Physics Flashcards: Optimize Designs For Collision Safety

What this deck covers

This deck focuses on Optimize Designs For Collision Safety, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.

How to use these flashcards

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.

Flashcard 1: Which collision type has the smallest rebound and typically reduces peak forces: more elastic or more inelastic?

Answer: More inelastic. Inelastic collisions absorb more energy, reducing rebound.

Flashcard 2: Which quantity is reduced by spreading the same force over a larger contact area: pressure $P$ or impulse $J$?

Answer: Pressure $P$. Pressure decreases with larger area; impulse stays constant.

Flashcard 3: For the same impulse $J$, which force profile has the lower peak force: longer, lower force or shorter, higher force?

Answer: Longer, lower force. Extended time spreads impulse, reducing peak force.

Flashcard 4: State the work–energy relation for energy absorbed by a stopping force over distance $d$.

Answer: $W=\Delta K$. Work done equals the change in kinetic energy.

Flashcard 5: State the linear momentum formula for a moving object of mass $m$ and velocity $v$.

Answer: $p=mv$. Momentum is the product of mass and velocity.

Flashcard 6: Which collision generally produces a larger peak force: elastic or inelastic, for similar $\Delta t$?

Answer: Elastic. Elastic collisions have higher rebound velocities, creating larger momentum changes.

Flashcard 7: What is the coefficient of restitution in terms of relative speeds along the line of impact?

Answer: $e=\frac{v_{\text{sep}}}{v_{\text{app}}}$. Ratio of separation to approach speeds measures collision elasticity.

Flashcard 8: Which design change most directly reduces peak force for the same momentum change: increase or decrease $\Delta t$?

Answer: Increase $\Delta t$. Longer collision time reduces average force for same momentum change.

Flashcard 9: What is the pressure formula relating force to contact area?

Answer: $P=\frac{F}{A}$. Pressure equals force divided by contact area.

Flashcard 10: What is $F_{\text{avg}}$ if $\Delta p=6000\,\text{kg}\cdot\text{m/s}$ and $\Delta t=0.20\,\text{s}$?

Answer: $3.0\times10^4\,\text{N}$. Direct application of $F_{avg}=\frac{\Delta p}{\Delta t}$.

Flashcard 11: State the average force in a collision in terms of momentum change $\Delta p$ and time $\Delta t$.

Answer: $F_{avg}=\frac{\Delta p}{\Delta t}$. Rearranged impulse-momentum theorem solving for force.

Flashcard 12: If the car in the previous card stops in $0.10\ \text{s}$, what is $F_{avg}$ using $F_{avg}=\frac{\Delta p}{\Delta t}$?

Answer: $F_{avg}=-2.0\times 10^5\ \text{N}$. $F_{avg} = \frac{-20000}{0.10} = -200000\ \text{N}$

Flashcard 13: What formula gives average deceleration when an object stops from speed $v$ over distance $d$?

Answer: $a_{\text{avg}}=\frac{v^2}{2d}$. Derived from kinematic equation $v^2 = v_0^2 + 2ad$ with final velocity zero.

Flashcard 14: A $1000\ \text{kg}$ car slows from $20$ to $0\ \text{m/s}$; what is $\Delta p$ in $\text{kg}\cdot\text{m/s}$?

Answer: $\Delta p=-2.0\times 10^4\ \text{kg}\cdot\text{m/s}$. $\Delta p = m(v_f - v_i) = 1000(0 - 20) = -20000$

Flashcard 15: What is $F_{\text{avg}}$ if a $1200\,\text{kg}$ car stops from $15\,\text{m/s}$ in $0.30\,\text{s}$?

Answer: $6.0\times10^4\,\text{N}$. Using $F_{avg}=\frac{m\Delta v}{\Delta t}=\frac{1200(15)}{0.30}$.

Flashcard 16: Which option gives the smaller average force for the same $\Delta p$: $\Delta t=0.10\,\text{s}$ or $\Delta t=0.40\,\text{s}$?

Answer: $\Delta t=0.40\,\text{s}$. Longer time interval reduces average force per impulse-momentum theorem.

Flashcard 17: What safety device increases collision time by stretching while restraining the passenger?

Answer: Seat belt. Elastic restraint extends collision time through stretching.

Flashcard 18: State the impulse formula in terms of average force $F_{avg}$ and collision time $\Delta t$.

Answer: $J=F_{avg}\Delta t$. Impulse is the product of average force and time duration.

Flashcard 19: State the constant-force work formula used for crash stopping distance $d$.

Answer: $W=Fd$. Work equals force times distance for constant force.

Flashcard 20: What is $F_{\text{avg}}$ if a $1500\,\text{kg}$ car stops from $20\,\text{m/s}$ over $4.0\,\text{m}$?

Answer: $7.5\times10^4\,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(4.0)}$.

Flashcard 21: What formula gives average stopping force for mass $m$ stopping from speed $v$ over distance $d$?

Answer: $F_{\text{avg}}=\frac{mv^2}{2d}$. Combines $F=ma$ with $a=\frac{v^2}{2d}$ from kinematics.

Flashcard 22: What safety device increases collision time and spreads force over a larger area in front impacts?

Answer: Airbag. Inflatable cushion extends time and distributes force.

Flashcard 23: What is the translational kinetic energy of a mass $m$ moving at speed $v$?

Answer: $K=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

Flashcard 24: State the pressure equation relating force $F$ and contact area $A$.

Answer: $P=\frac{F}{A}$. Pressure is force divided by contact area.

Flashcard 25: In a collision, which quantity is the area under a force–time graph: impulse $J$ or kinetic energy $K$?

Answer: Impulse $J$. Area under F-t graph gives total impulse delivered.

Flashcard 26: Identify the safest choice: for fixed $m$ and $v$, should a design maximize or minimize crush distance $d$?

Answer: Maximize $d$. Larger crush distance reduces average force, improving safety.

Flashcard 27: What is the momentum of an object of mass $m$ moving at speed $v$?

Answer: $p=mv$. Momentum is the product of mass and velocity.

Flashcard 28: What is the impulse–momentum theorem relating impulse $J$ to momentum change $\Delta p$?

Answer: $J=\Delta p$. Impulse equals the change in momentum for any collision.

Flashcard 29: What is the pressure if a force of $2000\,\text{N}$ acts over an area of $0.020\,\text{m}^2$?

Answer: $1.0\times10^5\,\text{Pa}$. Using $P=\frac{F}{A}=\frac{2000}{0.020}$.

Flashcard 30: Which option reduces pressure on contact for the same force: increase or decrease contact area $A$?

Answer: Increase $A$. Larger contact area distributes force, reducing pressure.

Flashcard 31: Which design change reduces average impact force for the same $\Delta p$: increase $\Delta t$ or decrease $\Delta t$?

Answer: Increase $\Delta t$. Longer collision time reduces force for same momentum change.

Flashcard 32: What is $F_{\text{avg}}$ if a $1500\,\text{kg}$ car stops from $20\,\text{m/s}$ over $2.0\,\text{m}$?

Answer: $1.5\times10^5\,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(2.0)}$.

Flashcard 33: For fixed crash energy $\Delta K$, which reduces average stopping force: larger $d$ or smaller $d$?

Answer: Larger $d$. Greater stopping distance reduces force for same energy.

Flashcard 34: What is the impulse-momentum theorem relating impulse to change in momentum?

Answer: $J=\Delta p=F_{\text{avg}}\Delta t$. Impulse equals momentum change and also equals average force times time interval.

Flashcard 35: What is the work-energy relation that connects stopping distance $d$ to average stopping force?

Answer: $W=F_{\text{avg}}d=\Delta K$. Work done by average force equals change in kinetic energy.

Flashcard 36: Identify the design feature that increases stopping distance by controlled deformation in a car front end.

Answer: Crumple zone. Deformable front structure extends stopping distance.

Flashcard 37: What equation links average impact force to change in momentum and collision time?

Answer: $F_{\text{avg}}=\frac{\Delta p}{\Delta t}$. Rearranging impulse-momentum theorem to solve for average force.

Flashcard 38: What is the kinetic energy formula used to estimate crash energy that must be absorbed?

Answer: $K=\frac{1}{2}mv^2$. Kinetic energy depends on mass and velocity squared.