Apply Physics to Collision Design - Physics

  1. What quantity equals the loss of kinetic energy in an inelastic collision (as heat, sound, deformation)?
    ΔK=Kafter−Kbefore<0\Delta K=K_{\text{after}}-K_{\text{before}}<0. Energy is lost to heat, sound, and permanent deformation.
  2. What is the impulse–momentum theorem relating impulse J⃗\vec{J} to momentum change?
    J⃗=Δp⃗\vec{J}=\Delta \vec{p}. Impulse equals the change in momentum by Newton's second law.
  3. Which design change reduces peak force for the same Δp\Delta p: increase or decrease Δt\Delta t?
    Increase Δt\Delta t. Since J⃗=F⃗avgΔt=Δp⃗\vec{J}=\vec{F}_{\text{avg}}\Delta t=\Delta \vec{p}, larger Δt\Delta t reduces F⃗avg\vec{F}_{\text{avg}}.
  4. In a 1D collision, what does e=0e=0 indicate about the post-collision motion?
    Objects move together: v1f=v2fv_{1f}=v_{2f}. Zero restitution means perfectly inelastic collision.
  5. Find FavgF_{\text{avg}} if Δp=12 N⋅s\Delta p=12\,\text{N}\cdot\text{s} and Δt=0.06 s\Delta t=0.06\,\text{s}.
    Favg=200 NF_{\text{avg}}=200\,\text{N}. Favg=ΔpΔt=120.06=200 NF_{\text{avg}}=\frac{\Delta p}{\Delta t}=\frac{12}{0.06}=200\,\text{N}.
  6. Which option increases stopping distance and reduces peak force: rigid bumper or crumple zone?
    Crumple zone. Crumple zones deform to extend collision time and reduce peak force.
  7. State the formula for linear momentum of an object in collision analysis.
    p⃗=mv⃗\vec{p}=m\vec{v}. Momentum equals mass times velocity vector.
  8. Which collision type has e=0e=0 with objects sticking together after impact?
    Perfectly inelastic collision. Maximum energy loss, objects move together after impact.
  9. Identify the correct relation between force and momentum change during a collision.
    F⃗net=Δp⃗Δt\vec{F}_{\text{net}}=\frac{\Delta \vec{p}}{\Delta t}. Average force equals momentum change divided by time interval.
  10. What principle states that total momentum of an isolated collision system stays constant?
    Conservation of momentum: ∑p⃗before=∑p⃗after\sum \vec{p}_{\text{before}}=\sum \vec{p}_{\text{after}}. In isolated systems, no external forces change the total momentum.
  11. What design change reduces average impact force for the same momentum change?
    Increase collision time Δt\Delta t. Longer impact time reduces force by F=ΔpΔtF=\frac{\Delta p}{\Delta t}.
  12. Find the average deceleration magnitude if a car slows from 2525 to 0 m/s0\,\text{m/s} in 0.50 s0.50\,\text{s}.
    ∣a∣=50 m/s2|a|=50\,\text{m/s}^2. a=ΔvΔt=0−250.50=−50 m/s2a=\frac{\Delta v}{\Delta t}=\frac{0-25}{0.50}=-50\,\text{m/s}^2.
  13. Find the stopping distance if a 1000 kg1000\,\text{kg} car at 20 m/s20\,\text{m/s} is stopped by Favg=2.0×105 NF_{\text{avg}}=2.0\times 10^5\,\text{N}.
    d=1.0 md=1.0\,\text{m}. W=Fd=ΔKW=Fd=\Delta K; d=12mv2F=200000200000=1.0 md=\frac{\frac{1}{2}mv^2}{F}=\frac{200000}{200000}=1.0\,\text{m}.
  14. State the kinetic energy formula used in collision energy calculations.
    K=12mv2K=\frac{1}{2}mv^2. Kinetic energy is half mass times velocity squared.
  15. A 2.0 kg2.0\,\text{kg} cart at 3.0 m/s3.0\,\text{m/s} sticks to a 1.0 kg1.0\,\text{kg} cart at rest; find vfv_f.
    2.0 m/s2.0\,\text{m/s}. Using conservation: (2.0)(3.0)+(1.0)(0)=(2.0+1.0)vf(2.0)(3.0)+(1.0)(0)=(2.0+1.0)v_f, so vf=2.0 m/sv_f=2.0\,\text{m/s}.
  16. State the impulse-momentum theorem used to design collision cushioning.
    J⃗=Δp⃗=F⃗avgΔt\vec{J}=\Delta \vec{p}=\vec{F}_{\text{avg}}\Delta t. Impulse equals momentum change and force times time.
  17. Find the kinetic energy lost when m1=1 kgm_1=1\,\text{kg} at 4 m/s4\,\text{m/s} sticks to m2=3 kgm_2=3\,\text{kg} at rest.
    ΔK=6 J\Delta K=6\,\text{J} lost. Ki=8 JK_i=8\,\text{J}, Kf=2 JK_f=2\,\text{J}, so ΔK=−6 J\Delta K=-6\,\text{J}.
  18. If the stopping distance is doubled at the same initial speed, how does average force change?
    Average force halves: Favg∝1dF_{\text{avg}}\propto \frac{1}{d}. Work-energy theorem: Favgd=ΔKF_{\text{avg}}d=\Delta K, so doubling dd halves FavgF_{\text{avg}}.
  19. Which collision type has e=1e=1 and conserves kinetic energy (idealized)?
    Perfectly elastic collision. No energy lost, objects bounce apart with e=1e=1.
  20. Which option best reduces peak force in a bumper: larger Δt\Delta t or smaller Δt\Delta t for the same Δp\Delta p?
    Larger Δt\Delta t. Larger Δt\Delta t reduces force since F=ΔpΔtF=\frac{\Delta p}{\Delta t}.
  21. What is the work–energy relation used to estimate average stopping force over distance dd?
    Favgd=ΔKF_{\text{avg}}d=\Delta K. Work done by stopping force equals change in kinetic energy.
  22. What collision type has objects stick together and share a final velocity?
    Perfectly inelastic collision. Maximum kinetic energy is lost when objects stick together.
  23. Find the final velocity of a 2 kg2\,\text{kg} cart initially at rest hit by a 1 kg1\,\text{kg} cart at 6 m/s6\,\text{m/s}; they stick.
    vf=2 m/sv_f=2\,\text{m/s}. (1+2)vf=1(6)+2(0)(1+2)v_f=1(6)+2(0); 3vf=63v_f=6; vf=2 m/sv_f=2\,\text{m/s}.
  24. State the formula for linear momentum of an object moving with velocity v⃗\vec{v}.
    p⃗=mv⃗\vec{p}=m\vec{v}. Momentum equals mass times velocity, a vector quantity.
  25. State the coefficient of restitution formula using relative speeds along the line of impact.
    e=∣v2f−v1f∣∣v1i−v2i∣e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}. Ratio of relative separation speed to relative approach speed.
  26. Find ee if approach speed is 10 m/s10\,\text{m/s} and separation speed is 3 m/s3\,\text{m/s} along the impact line.
    e=0.30e=0.30. e=separation speedapproach speed=310=0.30e=\frac{\text{separation speed}}{\text{approach speed}}=\frac{3}{10}=0.30.
  27. What is the kinetic energy formula used to check energy changes in collisions?
    K=12mv2K=\frac{1}{2}mv^2. Kinetic energy is half mass times velocity squared.
  28. State the work-energy relation often used to size a crumple zone distance.
    W=ΔKW=\Delta K. Work done equals the change in kinetic energy.
  29. What is the coefficient of restitution ee in terms of relative speeds along the impact line?
    e=∣v2f−v1f∣∣v1i−v2i∣e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}. Ratio of separation to approach speeds along impact line.
  30. Find the impulse when a 2 kg2\,\text{kg} cart changes velocity from +3+3 to −1 m/s-1\,\text{m/s} in 1D.
    J⃗=Δp⃗=−8 N⋅s\vec{J}=\Delta \vec{p}=-8\,\text{N}\cdot\text{s}. Δp=mΔv=2((−1)−(+3))=−8 N⋅s\Delta p=m\Delta v=2((-1)-(+3))=-8\,\text{N}\cdot\text{s}.
  31. In a force–time graph, what physical quantity equals the area under the curve during impact?
    Impulse J⃗\vec{J}. Area under force-time curve equals impulse by definition.
  32. What collision type conserves both momentum and kinetic energy (ideal case)?
    Perfectly elastic collision. No kinetic energy is lost to heat or deformation in ideal elastic collisions.
  33. Two equal masses collide head-on elastically; one is initially at rest. What happens to their speeds?
    They exchange speeds (moving one stops). For equal masses in elastic collision, velocities are exchanged.
  34. State the formula for impulse from an average force F⃗avg\vec{F}_{\text{avg}} over time Δt\Delta t.
    J⃗=F⃗avgΔt\vec{J}=\vec{F}_{\text{avg}}\Delta t. Impulse is the product of average force and time duration.
  35. State the 1D momentum conservation equation for two objects with masses m1,m2m_1,m_2 and speeds vv.
    m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}. Total momentum before equals total momentum after in 1D collisions.
  36. What condition must be met to treat a collision system as isolated for momentum conservation?
    Net external impulse is negligible: J⃗ext≈0\vec{J}_{\text{ext}}\approx 0. External forces like friction must be negligible compared to collision forces.
  37. What condition must be true to treat a collision system as isolated for momentum analysis?
    Net external impulse is negligible: J⃗ext≈0\vec{J}_{\text{ext}}\approx 0. External forces must be negligible compared to collision forces.
  38. Find the common final velocity when m1=1 kgm_1=1\,\text{kg} at 4 m/s4\,\text{m/s} sticks to m2=3 kgm_2=3\,\text{kg} at rest.
    vf=1 m/sv_f=1\,\text{m/s}. (m1+m2)vf=m1v1(m_1+m_2)v_f=m_1v_1; 4vf=44v_f=4; vf=1 m/sv_f=1\,\text{m/s}.

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