Physics Quiz: Optimize Designs For Collision Safety

What this quiz covers

This quiz focuses on Optimize Designs For Collision Safety, 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.

Question 1

1. Design A, because smaller $d$ always reduces force
2. Design A, because Design B violates the packaging constraint
3. Design B, because it produces a lower force and still satisfies $d\le 0.85\ \text{m}$ (correct answer)
4. Neither design meets the force limit

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1000 kg impacting at v = 18 m/s, the kinetic energy is KE = ½mv² = 0.51000324 = 162000 J. To keep force below F_max = 250000 N, the minimum distance is d_min = KE/F_max = 162000/250000 = 0.648 m. Design A provides d=0.50 m which is less than this minimum, so F_avg = 162000/0.50 = 324000 N > threshold; Design B provides d=0.80 m which is greater than minimum, so F_avg = 162000/0.80 = 202500 N < threshold. Choice C is correct because it identifies Design B as meeting the safety threshold with lower force while satisfying the packaging constraint. Choice A is wrong because smaller d increases force, using wrong assumption. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 2

1. Feasible; $\Delta t_{\min}=0.23\ \text{s}$ so $\Delta t_{\min}\le 0.40\ \text{s}$
2. Not feasible; $\Delta t_{\min}=0.45\ \text{s}$ so $\Delta t_{\min}>0.40\ \text{s}$ (correct answer)
3. Feasible; $\Delta t_{\min}=0.05\ \text{s}$ so $\Delta t_{\min}\le 0.40\ \text{s}$
4. Not feasible; $\Delta t_{\min}=0.10\ \text{s}$ so $\Delta t_{\min}>0.40\ \text{s}$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 75 kg impacting at v = 9.0 m/s, the momentum change is Δp = mv = 75*9 = 675 kg⋅m/s. To keep force below F_max = 1500 N, the minimum collision time is Δt_min = Δp/F_max = 675/1500 = 0.45 s. The material limit allows up to Δt = 0.40 s which is less than this minimum, so not feasible. Choice B is correct because it correctly calculates Δt_min using the appropriate formula and properly evaluates that it is not feasible within the constraint. Choice A makes calculation error by underestimating Δt_min, incorrectly concluding feasibility. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 3

1. $\Delta t_{\min}=0.20\ \text{s}$; feasible because $0.20\le 0.60$
2. $\Delta t_{\min}=0.40\ \text{s}$; feasible because $0.40\le 0.60$ (correct answer)
3. $\Delta t_{\min}=0.80\ \text{s}$; not feasible because $0.80>0.60$
4. $\Delta t_{\min}=0.05\ \text{s}$; feasible because $0.05\le 0.60$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 70 kg impacting at v = 20 m/s, the momentum change is Δp = mv = 70*20 = 1400 kg⋅m/s. To keep force below F_max = 3500 N, the minimum collision time is Δt_min = Δp/F_max = 1400/3500 = 0.4 s. The packaging limit allows up to Δt = 0.60 s which is greater than this minimum, so F_avg can be kept < threshold. Choice B is correct because it correctly calculates Δt_min using the appropriate formula and properly evaluates against both safety goal and constraints. Choice A makes calculation error by underestimating Δt_min, resulting in incorrect feasibility assessment. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 4

1. Yes; $d_{\min}=0.15\ \text{m}$, which is $\le 0.50\ \text{m}$
2. Yes; $d_{\min}=0.30\ \text{m}$, which is $\le 0.50\ \text{m}$
3. No; $d_{\min}=1.50\ \text{m}$, which is $>0.50\ \text{m}$ (correct answer)
4. No; $d_{\min}=0.05\ \text{m}$, which is $>0.50\ \text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 90 kg impacting at v = 10 m/s, the kinetic energy is KE = ½mv² = 0.590100 = 4500 J. To keep force below F_max = 3000 N, the minimum deformation distance is d_min = KE/F_max = 4500/3000 = 1.5 m. The width limit allows up to d = 0.50 m which is less than this minimum, so F_avg would > threshold. Choice C is correct because it correctly calculates d_min using the appropriate formula and properly evaluates that it is not possible within the constraint. Choice A makes calculation error by underestimating d_min, incorrectly concluding feasibility. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 5

1. Feasible; $d_{\min}=0.78 \text{m}$ so the design can meet $d \le 1.0 \text{m}$
2. Not feasible; $d_{\min}=1.56 \text{m}$ so the design cannot meet $d \le 1.0 \text{m}$ (correct answer)
3. Feasible; $d_{\min}=0.26 \text{m}$ so the design can meet $d \le 1.0 \text{m}$
4. Not feasible; $d_{\min}=0.10 \text{m}$ so the design cannot meet $d \le 1.0 \text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change $Δp = mv$, solve $F_{\max} = Δp/Δt_{\min}$ for minimum time: $Δt_{\min} = mv/F_{\max}$, or for given energy $KE = \frac{1}{2}mv^2$, solve $F_{\max} = KE/d_{\min}$ for minimum distance: $d_{\min} = (\frac{1}{2}mv^2)/F_{\max}$—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with $m = 1500$ kg impacting at $v = 25$ m/s, the kinetic energy is $KE = \frac{1}{2}mv^2 = 0.5*1500*625 = 468750$ J. To keep force below $F_{\max} = 300000$ N, the minimum crumple distance is $d_{\min} = KE/F_{\max} = 468750/300000 = 1.5625$ m. The layout limit allows up to $d = 1.0$ m which is less than this minimum, so not feasible. Choice B is correct because it correctly calculates $d_{\min}$ using the appropriate formula and properly evaluates against both safety goal and constraints. Choice A makes calculation error by underestimating $d_{\min}$, incorrectly concluding feasibility. Optimization strategy: (1) calculate minimum parameter needed ($Δt_{\min} = mv/F_{\max}$ or $d_{\min} = KE/F_{\max}$), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 6

1. $d_{\min}=0.22\ \text{m}$, fits within $0.80\ \text{m}$
2. $d_{\min}=2.16\ \text{m}$, does not fit within $0.80\ \text{m}$ (correct answer)
3. $d_{\min}=1.20\ \text{m}$, does not fit within $0.80\ \text{m}$
4. $d_{\min}=0.80\ \text{m}$, fits exactly within $0.80\ \text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1200 kg impacting at v = 18 m/s, the kinetic energy is KE = ½120018² = 194400 J. To keep force below F_max = 90000 N, the minimum collision distance is d_min = 194400/90000 = 2.16 m. The available space provides d = 0.80 m which is less than this minimum, so F_avg = 194400/0.80 = 243000 N > threshold. Choice B is correct because it correctly calculates d_min using the appropriate formula and properly evaluates against both safety goal and constraints. Choice A makes calculation error by using incorrect values or formula, resulting in insufficient d_min and wrong feasibility. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 7

1. $\Delta t_{\min}=0.14\,\text{s}$, satisfies ($0.14\le 0.50$)
2. $\Delta t_{\min}=0.35\,\text{s}$, satisfies ($0.35\le 0.50$) (correct answer)
3. $\Delta t_{\min}=0.70\,\text{s}$, violates ($0.70>0.50$)
4. $\Delta t_{\min}=0.07\,\text{s}$, satisfies

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 50 kg impacting at v = 14 m/s, the momentum change is Δp = mv = 50*14 = 700 kg⋅m/s. To keep force below F_max = 2000 N, the minimum collision time is Δt_min = Δp/F_max = 700/2000 = 0.35 s. Choice B provides Δt_min=0.35 s which is less than or equal to the constraint of 0.50 s, so satisfies with F_avg ≤ threshold. Choice B is correct because it correctly calculates Δt_min using appropriate formula and properly evaluates against both safety goal and constraints. Choice C makes calculation error by overestimating Δt_min to 0.70 s, incorrectly claiming violates constraint. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 8

1. $d_{\min}=0.61\,\text{m}$; current $0.70\,\text{m}$ already meets the force limit (correct answer)
2. $d_{\min}=1.21\,\text{m}$; not possible within the $0.80\,\text{m}$ limit
3. $d_{\min}=0.30\,\text{m}$; current design is excessive and unsafe
4. $d_{\min}=0.80\,\text{m}$; must increase to exactly $0.80\,\text{m}$ to be safe

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1000 kg impacting at v = 22 m/s, the kinetic energy is KE = ½mv² = 0.51000484 = 242000 J. To keep force below F_max = 400000 N, the minimum distance is d_min = KE/F_max = 242000/400000 ≈ 0.605 m. The current design provides d = 0.70 m which is greater than this minimum, so F_avg ≈ 345714 N < threshold. Choice A is correct because it correctly calculates d_min using appropriate formula and identifies that current design meets safety threshold within constraints. Choice B makes calculation error by overestimating d_min to 1.21 m, incorrectly claiming not possible. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 9

1. Feasible, because $d_{\min}=0.78\,\text{m} \le 0.90\,\text{m}$
2. Not feasible, because $d_{\min}=1.56\,\text{m} > 0.90\,\text{m}$ (correct answer)
3. Feasible, because $d_{\min}=0.39\,\text{m} \le 0.90\,\text{m}$
4. Not feasible, because $d_{\min}=0.12\,\text{m} > 0.90\,\text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1500 kg impacting at v = 25 m/s, the kinetic energy is KE = ½mv² = 0.51500625 = 468750 J. To keep force below F_max = 300000 N, the minimum distance is d_min = KE/F_max = 468750/300000 = 1.5625 m. Choice B identifies d_min ≈1.56 m which exceeds the constraint of 0.90 m, so not feasible with F_avg > threshold if d=0.90 m. Choice B is correct because it correctly calculates d_min using appropriate formula and properly evaluates against both safety goal and constraints. Choice A makes calculation error by underestimating d_min to 0.78 m, incorrectly claiming feasible. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 10

1. $\Delta t_{\min}=0.20\,\text{s}$ (feasible within $0.60\,\text{s}$)
2. $\Delta t_{\min}=0.40\,\text{s}$ (feasible within $0.60\,\text{s}$) (correct answer)
3. $\Delta t_{\min}=2.5\,\text{s}$ (not feasible within $0.60\,\text{s}$)
4. $\Delta t_{\min}=0.05\,\text{s}$ (feasible within $0.60\,\text{s}$)

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 70 kg impacting at v = 20 m/s, the momentum change is Δp = mv = 70*20 = 1400 kg⋅m/s. To keep force below F_max = 3500 N, the minimum collision time is Δt_min = Δp/F_max = 1400/3500 = 0.4 s. Choice B provides Δt = 0.4 s which equals this minimum, so F_avg = 3500 N [= threshold]. Choice B is correct because it correctly calculates Δt_min using appropriate formula and properly evaluates against both safety goal and constraints. Choice A makes calculation error by underestimating Δt_min, resulting in incorrect feasibility assessment. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 11

1. $d_{\min}\approx 1.56\,\text{m}$, so the force limit cannot be met with $d\le 0.80\,\text{m}$. (correct answer)
2. $d_{\min}\approx 0.78\,\text{m}$, so the force limit cannot be met because $0.78<0.80$.
3. $d_{\min}\approx 0.39\,\text{m}$, so the force limit cannot be met unless $d$ is smaller.
4. $d_{\min}\approx 1.56\,\text{m}$, so the force limit is met as long as $d\le 0.80\,\text{m}$.

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum or work-energy to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max, or for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1500 kg impacting at v = 25 m/s, the kinetic energy is KE = ½mv² = ½(1500)(25²) = 468,750 J. To keep force below F_max = 3.0×10⁵ N, the minimum crumple distance is d_min = KE/F_max = 468,750/300,000 = 1.5625 m ≈ 1.56 m. Design constraint limits d to 0.80 m maximum, which is less than d_min = 1.56 m, so the force limit cannot be met. Choice A is correct because it states d_min ≈ 1.56 m and correctly concludes the force limit cannot be met with d ≤ 0.80 m. Choice B incorrectly states d_min ≈ 0.78 m (half the correct value). Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max or d_min = KE/F_max), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (closest to minimum while still safe), (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 12

1. $\Delta t_{\min}=0.28\ \text{s}$; achievable because $0.28\le 0.45$
2. $\Delta t_{\min}=0.56\ \text{s}$; not achievable because $0.56>0.45$ (correct answer)
3. $\Delta t_{\min}=0.45\ \text{s}$; achievable only if exactly 0.45 s is used
4. $\Delta t_{\min}=0.07\ \text{s}$; achievable because $0.07\le 0.45$

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max—designs must provide at least this minimum to keep forces below safety thresholds. For this scenario with m = 75 kg impacting at v = 9 m/s, the momentum change is Δp = mv = 75 × 9 = 675 kg⋅m/s. To keep force below F_max = 1200 N, the minimum collision time is Δt_min = Δp/F_max = 675/1200 = 0.5625 s ≈ 0.56 s. Since the airbag system allows at most 0.45 s, this requirement is not achievable because 0.56 > 0.45, meaning F_avg would exceed 1200 N threshold. Choice B is correct because it correctly calculates Δt_min = 0.56 s and identifies that this exceeds the 0.45 s constraint, making the design infeasible. Choice A incorrectly calculates Δt_min = 0.28 s; Choice C suggests using exactly 0.45 s but this would result in F = 675/0.45 = 1500 N > 1200 N; Choice D gives an unrealistically short time. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max), (2) check against maximum allowed time, (3) if minimum exceeds maximum, the design is infeasible without changing other parameters. The airbag system needs redesign to allow longer deployment times or reduce initial relative velocity.

Question 13

1. $d_{\min}=0.60\ \text{m}$; feasible because $0.60\le 0.80$ (correct answer)
2. $d_{\min}=0.40\ \text{m}$; feasible because $0.40\le 0.80$
3. $d_{\min}=1.20\ \text{m}$; not feasible because $1.20>0.80$
4. $d_{\min}=0.20\ \text{m}$; feasible because $0.20\le 0.80$

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = KE/F_max—designs must provide at least this minimum to keep forces below safety thresholds. For this scenario with m = 1200 kg impacting at v = 20 m/s, the kinetic energy is KE = ½mv² = ½(1200)(20²) = ½(1200)(400) = 240,000 J. To keep force below F_max = 4.0×10⁵ N, the minimum crumple distance is d_min = KE/F_max = 240,000/400,000 = 0.60 m. Since the crumple zone allows up to 0.80 m, this design is feasible because 0.60 ≤ 0.80, so F_avg = 400,000 N ≤ threshold. Choice A is correct because it correctly calculates d_min = 0.60 m using the work-energy formula and identifies that this meets the safety threshold within the 0.80 m constraint. Choice B gives d_min = 0.40 m which would result in F = 240,000/0.40 = 600,000 N > 400,000 N threshold; Choice C gives 1.20 m which exceeds the 0.80 m constraint; Choice D gives 0.20 m resulting in excessive force. Optimization strategy: (1) calculate minimum parameter needed (d_min = KE/F_max), (2) check against maximum allowed distance, (3) select design meeting safety requirement with least excess. The 0.60 m minimum exactly meets the force requirement while fitting within the 0.80 m constraint.

Question 14

1. Yes; $d_{\min}=0.40\,\text{m}$, which is within $1.0\,\text{m}$
2. No; $d_{\min}=4.0\,\text{m}$, which exceeds $1.0\,\text{m}$ (correct answer)
3. Yes; $d_{\min}=0.04\,\text{m}$, which is within $1.0\,\text{m}$
4. No; because $F$ depends only on momentum, not on stopping distance

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 75 kg impacting at v = 16 m/s, the kinetic energy is KE = ½(75)(16²) = ½(75)(256) = 9600 J. To keep force below F_max = 2400 N, the minimum stopping distance is d_min = KE/F_max = 9600/2400 = 4.0 m. Since the constraint allows at most d ≤ 1.0 m and we need d_min = 4.0 m, this exceeds the available stretch distance. Choice B is correct because it correctly calculates d_min = 4.0 m and identifies that this exceeds the 1.0 m constraint, making it not possible to meet the force limit. Choice A incorrectly calculates d_min as 0.40 m, which would give F = 9600/0.40 = 24,000 N >> 2400 N. Optimization strategy: (1) calculate minimum parameter needed (d_min = KE/F_max), (2) check against constraint, (3) conclude feasibility. The required stopping distance far exceeds what the net can provide, making the design infeasible.

Question 15

1. $d=0.50\,\text{m}$
2. $d=0.60\,\text{m}$
3. $d=0.68\,\text{m}$ (correct answer)
4. $d=0.80\,\text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1400 kg impacting at v = 22 m/s, the kinetic energy is KE = ½(1400)(22²) = ½(1400)(484) = 338,800 J. To keep force below F_max = 500,000 N, the minimum crumple distance is d_min = KE/F_max = 338,800/500,000 = 0.6776 m ≈ 0.68 m. Since the constraint requires d ≤ 0.75 m, we need to select from the given options the smallest d that meets d ≥ 0.68 m. Choice C is correct because d = 0.68 m equals the minimum required distance, thus meeting the force requirement with no excess while satisfying the constraint (0.68 < 0.75). Choices A (0.50 m) and B (0.60 m) would result in forces exceeding 500,000 N. Optimization strategy: (1) calculate minimum parameter needed (d_min = KE/F_max = 0.68 m), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (0.68 m exactly meets requirement), (4) verify constraints satisfied. This represents optimal design with zero waste.

Question 16

1. $\Delta t_{\min}=0.30\,\text{s}$; not achievable because $0.30>0.20$ (correct answer)
2. $\Delta t_{\min}=0.03\,\text{s}$; achievable because $0.03<0.20$
3. $\Delta t_{\min}=0.20\,\text{s}$; not achievable because it equals the limit
4. $\Delta t_{\min}=3.0\,\text{s}$; not achievable because it is too long

Explanation: This question tests understanding of optimizing collision safety designs using impulse-momentum to minimize forces while meeting constraints. To minimize force for a given momentum change Δp = mv, solve F_max = Δp/Δt_min for minimum time: Δt_min = mv/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 90 kg impacting at v = 10 m/s, the momentum change is Δp = mv = (90)(10) = 900 kg⋅m/s. To keep force below F_max = 3000 N, the minimum collision time is Δt_min = Δp/F_max = 900/3000 = 0.30 s. The constraint allows at most Δt = 0.20 s, which is less than the required 0.30 s, so the design is not achievable. Choice A is correct because it correctly calculates Δt_min = 0.30 s and identifies that this exceeds the 0.20 s limit, making it not achievable. Choice B incorrectly calculates Δt_min as 0.03 s, which would give F = 900/0.03 = 30,000 N >> 3000 N. Optimization strategy: (1) calculate minimum parameter needed (Δt_min = mv/F_max), (2) check available designs against this minimum, (3) verify constraints satisfied. Designs below minimum fail safety criteria; this system cannot provide sufficient collision time.

Question 17

1. $d_{\min}=0.80\,\text{m}$; feasible because $0.80\,\text{m}<0.90\,\text{m}$ (correct answer)
2. $d_{\min}=1.60\,\text{m}$; not feasible because $1.60\,\text{m}>0.90\,\text{m}$
3. $d_{\min}=0.40\,\text{m}$; feasible because $0.40\,\text{m}<0.90\,\text{m}$
4. $d_{\min}=0.90\,\text{m}$; not feasible because it uses the maximum allowed distance

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1200 kg impacting at v = 20 m/s, the kinetic energy is KE = ½(1200)(20²) = 240,000 J. To keep force below F_max = 300,000 N, the minimum crumple distance is d_min = 240,000/300,000 = 0.80 m. Design with d = 0.80 m provides exactly the minimum needed, so F_avg = 240,000/0.80 = 300,000 N = threshold. Choice A is correct because it correctly calculates d_min = 0.80 m using the work-energy formula and identifies that this is feasible within the 0.90 m limit. Choice B incorrectly calculates d_min = 1.60 m, which would give F = 240,000/1.60 = 150,000 N (half the threshold). Optimization strategy: (1) calculate minimum parameter needed (d_min = KE/F_max = 0.80 m), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess, (4) verify constraints satisfied. Designs significantly exceeding minimum waste resources; designs below minimum fail safety criteria.

Question 18

1. $d_{\min}=0.78\,\text{m}$, so $0.60\,\text{m}$ is unsafe but $0.80\,\text{m}$ would be safe (correct answer)
2. $d_{\min}=0.39\,\text{m}$, so $0.60\,\text{m}$ is already unsafe because it is larger than needed
3. $d_{\min}=1.56\,\text{m}$, so even $0.80\,\text{m}$ cannot meet the force limit
4. $d_{\min}=0.60\,\text{m}$ exactly, so any smaller or larger distance gives the same force

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1000 kg impacting at v = 25 m/s, the kinetic energy is KE = ½(1000)(25²) = 312,500 J. To keep force below F_max = 400,000 N, the minimum crumple distance is d_min = 312,500/400,000 = 0.78125 m ≈ 0.78 m. Design with d = 0.60 m would result in F_avg = 312,500/0.60 ≈ 520,833 N > 400,000 N threshold, making it unsafe. Choice A is correct because it correctly calculates d_min = 0.78 m and identifies that 0.60 m is unsafe while 0.80 m would be safe (giving F = 312,500/0.80 = 390,625 N < 400,000 N). Choice C incorrectly calculates d_min = 1.56 m, which would give F = 312,500/1.56 ≈ 200,321 N (half the threshold). Optimization strategy: (1) calculate minimum parameter needed (d_min = KE/F_max = 0.78 m), (2) check proposed designs against this minimum, (3) select design meeting safety requirement with least excess. The 0.80 m maximum design barely meets the requirement.

Question 19

1. Design (1) $d=0.60\,\text{m}$
2. Design (2) $d=0.80\,\text{m}$
3. Design (3) $d=0.90\,\text{m}$ (correct answer)
4. Design (4) $d=1.20\,\text{m}$

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d_min for minimum distance: d_min = (½mv²)/F_max—designs must provide at least these minimums to keep forces below safety thresholds. For this scenario with m = 1800 kg impacting at v = 18 m/s, the kinetic energy is KE = ½(1800)(18²) = 291,600 J. To keep force below F_max = 350,000 N, the minimum crumple distance is d_min = 291,600/350,000 = 0.833 m. Design (3) with d = 0.90 m provides the minimum distance that exceeds this requirement, so F_avg = 291,600/0.90 = 324,000 N < 350,000 N threshold. Choice C is correct because it identifies design (3) as the minimum deformation distance that still meets the force requirement. Choices A and B (0.60 m and 0.80 m) would result in forces of 486,000 N and 364,500 N respectively, both exceeding the limit. Optimization strategy: (1) calculate minimum parameter needed (d_min = 0.833 m), (2) check available designs against this minimum, (3) select design meeting safety requirement with least excess (0.90 m is closest to 0.833 m while still safe), (4) verify constraints satisfied. Design (4) at 1.20 m would work but wastes resources.

Question 20

1. $F\approx 1.1\times 10^5\ \text{N}$; meets the requirement
2. $F\approx 2.8\times 10^5\ \text{N}$; does not meet the requirement (correct answer)
3. $F\approx 2.3\times 10^5\ \text{N}$; meets the requirement
4. $F\approx 3.6\times 10^5\ \text{N}$; meets the requirement

Explanation: This question tests understanding of optimizing collision safety designs using work-energy to minimize forces while meeting constraints. To minimize force for given energy KE = ½mv², solve F_max = KE/d for the actual force when using a specific distance d—we need to check if this meets the safety threshold. For this scenario with m = 1400 kg impacting at v = 18 m/s, the kinetic energy is KE = ½mv² = ½(1400)(18²) = ½(1400)(324) = 226,800 J. Using the full available crumple distance d = 0.80 m, the average force is F = KE/d = 226,800/0.80 = 283,500 N ≈ 2.84×10⁵ N. Since this exceeds F_max = 2.5×10⁵ N, the design does not meet the requirement. Choice B is correct because it correctly calculates F ≈ 2.8×10⁵ N (rounded from 2.835×10⁵ N) and identifies that this exceeds the 2.5×10⁵ N threshold. Choice A incorrectly calculates F ≈ 1.1×10⁵ N; Choices C and D give wrong force values and incorrect conclusions. Optimization strategy: (1) calculate kinetic energy KE = ½mv², (2) determine force using available distance F = KE/d, (3) compare to threshold. Since 2.84×10⁵ N > 2.5×10⁵ N, the design fails—either more crumple distance is needed (d_min = 226,800/250,000 = 0.907 m) or impact energy must be reduced.