Design B gives F = 98/0.20 = 490 N, Design C gives F = 98/0.10 = 980 N, Design D gives F = 98/0.02 = 4900 N. Choice B is correct because it identifies that only Design B with d = 0.20 m produces F_avg = 490 N < 500 N. Choice C incorrectly includes both Designs B and C, but Design C produces 980 N which exceeds the 500 N limit, making it unsuitable for protecting the fragile instrument. When evaluating package cushioning: (1) calculate potential energy mgh from drop height, (2) for each design, identify compression distance d, (3) calculate F_avg = mgh/d, (4) compare to damage threshold, and (5) select designs below the force limit. Greater cushioning thickness allows more compression distance and lower forces.

Question 17

A sports helmet must manage the same impact (same $\Delta p$ ) but designers can change padding to increase collision time $\Delta t$ . The budget is $\le\$ 40 and added mass must be $\le 0.30\ \text{kg}$ . The safety requirement is $F_{\text{avg}}<2000\ \text{N}$ . In testing, the headform’s momentum change is $\Delta p=240\ \text{N\,s}$ .

Design options:

Design A: $\Delta t=0.10\ \text{s}$ , cost $\$ 20, mass $0.10\ \text{kg}$ .
Design B: $\Delta t=0.14\ \text{s}$ , cost $\$ 35, mass $0.25\ \text{kg}$ .
Design C: $\Delta t=0.18\ \text{s}$ , cost $\$ 55, mass $0.28\ \text{kg}$ .
Design D: $\Delta t=0.12\ \text{s}$ , cost $\$ 30, mass $0.35\ \text{kg}$ .

Using $F_{\text{avg}}=\Delta p/\Delta t$ , which design meets all constraints (force, cost, and mass)?

Design A
Design B (correct answer)
Design C
Design D

Explanation: This question tests understanding of evaluating collision protection designs using impulse-momentum principles with multiple constraints (force, cost, and mass). The impulse-momentum theorem F_avg = Δp/Δt shows that extending collision time reduces average force—designs must balance safety performance with practical constraints of cost and added weight. Design B provides Δt = 0.14 s, which gives F_avg = 240/0.14 = 1714 N < 2000 N, costs $35 ≤$ 40, and adds 0.25 kg ≤ 0.30 kg mass, meeting all three constraints. Checking all designs: Design A gives F = 240/0.10 = 2400 N (exceeds force limit), Design B gives F = 240/0.14 = 1714 N with cost $35 and mass 0.25 kg (all OK), Design C gives F = 240/0.18 = 1333 N but costs$ 55 > $40 (over budget), Design D gives F = 240/0.12 = 2000 N with mass 0.35 kg > 0.30 kg (too heavy). Choice B is correct because Design B is the only option meeting all constraints: F = 1714 N < 2000 N, cost =$ 35 ≤ $40, and mass = 0.25 kg ≤ 0.30 kg. Choice C fails the cost constraint at$ 55, while Choices A and D fail force and mass constraints respectively. When evaluating multi-constraint designs: (1) calculate F_avg = Δp/Δt for each option, (2) check against force limit, (3) verify cost within budget, (4) confirm mass within allowance, and (5) select only designs meeting ALL requirements. Real-world design requires balancing multiple competing constraints.

Question 18

A shipping company is choosing a packaging design for a $m=20\ \text{kg}$ device that may experience an impact speed of $v=4.0\ \text{m/s}$ during handling. The device must experience an average stopping force below $F_{\max}=800\ \text{N}$ . Assume the cushioning provides a constant stopping force over compression distance $d$ , so $F_{\text{avg}}\approx \text{KE}/d$ with $\text{KE}=\tfrac12 mv^2$ .

Designs (all fit in the box):

Design A: $d=0.10\ \text{m}$ , cost $\$ 6.
Design B: $d=0.20\ \text{m}$ , cost $\$ 9.
Design C: $d=0.30\ \text{m}$ , cost $\$ 14.
Design D: $d=0.12\ \text{m}$ , cost $\$ 4.

Budget constraint: cost $\le \$ 10. Which design meets both safety and budget constraints?

Design A
Design B (correct answer)
Design C
Design D

Explanation: This question tests understanding of evaluating collision protection designs using work-energy principles with both safety and budget constraints. The kinetic energy KE = ½mv² must be absorbed through work W = Fd, so F_avg = KE/d—packaging with greater compression distance d reduces average force while meeting cost constraints. Design B provides d = 0.20 m compression distance, which using F_avg = KE/d = ½(20)(4.0)²/0.20 = ½(20)(16)/0.20 = 160/0.20 = 800 N, exactly meets the 800 N threshold at a cost of $9, which is under the$ 10 budget. Checking all designs: Design A gives F = 160/0.10 = 1600 N at $6, Design B gives F = 160/0.20 = 800 N at$ 9, Design C gives F = 160/0.30 = 533 N at $14 (over budget), Design D gives F = 160/0.12 = 1333 N at$ 4. Choice B is correct because Design B meets both the safety requirement (F = 800 N ≤ 800 N) and budget constraint ( $9 ≤$ 10). Choice A fails because although under budget, it produces 1600 N which exceeds the safety limit, while Choice C exceeds the budget despite better safety performance. When evaluating packaging designs: (1) calculate kinetic energy KE = ½mv², (2) for each design, calculate F_avg = KE/d, (3) check against force limit, (4) verify cost meets budget, and (5) select designs meeting all constraints. Balancing safety and cost often requires optimizing cushioning thickness.

Question 19

A bicycle helmet must limit the wearer’s head deceleration to less than $100\ g$ during an impact. A test headform of mass $m=5.0\ \text{kg}$ hits a barrier at $v_i=6.0\ \text{m/s}$ and comes to rest. Assume the helmet padding provides an approximately constant stopping distance equal to its compressible thickness $d$ (ignore shell deformation).

Helmet options:

Design A: soft foam, $d=1\ \text{cm}=0.01\ \text{m}$ .
Design B: multi-density foam, $d=3\ \text{cm}=0.03\ \text{m}$ .
Design C: air-pocket liner, $d=5\ \text{cm}=0.05\ \text{m}$ .
Design D: minimal liner, $d=0.5\ \text{cm}=0.005\ \text{m}$ .

Using $v^2\approx 2ad$ (constant deceleration), which design(s) meet the requirement $a<100g$ ? Select the best answer.

Only Design C meets $a<100g$ .
Designs B and C meet $a<100g$ . (correct answer)
Designs A, B, and C meet $a<100g$ because thicker padding always reduces deceleration below any limit.
None of the designs meet $a<100g$ .

Explanation: This question tests understanding of evaluating collision protection designs using work-energy principles applied to helmet safety. The work-energy relationship combined with kinematics gives us v² = 2ad, so for a given impact speed v, increasing stopping distance d reduces deceleration a—helmet designs with thicker compressible padding allow greater deformation and thus lower decelerations. Design B provides 0.03 m stopping distance, which using a = v²/(2d) = 6.0²/(2×0.03) = 36/0.06 = 600 m/s². Since 100g = 100×9.8 = 980 m/s², Design B at 600 m/s² meets the requirement. Checking all designs: Design A gives a = 36/(2×0.01) = 1800 m/s² (184g), Design B gives a = 36/0.06 = 600 m/s² (61g), Design C gives a = 36/0.10 = 360 m/s² (37g), Design D gives a = 36/0.01 = 3600 m/s² (367g). Choice B is correct because it correctly identifies that both Design B (61g) and Design C (37g) meet the 100g requirement. Choice A incorrectly states only Design C meets the requirement, missing that Design B also produces deceleration below 100g at 61g. When evaluating helmet designs: (1) identify impact speed v, (2) for each design, determine stopping distance d (padding thickness), (3) calculate a = v²/(2d), (4) convert to g's by dividing by 9.8 m/s², and (5) compare to safety threshold. Designs with thicker compressible padding provide longer stopping distances and lower decelerations.

Question 20

A bicycle helmet must keep head deceleration below $100\,g$ during a crash. A rider’s head is modeled as $m=5.0\,\text{kg}$ impacting at $v=6.0\,\text{m/s}$ and coming to rest. Three helmet padding designs are proposed; assume each provides approximately constant deceleration over its crush distance $d$ :

Design A: soft foam, $d=0.010\,\text{m}$
Design B: multi-density foam, $d=0.030\,\text{m}$
Design C: air-pocket liner, $d=0.050\,\text{m}$

Using $v^2=2ad$ (so $a=\tfrac{v^2}{2d}$ ), which design(s) satisfy $a<100g$ (take $g\approx 9.8\,\text{m/s}^2$ )?

Only Design A.
Designs B and C only. (correct answer)
Only Design C.
All three designs.

Explanation: This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship combined with kinematics shows that for constant deceleration, v² = 2ad gives a = v²/(2d), so increasing deformation distance d reduces deceleration a—designs allowing greater deformation absorb the same energy with smaller accelerations and forces. Design C provides greatest deformation distance of 0.050 m compared to Design B's 0.030 m and Design A's 0.010 m, which reduces the deceleration significantly. Using a = v²/(2d) where v = 6.0 m/s, the deceleration in Design A is a = 36/(2×0.010) = 1800 m/s² = 184g which exceeds 100g, Design B gives a = 36/(2×0.030) = 600 m/s² = 61g which is below 100g, and Design C produces a = 36/(2×0.050) = 360 m/s² = 37g which also meets the limit. This demonstrates that greater distance is critical for keeping accelerations within safe ranges. Choice B is correct because it correctly identifies that Designs B and C meet the 100g requirement while Design A exceeds it. Choice A incorrectly selects only Design A, which has the shortest deformation distance and therefore the highest deceleration, making it fail the safety requirement. When evaluating collision protection designs: (1) identify impact velocity v, (2) for each design, determine deformation distance d, (3) calculate a = v²/(2d) for each design, (4) compare to safety threshold in appropriate units (convert to g's by dividing by 9.8), and (5) consider other constraints. Designs that extend time and increase deformation distance provide the best protection.

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Physics Quiz

Physics Quiz: Evaluate Collision Design Solutions

Practice Evaluate Collision Design Solutions in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

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?

What this quiz covers

This quiz focuses on Evaluate Collision Design Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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?

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.
Designs B and C, because doubling or tripling area cuts pressure and both fall below 50 kPa.
None, because even Design C gives $P\approx 3000/0.009=3.3\times 10^5\,\text{Pa}=330\,\text{kPa}>50\,\text{kPa}$ . (correct answer)
Design A only, because smaller area concentrates force and therefore lowers pressure.

Question 2

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?

Design C only, because $F=(60\cdot 15)/0.200=4500\,\text{N}<5000\,\text{N}$ , while A and B are above 5000 N. (correct answer)
Designs B and C, because both have longer stopping times than A and therefore both must be below 5000 N.
All designs, because the passenger mass is moderate and the speed is only 15 m/s.
Design A only, because a shorter stopping time reduces the impulse and therefore reduces force.

Question 3

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?

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. (correct answer)
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.
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).

Question 4

Design A: no airbag (seat belt only), stopping time $\Delta t=0.040\,\text{s}$ , contact area on chest $A=0.020\,\text{m}^2$
Design B: airbag, stopping time $\Delta t=0.120\,\text{s}$ , contact area on chest $A=0.080\,\text{m}^2$

Design A meets both limits because it has the smaller contact area, which lowers pressure.
Design B meets both limits because it increases $\Delta t$ (reducing force) and increases $A$ (reducing pressure). (correct answer)
Design B fails the force limit because a longer $\Delta t$ increases impulse and therefore increases force.
Design A meets the force limit since $F_{\text{avg}}=\frac{(40)(10)}{0.040}=1000\,\text{N}<5000\,\text{N}$ .

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 0.120 s compared to Design A's 0.040 s, which reduces the average force by a factor of 3. Using F = Δp/Δt, the force in Design B is F = (40×10)/0.120 ≈3333 N, which is below the safety threshold of 5000 N with P ≈41.7 kPa below 100 kPa, while Design A produces F = (40×10)/0.040 = 10000 N which exceeds the force limit and P=500 kPa exceeds pressure limit. This demonstrates that longer time is critical for keeping forces within safe ranges. Choice B is correct because it accurately explains the physics principle: extending time or distance reduces force and properly evaluates both force and pressure meeting thresholds. Choice D makes a calculation error by stating F=1000 N for Design A, when it's actually 10000 N, leading to incorrect force estimate that doesn't reflect actual performance. When evaluating collision protection designs: (1) identify momentum change Δp = mΔv, (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 (pressure). 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).

Question 5

Two proposed restraint designs:

Design A, because larger area always reduces force and pressure.
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.
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). (correct answer)

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).

Question 6

Design A, because its longer stopping time gives the lowest force and it is under the $\$ 700 budget.
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 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. (correct answer)

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).

Question 7

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

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). (correct answer)

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).

Question 8

Assume constant average stopping force so $F\approx \frac{\tfrac12 mv^2}{d}$ . Which design meets both the force and mass constraints?

Design A, because it is under the 1.5 kg mass limit and stiffer foam reduces the force.
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.
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. (correct answer)

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).

Question 9

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 A meets the requirement because it stops the head fastest.
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.
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 $980\ \text{m/s}^2$ . (correct answer)
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).

Question 10

Proposed designs:

Using impulse-momentum ( $F_{\text{avg}}=\Delta p/\Delta t$ ) and the distance constraint, which design meets the requirements?

Design A, because $d$ is smallest so it fits best and therefore reduces force.
Design B, because $F\approx \frac{1200\cdot 8}{0.10}=96{,}000\ \text{N}<150{,}000\ \text{N}$ and $d=0.35\ \text{m}\le 0.40\ \text{m}$ . (correct answer)
Design C, because it has the longest stopping time and thus the smallest force, and it also fits within $0.40\ \text{m}$ .
None, because $F_{\text{avg}}$ depends on crush distance, not on stopping time.

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 0.10 s compared to Design A's 0.05 s, which reduces the average force by a factor of 2; using F = Δp/Δt, the force in Design B is F = 9,600 / 0.10 = 96,000 N, which is below the safety threshold of 150,000 N, while Design C produces F = 9,600 / 0.14 ≈ 68,571 N but exceeds the 0.40 m space limit with d=0.45 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 appropriate Δt that produces forces below the limit while fitting the space constraint. Choice C selects the design with longer collision time but ignores the stated space criterion, selecting based on irrelevant factor of lowest force without checking d <= 0.40 m. 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).

Question 11

Design A: $d=0.010\,\text{m}$ (1 cm)
Design B: $d=0.030\,\text{m}$ (3 cm)
Design C: $d=0.050\,\text{m}$ (5 cm)

Safety target: $F_{\text{avg}}<2500\,\text{N}$ . Which design(s) meet the target?

Only Design A.
Only Design B.
Only Design C. (correct answer)
Designs B and C.

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 average force F—designs allowing greater deformation (thicker padding, longer crumple zones) absorb the same energy with smaller forces. Design C provides greatest deformation distance of 0.050 m compared to Design B's 0.030 m, which reduces the average force by a factor of about 1.67. Using W = Fd, the force in Design C is F = (0.5×5.0×7.0²)/0.050 = 2450 N, which is below the safety threshold of 2500 N, while Design B produces F = (0.5×5.0×7.0²)/0.030 ≈4083 N which exceeds the limit; Design A has F ≈12250 N exceeding the limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice C is correct because it correctly identifies the design with greatest d that produces lowest forces and meets the safety target. Choice A selects the design with smaller deformation distance, which actually produces higher forces (F = KE/d means smaller d gives larger F), making it less effective for protection. When evaluating collision protection designs: (1) identify energy to absorb KE = ½mv², (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. 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).

Question 12

Design A: $d=0.015\,\text{m}$ , mass of helmet 0.30 kg Design B: $d=0.030\,\text{m}$ , mass of helmet 0.45 kg Design C: $d=0.050\,\text{m}$ , mass of helmet 0.70 kg

Requirements: $F_{\text{avg}}<1000\,\text{N}$ and helmet mass $\le 0.50\,\text{kg}$ .

Which design meets BOTH requirements?

Design A, because it is lightest and therefore reduces impact force the most.
Design B, because $\text{KE}=\tfrac12(4)(25)=50\,\text{J}$ and $F\approx 50/0.030\approx 1.67\times 10^3\,\text{N}<1000\,\text{N}$ with mass under 0.50 kg.
Design C, because $F\approx 50/0.050=1000\,\text{N}$ but it fails the mass limit.
None, because A and B exceed 1000 N and C exceeds the mass limit (and is not below 1000 N). (correct answer)

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 deformation distance of 0.050 m compared to Design B's 0.030 m, which reduces the peak force by a factor of about 1.67. Using F = KE/d, the force in Design C is F = 50/0.050 = 1000 N, which is not below the safety threshold of 1000 N, while Design B produces F = 50/0.030 ≈ 1667 N which exceeds the limit, and C exceeds mass limit. This demonstrates that greater distance is critical for keeping forces within safe ranges. Choice D is correct because it properly calculates forces showing no design meets both the force and mass thresholds. Choice C ignores the stated mass criterion, selecting based on force, when the primary goal is to minimize weight while meeting force limit, and C does not strictly meet <1000 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).

Question 13

Design A: rubber mat, $d=0.15\,\text{m}$
Design B: wood chips, $d=0.30\,\text{m}$
Design C: sand, $d=0.45\,\text{m}$

Use $v=\sqrt{2gh}$ to find impact speed and then $a\approx \dfrac{v^2}{2d}$ to estimate average deceleration. Which surfaces meet the $100\,g$ requirement?

Only Design A.
Designs B and C only.
Only Design C.
All three designs. (correct answer)

Question 14

A 75 kg occupant in a crash goes from $20\,\text{m/s}$ to $0\,\text{m/s}$ . Two airbag designs produce the same stopping time $\Delta t=0.15\,\text{s}$ but different contact areas on the chest:

Design A: contact area $A=0.030\,\text{m}^2$
Design B: contact area $A=0.090\,\text{m}^2$

Assuming the average force is the same for both (set by $\Delta p/\Delta t$ ), which statement best explains why one design reduces injury risk more effectively?

Design A is safer because smaller area increases pressure and helps stop the body faster.
Design B is safer because larger area reduces pressure ( $P=F/A$ ) for the same average force. (correct answer)
Both are equally safe because pressure does not depend on area.
Design A is safer because pressure is $P=A/F$ , so smaller $A$ gives smaller pressure.

Question 15

Designs:

Design A: small contact area $A=0.010\ \text{m}^2$ .
Design B: medium contact area $A=0.020\ \text{m}^2$ .
Design C: large contact area $A=0.040\ \text{m}^2$ .
Design D: very small contact area $A=0.005\ \text{m}^2$ .

Which design best reduces pressure on the skull, and why?

Design A, because smaller area concentrates the force and reduces pressure.
Design C, because larger area reduces pressure for the same force ( $P=F/A$ ). (correct answer)
Design D, because the smallest area gives the smallest pressure.
Design B, because pressure depends only on momentum change, not area.

Question 16

Cushioning designs:

Design A: bubble wrap, $d=0.05\ \text{m}$ .
Design B: foam blocks, $d=0.20\ \text{m}$ .
Design C: air pillows, $d=0.10\ \text{m}$ .
Design D: thin cardboard inserts, $d=0.02\ \text{m}$ .

Which design keeps $F_{\text{avg}}<500\ \text{N}$ ? (Use $g\approx 9.8\ \text{m/s}^2$ .)

Design A only
Design B only (correct answer)
Designs B and C
Designs A, B, and C

Question 17

Design options:

Design A: $\Delta t=0.10\ \text{s}$ , cost $\$ 20, mass $0.10\ \text{kg}$ .
Design B: $\Delta t=0.14\ \text{s}$ , cost $\$ 35, mass $0.25\ \text{kg}$ .
Design C: $\Delta t=0.18\ \text{s}$ , cost $\$ 55, mass $0.28\ \text{kg}$ .
Design D: $\Delta t=0.12\ \text{s}$ , cost $\$ 30, mass $0.35\ \text{kg}$ .

Using $F_{\text{avg}}=\Delta p/\Delta t$ , which design meets all constraints (force, cost, and mass)?

Design A
Design B (correct answer)
Design C
Design D

Question 18

Designs (all fit in the box):

Design A: $d=0.10\ \text{m}$ , cost $\$ 6.
Design B: $d=0.20\ \text{m}$ , cost $\$ 9.
Design C: $d=0.30\ \text{m}$ , cost $\$ 14.
Design D: $d=0.12\ \text{m}$ , cost $\$ 4.

Budget constraint: cost $\le \$ 10. Which design meets both safety and budget constraints?

Design A
Design B (correct answer)
Design C
Design D

Question 19

Helmet options:

Design A: soft foam, $d=1\ \text{cm}=0.01\ \text{m}$ .
Design B: multi-density foam, $d=3\ \text{cm}=0.03\ \text{m}$ .
Design C: air-pocket liner, $d=5\ \text{cm}=0.05\ \text{m}$ .
Design D: minimal liner, $d=0.5\ \text{cm}=0.005\ \text{m}$ .

Using $v^2\approx 2ad$ (constant deceleration), which design(s) meet the requirement $a<100g$ ? Select the best answer.

Only Design C meets $a<100g$ .
Designs B and C meet $a<100g$ . (correct answer)
Designs A, B, and C meet $a<100g$ because thicker padding always reduces deceleration below any limit.
None of the designs meet $a<100g$ .

Question 20

Design A: soft foam, $d=0.010\,\text{m}$
Design B: multi-density foam, $d=0.030\,\text{m}$
Design C: air-pocket liner, $d=0.050\,\text{m}$

Using $v^2=2ad$ (so $a=\tfrac{v^2}{2d}$ ), which design(s) satisfy $a<100g$ (take $g\approx 9.8\,\text{m/s}^2$ )?

Only Design A.
Designs B and C only. (correct answer)
Only Design C.
All three designs.

Explanation: This question tests understanding of evaluating collision protection designs using impulse-momentum and work-energy principles. The work-energy relationship combined with kinematics shows that for constant deceleration, v² = 2ad gives a = v²/(2d), so increasing deformation distance d reduces deceleration a—designs allowing greater deformation absorb the same energy with smaller accelerations and forces. Design C provides greatest deformation distance of 0.050 m compared to Design B's 0.030 m and Design A's 0.010 m, which reduces the deceleration significantly. Using a = v²/(2d) where v = 6.0 m/s, the deceleration in Design A is a = 36/(2×0.010) = 1800 m/s² = 184g which exceeds 100g, Design B gives a = 36/(2×0.030) = 600 m/s² = 61g which is below 100g, and Design C produces a = 36/(2×0.050) = 360 m/s² = 37g which also meets the limit. This demonstrates that greater distance is critical for keeping accelerations within safe ranges. Choice B is correct because it correctly identifies that Designs B and C meet the 100g requirement while Design A exceeds it. Choice A incorrectly selects only Design A, which has the shortest deformation distance and therefore the highest deceleration, making it fail the safety requirement. When evaluating collision protection designs: (1) identify impact velocity v, (2) for each design, determine deformation distance d, (3) calculate a = v²/(2d) for each design, (4) compare to safety threshold in appropriate units (convert to g's by dividing by 9.8), and (5) consider other constraints. Designs that extend time and increase deformation distance provide the best protection.