Optimize Designs for Collision Safety - Grade 11 Physics

1. If all four patterns fit within the available occupant travel distance, which pattern best minimizes the peak force while still providing the required impulse?
   Answer: Pattern C, because it delivers the required impulse with a peak force of only $9\ \text{kN}$, the lowest of all four patterns. When comparing restraint-force patterns, your goal is to find which one delivers the required impulse while keeping the peak force as low as possible — because peak force is what causes injury. Impulse equals force multiplied by time, so extending the duration of a force allows you to achieve the same impulse with a lower peak. Pattern C is the correct choice. It applies a constant $9\ \text{kN}$ force over $0.20\ \text{s}$, giving an impulse of $9{,}000 \times 0.20 = 1{,}800\ \text{N}\cdot\text{s}$ ✓. Crucially, its peak force of $9\ \text{kN}$ is the lowest of all four patterns, making it the safest for the dummy. Here's why the other options fall short. Pattern A has a constant $12\ \text{kN}$ force — higher than C's peak, so it exposes the dummy to more force even though the profile is "smooth." The constant shape doesn't help if the magnitude is larger. Pattern B peaks at $24\ \text{kN}$, which is the highest of all patterns. Short duration does not reduce peak force — it actually requires a larger peak to deliver the same impulse in less time. Pattern D rises linearly to $15\ \text{kN}$, which is still higher than C's $9\ \text{kN}$. A gradual rise sounds gentler, but the peak value is what matters for injury risk, and D's peak exceeds C's. A useful strategy: when a question asks about minimizing peak force, directly compare the maximum force value in each pattern — don't be distracted by profile shape or duration alone. Longer duration helps only if the peak is actually reduced.
2. What average pressure should the redesigned airbag provide, and what is the safety advantage?
   Answer: About $30\ \text{kPa}$, distributing the required force over a larger area of the occupant. When a question links force, pressure, and area, your go-to relationship is $P = \frac{F}{A}$. The key insight here is that the required force is fixed — the airbag must still deliver $13.5\ \text{kN}$ to stop the occupant in the same time. Only the contact area changes, so pressure must adjust accordingly. Plugging in the new area: $P = \frac{F}{A} = \frac{13{,}500\ \text{N}}{0.45\ \text{m}^2} = 30{,}000\ \text{Pa} = 30\ \text{kPa}$. This confirms A — the redesigned airbag needs only about $30\ \text{kPa}$. The safety advantage is real and meaningful: the same force is now spread across a larger surface area of the occupant's body, reducing the pressure any single region of skin or tissue must absorb, which lowers the risk of localized injury. B is wrong because keeping pressure at $45\ \text{kPa}$ while increasing area would actually increase the total force ($F = PA = 45{,}000 \times 0.45 = 20{,}250\ \text{N}$), which contradicts the fixed-force requirement. C goes in the opposite direction entirely — increasing pressure proportionally with area would mean the force scales up dramatically, which is both physically incorrect and dangerous. D introduces stopping time as a variable, but the problem explicitly states stopping time is unchanged; this distractor is designed to lure you into confusing impulse with pressure. Remember: when force is constant and area increases, pressure decreases. Watch for distractors that reverse this relationship or sneak in extra variables the problem hasn't changed.
3. Which proposed change produces the greatest percentage reduction in the kinetic energy that must be absorbed during impact?
   Answer: Reduce the initial speed by $10\%$ while keeping the original braking acceleration and braking distance unchanged. Whenever you see a question about impact energy reduction, your tool is the work-energy theorem: the kinetic energy remaining at impact equals the initial KE minus the work done by braking. Specifically, $KE_{impact} = \frac{1}{2}mv_0^2 - F_{brake} \cdot d$, where $F_{brake} = ma$ and $d = 30.0\ \text{m}$. Start by finding the baseline impact KE for a 1000 kg car (a convenient reference mass): $KE_0 = \frac{1}{2}(1000)(22.0)^2 - (1000)(6.0)(30.0) = 242{,}000 - 180{,}000 = 62{,}000\ \text{J}$. Option D reduces initial speed by 10% to $19.8\ \text{m/s}$: $KE_D = \frac{1}{2}(1000)(19.8)^2 - 180{,}000 = 196{,}020 - 180{,}000 = 16{,}020\ \text{J}$. That's a 74% reduction — by far the largest. Option A reduces mass by 10% to 900 kg: $KE_A = \frac{1}{2}(900)(22.0)^2 - (900)(6.0)(30.0) = 217{,}800 - 162{,}000 = 55{,}800\ \text{J}$, only a 10% reduction. Option B increases crush distance — this changes how energy is absorbed, not how much impact KE exists. It doesn't reduce the KE that must be absorbed at all; it's a trap that confuses energy absorption capacity with energy reduction. Option C increases braking acceleration to $7.0\ \text{m/s}^2$: $KE_C = 242{,}000 - (1000)(7.0)(30.0) = 242{,}000 - 210{,}000 = 32{,}000\ \text{J}$, a 48% reduction — better than A or B, but still less than D. The key study tip: since KE scales with $v^2$, reducing speed has a squared effect on the initial energy term, making speed reduction disproportionately powerful compared to mass reduction, which only scales linearly.
4. Neglecting other horizontal forces, which conclusion best supports choosing the load-limiting belt?
   Answer: It reduces the average force from about $14.0\ \text{kN}$ to $8.4\ \text{kN}$ while producing the same impulse. Whenever you see a crash or collision question, anchor yourself to two key equations: impulse equals force times time ($J = F \Delta t$) and impulse equals change in momentum ($J = \Delta p = m \Delta v$). The central insight is that impulse depends only on the mass and velocity change — not on how long the stop takes. In both tests, the dummy goes from $18\ \text{m/s}$ to zero, so the change in momentum is identical: $\Delta p = 70 \times 18 = 1260\ \text{N}\cdot\text{s}$. Since impulse equals $\Delta p$, both belts deliver the same impulse. The difference is how that impulse is spread over time. The original belt: $F = \frac{1260}{0.090} \approx 14{,}000\ \text{N} = 14.0\ \text{kN}$. The load-limiting belt: $F = \frac{1260}{0.150} \approx 8{,}400\ \text{N} = 8.4\ \text{kN}$. A longer stopping time means a smaller average force — and less injury risk. Choice A is correct because it accurately states both values and correctly notes that the impulse is unchanged. Choice B is wrong because it claims the impulse decreases — but impulse equals $m\Delta v$, and since neither mass nor velocity change differs between tests, the impulse must be the same in both cases. Choice C is wrong because the dummy's change in momentum is identical in both tests (same mass, same $\Delta v$), so it cannot decrease. Choice D is wrong on two counts: it reverses which belt produces the larger force, and longer stopping time actually reduces force, not increases it. Your study tip: remember that impulse is fixed by the momentum change — the only design lever is time. Longer time → smaller force → safer collision.
5. Which comparison best identifies the safer design if reducing average force is the main goal?
   Answer: Design P is safer because design Q produces a velocity change and average force that are $25\%$ larger. Whenever you see a collision question involving force, remember that impulse ties everything together: $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{m \Delta v}{\Delta t}$. Since both collisions last the same time and involve identical cars, the only thing determining which bumper exerts a larger force is the magnitude of the velocity change. For Design P, the car goes from $+8.0\ \text{m/s}$ to $0$, giving $|\Delta v| = 8.0\ \text{m/s}$. For Design Q, the car goes from $+8.0\ \text{m/s}$ to $-2.0\ \text{m/s}$, giving $|\Delta v| = 10.0\ \text{m/s}$. Design Q's velocity change is $10.0\ \text{m/s}$ versus Design P's $8.0\ \text{m/s}$ — that's $25\%$ larger, meaning Design Q produces $25\%$ more average force. So Design P is the safer choice, which confirms answer C. Answer A is wrong because it focuses on final speed rather than the change in velocity. The car rebounding adds to the total momentum change — it doesn't reduce it. Answer B is incorrect because kinetic energy and impulse are different quantities; removing the same kinetic energy does not mean the same force was exerted. Answer D makes a tempting but false claim: Design Q's $\Delta v$ is $10.0\ \text{m/s}$ compared to $8.0\ \text{m/s}$, which is $25\%$ more — not $100\%$ (double). D overstates the difference. A key strategy: whenever a problem involves rebound, always compute $\Delta v$ using sign conventions. A rebound adds to the velocity change, making impacts with bounce more forceful, not gentler.
6. With the added padding, what minimum average force rating should each strap have, and would straps rated at $1.8\ \text{kN}$ be sufficient?
   Answer: Each strap needs $1.44\ \text{kN}$, so the rated straps would be sufficient under the assumptions. When a moving object is brought to rest, you can find the stopping force using the work-energy theorem: the net work done equals the change in kinetic energy. Since the force is constant, $W = F \cdot d$, so $F \cdot d = \frac{1}{2}mv^2$. Solving for force: $F = \frac{mv^2}{2d}$. With the added padding, the stopping distance increases to $0.50\ \text{m}$. Plugging in the values: $F_{total} = \frac{(20)(12)^2}{2(0.50)} = \frac{2880}{1.0} = 2880\ \text{N} = 2.88\ \text{kN}$ Since the two straps share this equally, each strap must handle: $F_{strap} = \frac{2.88}{2} = 1.44\ \text{kN}$ Straps rated at $1.8\ \text{kN}$ exceed $1.44\ \text{kN}$, so they would be sufficient — confirming answer B. Answer A uses the original $0.30\ \text{m}$ stopping distance but then incorrectly halves the result twice, arriving at an unrealistically low $0.72\ \text{kN}$. Answer C appears to use a miscalculated total force and divides incorrectly, producing $2.40\ \text{kN}$ per strap. Answer D correctly calculates the total force ($2.88\ \text{kN}$) but forgets to divide by two straps — a classic "shared load" trap that inflates the per-strap requirement. Study tip: Whenever a problem says forces or loads are shared equally, your final step is always to divide the total by the number of supports. Don't let the total force be your final answer — check what the question is actually asking for.
7. Which redesign best addresses the greater velocity change experienced by the compact car?
   Answer: Use compatible, progressively deforming front structures on both vehicles to increase the compact car's stopping time. Whenever you see a collision safety question, anchor your thinking in the impulse-momentum theorem: $F = \Delta p / \Delta t$. To reduce the force on occupants, you either reduce the momentum change or increase the stopping time. Since we can't change what happened before impact, the key lever is extending $\Delta t$. First, find the post-collision velocity using conservation of momentum. Taking east as positive: $v_f = \frac{(1500)(12) + (900)(-18)}{1500 + 900} = \frac{18000 - 16200}{2400} = \frac{1800}{2400} = 0.75 \ \text{m/s (east)}$ The compact car changes from $-18$ m/s to $+0.75$ m/s — a change of $18.75$ m/s. The SUV only changes by $11.25$ m/s. The compact car experiences the larger velocity change, so its occupants face greater forces. Option D directly targets this by using progressive crumple zones on both vehicles, which stretches out the collision duration and lowers the peak force — without weakening the passenger compartment itself. A is wrong because a completely rigid front end eliminates crumpling, which actually decreases stopping time and increases force on occupants — the opposite of the goal. B is wrong because removing mass doesn't meaningfully reduce occupant forces; the velocity change is set by the collision dynamics, not the car's mass alone. C is wrong because stiffening only the SUV's front end would transfer energy faster to the compact car, shortening the stopping time and worsening forces on its occupants. Your study tip: in any safety redesign question, ask yourself "does this change increase or decrease stopping time?" Longer $\Delta t$ always means smaller force — that's the engineering goal behind every crumple zone.
8. Which evaluation of the two designs is physically correct and most useful for improving safety?
   Answer: Design Y gives the larger impulse, so its force should be reduced by making the redirection occur more gradually. Whenever you see a question involving collisions or barrier impacts, your first instinct should be to calculate impulse as the change in momentum — a vector quantity. Direction matters just as much as magnitude. Impulse is defined as $J = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i$. Since momentum is a vector, you must account for direction when finding the change. For Design X, the car stops completely: $\Delta p = 1000(0) - 1000(20) = -20{,}000 \ \text{N}\cdot\text{s}$, giving a magnitude of $20{,}000 \ \text{N}\cdot\text{s}$. For Design Y, the initial momentum is $20{,}000 \ \text{N}\cdot\text{s}$ east and the final momentum is $15{,}000 \ \text{N}\cdot\text{s}$ north — these are perpendicular vectors. The magnitude of the change is $\Delta p = \sqrt{20{,}000^2 + 15{,}000^2} = \sqrt{625{,}000{,}000} = 25{,}000 \ \text{N}\cdot\text{s}$. Design Y delivers the larger impulse, confirming answer C. Since the interaction time is the same (0.20 s), a larger impulse means a larger average force — which engineers should reduce by extending the collision duration. A is wrong because retaining some speed doesn't automatically mean less impulse — the direction change here creates a larger vector difference. B is wrong because it assumes stopping always produces the greatest momentum change, ignoring the geometry of perpendicular vectors. D is wrong because equal initial momentum does not mean equal change in momentum — the final direction and speed differ between designs. Your strategy tip: whenever vectors change direction, never subtract magnitudes alone — always use vector subtraction or the Pythagorean theorem when the angle is 90°.
9. Compared with the thinner liner, what is the approximate effect of the thicker liner on the average force exerted on the headform?
   Answer: The force decreases by about $43\%$ because the same energy is removed over a longer distance. Whenever you see a collision question involving stopping distance, reach for the work-energy theorem: the net work done on an object equals its change in kinetic energy. Here, the padding must remove all of the headform's kinetic energy, so $F \cdot d = \frac{1}{2}mv^2$. Since the right side is fixed (same mass, same speed), a longer stopping distance $d$ means a smaller average force. Calculate the kinetic energy once: $KE = \frac{1}{2}(5.0)(6.0)^2 = 90\ \text{J}$. For the thin liner: $F_1 = \frac{90}{0.020} = 4500\ \text{N}$. For the thick liner: $F_2 = \frac{90}{0.035} \approx 2571\ \text{N}$. The ratio gives $\frac{F_2}{F_1} = \frac{0.020}{0.035} \approx 0.57$, so the force drops by about $43\%$. That confirms D is correct. A is wrong because it reverses the logic — more energy absorbed doesn't mean more force; force depends on how that energy is spread over distance. B identifies the right direction (force decreases) but makes a critical error: it treats the 1.5 cm increase as a simple percentage of the force, ignoring that you must compare the ratio of distances ($0.020/0.035$), not the raw difference. C is tempting because the kinetic energy is indeed the same, but it misses the point — equal energy over unequal distances produces unequal forces. Study tip: Whenever stopping distance changes in a collision problem, immediately write $F = \frac{KE}{d}$. Force and stopping distance are inversely proportional when energy is constant — longer stop, smaller force. This relationship is exactly why crumple zones and thick padding save lives.
10. Approximately how does the redesign change the average crush force, and why is this change beneficial?
    Answer: It lowers the force by about $38\%$ because the same kinetic energy is removed over more distance. Whenever you see a question involving a force acting over a distance to stop a moving object, reach for the work-energy theorem: the net work done on an object equals its change in kinetic energy. Here, the crush force must remove all the car's kinetic energy, so $F \cdot d = \frac{1}{2}mv^2$. Since the kinetic energy is fixed (same mass, same speed), solving for force gives $F = \frac{mv^2}{2d}$, which means force and crush distance are inversely proportional. For the original design: $F_1 = \frac{(1200)(20)^2}{2(0.40)} = \frac{480{,}000}{0.80} = 600{,}000\ \text{N}$. For the redesign: $F_2 = \frac{480{,}000}{2(0.65)} = \frac{480{,}000}{1.30} \approx 369{,}000\ \text{N}$. The ratio is $\frac{369{,}000}{600{,}000} \approx 0.615$, meaning the force drops by about 38%. The benefit is that lower average force transmitted to the passenger compartment means less injury risk — this is exactly the engineering principle behind crumple zones. That confirms B. Choice A is wrong on two counts: the kinetic energy doesn't change between designs (same speed, same mass), and the percentage cited doesn't match the calculation. Choice C inverts the relationship — more crush distance reduces force, it doesn't raise it; a longer crush zone also doesn't increase impulse since impulse depends on the change in momentum, which is identical in both cases. Choice D correctly notes that momentum change is the same, but confuses impulse with force — the same impulse spread over more time (or more distance) means a smaller average force. Study tip: When a question involves stopping a moving object, ask yourself whether it's testing force-distance (use work-energy) or force-time (use impulse-momentum). These are two different tools, and mixing them up is the most common trap on these problems.

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