Apply Physics to Collision Design - Grade 11 Physics

1. What minimum average horizontal force must the barrier supports withstand during this test?
   Answer: Approximately $2.2\times10^5\ \text{N}$, based on the full velocity change. When a collision involves a direction reversal, impulse-momentum problems require careful attention to signs. The key concept here is the impulse-momentum theorem: the average force equals the total change in momentum divided by the collision time. The critical trap is forgetting that momentum is a vector — when the car bounces backward, its final velocity is negative relative to its initial direction. Taking the initial direction as positive, the car starts at $+18\ \text{m/s}$ and ends at $-4.0\ \text{m/s}$. The total velocity change is: $\Delta v = v_f - v_i = (-4.0) - (+18) = -22\ \text{m/s}$ The magnitude of the average force is therefore: $F = \frac{m|\Delta v|}{\Delta t} = \frac{1500 \times 22}{0.15} \approx 2.2 \times 10^5\ \text{N}$ This confirms C is correct. A is wrong because using only the rebound speed ($4.0\ \text{m/s}$) ignores the initial momentum entirely — that would apply to a ball thrown from rest, not one reversing direction. B incorrectly treats the scenario as simply stopping the car (using only $18\ \text{m/s}$), which underestimates the force because it ignores the additional momentum change caused by the rebound. D doubles both speeds, which has no physical basis — it seems to confuse elastic collision formulas with impulse calculations. Study tip: Whenever a collision involves a direction reversal, always add the magnitudes of the two speeds to find $|\Delta v|$ — the sign flip means the change is larger than either speed alone, not smaller.
2. What is the minimum effective stopping distance the barrier and vehicle structure should provide?
   Answer: Approximately $0.82\ \text{m}$, found using the constant-acceleration kinematic relation for stopping distance. When a vehicle decelerates uniformly from some initial speed to rest, the governing kinematic equation is $v^2 = v_0^2 - 2a\,d$. Setting the final velocity to zero and solving for stopping distance gives $d = \frac{v_0^2}{2a}$. That factor of 2 in the denominator is critical — it comes directly from the kinematics, not from intuition. Here, the maximum allowed deceleration is $25g = 25 \times 9.8 = 245\ \text{m/s}^2$, and the initial speed is $v_0 = 20\ \text{m/s}$. Plugging in: $d = \frac{(20)^2}{2 \times 245} = \frac{400}{490} \approx 0.82\ \text{m}$ This confirms B as the correct answer. A makes the classic error of dropping the factor of 2, computing $d = \frac{400}{490} \times 2 \approx 0.41\ \text{m}$... wait — more precisely, A uses $d = \frac{v_0^2}{a} = \frac{400}{245} \approx 1.63\ \text{m}$... actually A gives $0.41\ \text{m}$, which means it uses $\frac{v_0^2}{2a}$ but doubles the deceleration erroneously, or simply misapplies the formula. The trap is omitting the factor of 2 from the denominator of the correct formula. C doubles the correct answer, arriving at $\approx 1.63\ \text{m}$. This likely comes from forgetting to divide by 2, effectively computing $d = \frac{v_0^2}{a}$ instead of $\frac{v_0^2}{2a}$. D uses $g = 9.8\ \text{m/s}^2$ instead of $25g = 245\ \text{m/s}^2$, ignoring the "25g" specification entirely and dramatically overestimating the required stopping distance. Your study tip: always write $v^2 = v_0^2 + 2a\Delta x$ from memory before solving stopping-distance problems — that equation makes the factor of 2 impossible to forget.
3. What minimum tension rating should each strap have?
   Answer: Approximately $3.5\ \text{kN}$, accounting for each strap's backward component. Whenever you see a question involving forces from multiple angled straps, your job is to use vector components — specifically, to find how much of each strap's tension actually contributes in the direction you care about. Here, two straps each pull at $30^\circ$ from the backward direction. Each strap's backward component is $T\cos(30^\circ)$. Together, both straps must produce a total backward force of $6.0\ \text{kN}$: $2 \cdot T\cos(30^\circ) = 6.0\ \text{kN}$ Solving for $T$: $T = \frac{6.0}{2\cos(30^\circ)} = \frac{6.0}{2 \times 0.866} \approx \frac{6.0}{1.732} \approx 3.46\ \text{kN}$ That's approximately $3.5\ \text{kN}$, confirming B is correct. Choice A makes an appealing but wrong shortcut — simply dividing $6.0\ \text{kN}$ by 2 gives $3.0\ \text{kN}$, but this ignores the angle. Because each strap is angled, only part of its tension acts backward, so the actual tension must be higher than the split share, not equal to it. Choice C assigns the full $6.0\ \text{kN}$ to each strap individually, dramatically overestimating by treating both straps as if they independently carry the entire load with no geometry involved. Choice D adds both full tensions directly as scalars ($2 \times 3.5 \approx 6.9\ \text{kN}$), confusing the total rope tension with the resultant force — vector components don't add that way. Your key study tip: whenever straps, cables, or ropes are angled, always resolve forces into components first. The required tension is always greater than a naive equal-share calculation would suggest.
4. Which proposed airbag design satisfies both requirements?
   Answer: A stopping time of $0.14\ \text{s}$ and a contact area of $0.20\ \text{m}^2$. When a question gives you two separate engineering constraints, your job is to check both independently — a design only passes if it satisfies each one. The two formulas you need come from impulse-momentum and pressure: $F_{avg} = \frac{m \Delta v}{\Delta t} = \frac{(70)(14)}{\Delta t} = \frac{980}{\Delta t}$ $P_{avg} = \frac{F_{avg}}{A}$ The limits are $F_{avg} \leq 7000\ \text{N}$ and $P_{avg} \leq 35{,}000\ \text{Pa}$. For option D ($\Delta t = 0.14\ \text{s}$, $A = 0.20\ \text{m}^2$): $F_{avg} = \frac{980}{0.14} = 7000\ \text{N} \leq 7000\ \text{N}\ ✓$ $P_{avg} = \frac{7000}{0.20} = 35{,}000\ \text{Pa} \leq 35{,}000\ \text{Pa}\ ✓$ Both constraints are exactly met — D is correct. Option A ($\Delta t = 0.12\ \text{s}$, $A = 0.20\ \text{m}^2$): $F = \frac{980}{0.12} \approx 8167\ \text{N}$, which exceeds the 7 kN force limit. ✗ Option B ($\Delta t = 0.20\ \text{s}$, $A = 0.10\ \text{m}^2$): $F = \frac{980}{0.20} = 4900\ \text{N}\ ✓$, but $P = \frac{4900}{0.10} = 49{,}000\ \text{Pa}$, exceeding the pressure limit. ✗ Option C ($\Delta t = 0.14\ \text{s}$, $A = 0.10\ \text{m}^2$): $F = 7000\ \text{N}\ ✓$, but $P = \frac{7000}{0.10} = 70{,}000\ \text{Pa}$, far over the pressure limit. ✗ A common trap here is finding that the force is acceptable and stopping there — always check the second constraint. On multi-condition problems, eliminate any option that fails either requirement.
5. What average resistive force must the liner exert during compression?
   Answer: Approximately $4.0\ \text{kN}$, found by setting the work done by the liner equal to the headform's initial kinetic energy. When a question asks for a force during a compression or stopping event, your go-to tool is the work-energy theorem: the net work done on an object equals its change in kinetic energy. Here, the liner does negative work on the headform, bringing it from $v = 8.0\ \text{m/s}$ to rest. The headform's initial kinetic energy is: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(5.0)(8.0)^2 = 160\ \text{J}$ Setting the work done by the liner equal to this energy (the liner must remove all of it): $F \cdot d = KE \implies F = \frac{160\ \text{J}}{0.040\ \text{m}} = 4{,}000\ \text{N} \approx 4.0\ \text{kN}$ That confirms C is correct. Each wrong answer reflects a specific, common mistake. A divides momentum ($p = mv = 40\ \text{kg}\cdot\text{m/s}$) by displacement — this mixes up impulse-momentum with work-energy. Momentum divided by distance gives units of $\text{N}\cdot\text{s/m}$, not Newtons, so the approach is physically invalid. B uses the correct formula but accidentally halves the compression distance (using $0.020\ \text{m}$), doubling the force to $8{,}000\ \text{N}$ — wait, that's actually choice D's value. Choice B's $2.0\ \text{kN}$ result comes from applying the correct distance but doubling it instead, yielding half the right answer. D forgets the $\frac{1}{2}$ in kinetic energy, computing $F = \frac{mv^2}{d} = \frac{(5.0)(64)}{0.040} = 8{,}000\ \text{N}$, exactly double the correct answer. Study tip: Always write out $KE = \frac{1}{2}mv^2$ explicitly before plugging in numbers — the missing one-half is one of the most common arithmetic errors on energy problems.
6. Ignoring external horizontal impulse during the collision, what is the minimum total mass of ballast required?
   Answer: $2000\ \text{kg}$, because the combined mass must be three times the vehicle mass. Whenever you see a collision where objects stick together, your go-to tool is conservation of momentum. Since external horizontal forces are ignored, total momentum before equals total momentum after. Set up the equation: the vehicle's initial momentum must equal the combined momentum after impact. $m_v v_i = (m_v + m_b) v_f$ $1000 \times 15 = (1000 + m_b) \times 5.0$ $15000 = 5000 + 5m_b$ $m_b = \frac{10000}{5} = 2000\ \text{kg}$ So the minimum ballast mass is 2000 kg, which is exactly twice the vehicle mass — making the combined mass 3000 kg, or three times the vehicle mass. That confirms C is correct. A is wrong because equal masses (1000 kg ballast) would give a combined mass of 2000 kg, yielding $v_f = 15000/2000 = 7.5\ \text{m/s}$ — still too fast. Halving the speed requires doubling the total mass, not matching the vehicle mass. B is a tempting trap. Yes, the initial speed (15 m/s) is three times the final speed (5 m/s), but that ratio applies to the total mass, not just the ballast. The ballast needed is 2000 kg, not 1500 kg. D incorrectly states the ballast alone must be three times the vehicle mass (3000 kg). In reality, the combined mass must be three times the vehicle mass — so the ballast is only twice the vehicle mass (2000 kg). Study tip: In perfectly inelastic collision problems, always write out $m_1 v_1 = (m_1 + m_2)v_f$ explicitly and solve algebraically — ratio shortcuts are easy to misapply and cost you points.
7. What is the shortest possible duration of the triangular force pulse?
   Answer: $0.24\ \text{s}$, using the area of the triangular force-time pulse. When a question involves impulse and a shaped force-time graph, your key tool is the impulse-momentum theorem: impulse equals the area under the force-time curve. For a triangular pulse, that area is $J = \frac{1}{2} \times F_{\text{peak}} \times \Delta t$, not simply force times time. To find the shortest possible duration, you want to maximize the force — so you set $F_{\text{peak}}$ to its upper limit of $5.0\ \text{kN} = 5000\ \text{N}$. Solving for $\Delta t$: $\Delta t = \frac{2J}{F_{\text{peak}}} = \frac{2 \times 600}{5000} = 0.24\ \text{s}$ This confirms D is correct — it properly applies the triangular area formula and uses the maximum allowable force to minimize duration. Choice A arrives at $0.12\ \text{s}$ by using $J = F \times \Delta t$ (a rectangle, not a triangle), forgetting the factor of $\frac{1}{2}$. This halves the required time but only delivers half the needed impulse. Choice B gives $0.20\ \text{s}$, which actually produces $\frac{1}{2}(5000)(0.20) = 500\ \text{N}\cdot\text{s}$ — falling 100 N·s short of the required 600 N·s, so the restraint system fails. Choice C offers $0.30\ \text{s}$, which does satisfy the impulse requirement ($\frac{1}{2}(5000)(0.30) = 750\ \text{N}\cdot\text{s}$), but it's longer than necessary — it's not the shortest possible duration. Study tip: Whenever a force pulse is described as triangular, always use $J = \frac{1}{2}F_{\text{peak}}\Delta t$. Forgetting that factor of one-half is the most common trap on impulse problems with non-rectangular force profiles.
8. How does the average collision force on the car with Design Y compare with that for Design X?
   Answer: It is one-third as large, approximately $1.3\times10^5\ \text{N}$. Whenever you see a collision question comparing forces, your go-to tool is the impulse-momentum theorem: $F_{avg} = \frac{\Delta p}{\Delta t}$. The key insight is that if two scenarios produce the same momentum change, the force depends entirely on how long the collision lasts — longer time means smaller force. Here, both designs bring a 1200 kg car from 20 m/s to rest, so both produce the same impulse: $\Delta p = 1200 \times 20 = 24{,}000\ \text{N}\cdot\text{s}$. For Design X: $F_X = \frac{24{,}000}{0.060} = 4.0\times10^5\ \text{N}$. For Design Y: $F_Y = \frac{24{,}000}{0.18} = 1.33\times10^5\ \text{N}$. Since Design Y's stop time is three times longer (0.18 s vs. 0.060 s), its force is one-third as large — confirming answer A. Answer B is wrong because it suggests a factor of one-half, which would only be true if the time ratio were 2:1, not 3:1. Answer C inverts the relationship entirely — a longer stop time produces a smaller force, not a larger one; this reflects a fundamental misunderstanding of the impulse equation. Answer D is a clever trap: it's true that both designs produce the same momentum change, but that fact does not mean the forces are equal — it means the impulse (force × time) is equal, so a longer time necessarily implies a smaller force. As a study tip, remember this pattern: same impulse, longer time → smaller force. Crumple zones, airbags, and padded dashboards all exploit this principle. On exam questions, if the car stops completely in both cases, the impulse is always identical — focus on the time ratio to find the force ratio.
9. Which combination gives the speed of the coupled carts and the minimum energy the crush unit must absorb?
   Answer: A final speed of $2.0\ \text{m/s}$ and absorbed energy of $3.2\ \text{kJ}$. When two objects collide and stick together, you're dealing with a perfectly inelastic collision. The key principle is that momentum is always conserved, regardless of energy losses. Use this to find the final speed first, then calculate how much kinetic energy the crush unit must absorb. Finding final speed using conservation of momentum: $p_i = p_f \Rightarrow m_1 v_1 = (m_1 + m_2)v_f$ $800 \times 4.0 = (800 + 800) \times v_f$ $v_f = \frac{3200}{1600} = 2.0\ \text{m/s}$ This immediately eliminates choices A and D — a final speed of $1.0\ \text{m/s}$ or $4.0\ \text{m/s}$ violates conservation of momentum. Finding absorbed energy (the loss in kinetic energy): $KE_i = \frac{1}{2}(800)(4.0)^2 = 6400\ \text{J}$ $KE_f = \frac{1}{2}(1600)(2.0)^2 = 3200\ \text{J}$ $\Delta KE = 6400 - 3200 = 3200\ \text{J} = 3.2\ \text{kJ}$ So the correct answer is B: $2.0\ \text{m/s}$ and $3.2\ \text{kJ}$ absorbed. Choice A gets both values wrong — neither speed nor energy matches. Choice C correctly finds $2.0\ \text{m/s}$ but incorrectly states $6.4\ \text{kJ}$, which is actually the initial kinetic energy, not the energy lost. Choice D assumes the moving cart's speed is unchanged, ignoring the stationary cart entirely. Study tip: In collision problems, always solve for velocity using momentum conservation first, then calculate energy changes. Never assume momentum and kinetic energy are both conserved — only momentum is guaranteed.
10. Which beam placement best meets this goal, assuming the line of action of the impact force passes through the beam?
    Answer: Place the beam at $0.65\ \text{m}$ so the impact force acts through the center of mass, producing no torque about it. Whenever you see a question involving forces, rotation, and a center of mass, your first instinct should be to think about torque: $\tau = F \cdot d$, where $d$ is the perpendicular distance (lever arm) between the line of action of the force and the pivot point — here, the vehicle's center of mass. Rolling (tipping) during a side impact is a rotational motion. It begins when the impact force creates a net torque about the center of mass. The key insight is simple: if the force's line of action passes directly through the center of mass, the lever arm is zero, so torque is zero, and no rolling tendency is created. Placing the beam at $0.65\ \text{m}$ — exactly at the center of mass height — achieves this. The same horizontal impulse is delivered, but it produces pure translation, not rotation. That makes A the correct answer. B is wrong because placing the beam at $0.25\ \text{m}$ creates a lever arm of $0.65 - 0.25 = 0.40\ \text{m}$ below the center of mass. A large lever arm means a large torque — the opposite of the goal. C makes the same error in the opposite direction: a beam at $1.05\ \text{m}$ creates a $0.40\ \text{m}$ lever arm above the center of mass, again generating significant torque and worsening the roll tendency. D sounds clever, but it's physically incorrect — two forces applied at different heights don't cancel each other's torques; they each independently create torques that add, not subtract. Your study tip: when a question asks about reducing rotation, always ask yourself where the force acts relative to the center of mass. Zero lever arm = zero torque = no rotation.

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