What you’ll explore
Why a car that crumples saves the people inside. In this lab you why a car that crumples saves the people inside and test your intuition against a real physics simulation.
Challenge levels
- How far does it crush?. A 1,500 kg car hits a wall at 15.6 m/s (35 mph). Its front crushes at a steady 300 kN. How far does the front crush, in cm?
- How long is the crash?. Same car, same 300 kN crush force. How long does the car take to stop once the bumper touches the wall, in milliseconds?
- The person inside. The car stops over its 61 cm crush. The belt and airbag let the occupant move another 30 cm inside the cabin. What is the occupant’s deceleration, in g?
- Design the front. Tune the crush force so the occupant stays at or under 20 g, with a 70 cm crumple zone and 15 cm of belt ride-down. Too stiff hurts the occupant; too soft crushes into the passenger cell.
- What if it’s heavier?. Same front (300 kN crush force, 70 cm crumple zone), but now a 2,500 kg SUV hits the wall at 15.6 m/s. How many cm does the crush intrude into the passenger cell?
Key ideas
Show the explanations (spoilers: try the levels first)
- All of the car’s kinetic energy, ½ × 1,500 × 15.6² ≈ 183 kJ, is absorbed by the 300 kN crush force: d = 183,000 J ÷ 300,000 N ≈ 61 cm.
- The wall must remove m·v = 1,500 × 15.6 = 23,400 N·s of momentum at 300,000 N: 23,400 ÷ 300,000 = 0.078 s, about 78 ms. Same answer from 2d/v = 2 × 0.61 ÷ 15.6.
- The occupant goes from 15.6 m/s to 0 over 0.61 + 0.30 = 0.91 m: a = v² ÷ 2(d + s) ≈ 134 m/s² ≈ 13.7 g, while the car body takes about 20 g. Crumple zone and restraints add their distances.
- Stop within the 70 cm crumple zone (F ≥ ½mv²/L ≈ 261 kN) but keep d + s ≥ 62 cm so the occupant stays under 20 g (F ≤ about 388 kN). Real cars aim for a front soft enough to crush, backed by a passenger cell stiff enough not to.
- The SUV carries ½ × 2,500 × 15.6² ≈ 304 kJ. At 300 kN that takes 1.01 m of crush, 31 cm more than the 70 cm zone. Heavier cars need stiffer fronts (or longer ones) to stop in the same space.
About the model
- A car of mass m drives straight into a rigid, immovable wall (a full-width frontal barrier test; 15.6 m/s = 35 mph is the US NCAP speed). The clock starts 80 ms before contact, so the car is already moving at t = 0 by design.
- The front structure crushes at one steady force F, so the car decelerates uniformly: work–energy gives the crush distance d = m·v²/(2F), impulse–momentum gives the stop time T = m·v/F = 2d/v. Real crash pulses rise, peak and oscillate; a constant force is the idealized design target (a "square-wave" pulse), so the peak here equals the average.
- The occupant is held by an ideal belt + airbag that starts slowing them at contact and lets them move a ride-down distance s relative to the cabin, so they stop uniformly over d + s: a = v²/(2(d + s)). Real belts have slack and stretch; at very low speeds a real belt would not use its full ride-down.
- If the crush distance d is longer than the crumple zone L, the overrun d − L is shown as intrusion into the passenger cell; the model keeps the same crush force there for simplicity (a real passenger cell is much stiffer). Deceleration in g = a ÷ 9.81 m/s². The 20 g target in level 4 is illustrative, not a legal limit.
- The occupant is a crash-test form (no person): torso and head move together; neck, chest and bounce-back are not modeled. Tires, braking before impact and the wall’s give are ignored.
Every number on screen comes from the simulation, and each lab passes automated checks against the physics formulas. Spotted a mistake? Tell us.