Optimize Designs for Collision Safety - Physics

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    Seat belt. Elastic restraint extends collision time through stretching.

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

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

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

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

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

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

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

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

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

    Airbag. Inflatable cushion extends time and distributes force.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    Crumple zone. Deformable front structure extends stopping distance.

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

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

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

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

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