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This deck focuses on Evaluate Collision Design Solutions, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.
Study Evaluate Collision Design Solutions in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Which evaluation tool best organizes criteria, constraints, and scores across designs?
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A decision matrix (scoring table). Systematically compares designs against all criteria and constraints.
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This deck focuses on Evaluate Collision Design Solutions, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: A decision matrix (scoring table). Systematically compares designs against all criteria and constraints.
Answer: Option B (e=0.80). Higher e means more elastic (closer to 1).
Answer: Choose the highest weighted total score. Sum weighted scores for each design; pick the highest.
Answer: p=mv. Momentum equals mass times velocity.
Answer: W=ΔK. Work done equals change in kinetic energy for stopping analysis.
Answer: Δt=0.20s. Doubling time halves average force per Favg=ΔtΔp.
Answer: K=21mv2. Standard kinetic energy formula for calculating crash energies.
Answer: Increases by a factor of 4. Kinetic energy depends on v2, so doubling v quadruples K.
Answer: Favg=ΔtΔp. Rearranging impulse-momentum theorem to isolate average force.
Answer: e=0. Objects stick together with no relative velocity after impact.
Answer: p=mv. Linear momentum equals mass times velocity for collision analysis.
Answer: Elastic collision. Elastic collisions conserve kinetic energy; inelastic ones lose it.
Answer: Total momentum, ptotal, is conserved. In isolated systems, momentum before equals momentum after collision.
Answer: Design A (Δt=0.20s). Twice the time means half the average force.
Answer: Design A (B violates the 1000N constraint). B exceeds the maximum allowed force.
Answer: Design A (Design B violates m≤2.0kg). B exceeds the maximum allowed mass.
Answer: J=Δp=FavgΔt. Impulse equals momentum change and average force times time interval.
Answer: Objects share a final velocity: v1f=v2f. Objects stick together, moving with same final velocity.
Answer: e=1. Objects bounce apart with no kinetic energy loss.
Answer: Design A (ΔK=8J absorbed). More energy absorbed: 10−2=8J vs 10−7=3J.
Answer: Larger Δt. Reduces average force via Favg=ΔtΔp.
Answer: m1v1i+m2v2i=m1v1f+m2v2f. Initial total momentum equals final total momentum.
Answer: Minimize Fmax. Lower peak forces reduce injury risk to occupants.
Answer: e=∣v1i−v2i∣∣v2f−v1f∣. Ratio of relative separation to approach velocities.
Answer: Larger Δt gives smaller Favg. From Favg=ΔtJ, larger time gives smaller force.
Answer: Criterion: desired performance; constraint: nonnegotiable limit. Criteria are goals to optimize; constraints are absolute requirements.
Answer: K=21mv2. Kinetic energy equals half mass times velocity squared.
Answer: e=1. Unity restitution means perfect bounce with no energy loss.
Answer: J=FΔt=Δp. Impulse equals force times time, which equals change in momentum.
Answer: "Mass must be ≤2.0kg". Mass limit is a hard requirement (constraint), not a goal.
Answer: Total momentum is conserved. No external forces means system momentum stays constant.
Answer: e=0. Zero restitution means no relative separation after impact.
Answer: Design A (e=0.10). Lower e means less bounce-back velocity.
Answer: Favg=ΔtΔp. Longer collision time reduces force for same momentum change.
Answer: m1v1i+m2v2i=m1v1f+m2v2f. Total momentum before equals total momentum after.
Answer: Increase Δt. Longer collision time spreads force over more time, reducing peak.
Answer: Reject or redesign it; it is not acceptable. Constraint violations disqualify designs immediately.