4 ms·
E=mV^2/2 60 km/h car will kill you either way, 1 km/h car won't kill you either way, but somewhere in between there's a velocity range where a SUV will kill yo
by ajuc 4y ago
E=mV^2/2
60 km/h car will kill you either way, 1 km/h car won't kill you either way, but somewhere in between there's a velocity range where a SUV will kill you and a compact car won't - simply because of the mass difference. I'd argue it's the velocity range in which a lot of collisions happen (especially in city traffic).
- 2rsf 4y agoIt is not just the mass, the design is critical as well. Check pedestrian impact tests, on regular cars pedestrians can roll over while with tall cars or SUVs they are simply splashed against it.
- mc32 4y agoI think that band where the light car is not lethal is a narrow bad. I could be wrong but I have a feeling it's sitting up high in an SUV/truck that is causing some of this.
- vihren 4y agoIts the momentum that kills you p=mV, not the kinetic energy. So I would guess that mass ia much more important.
- Ma8ee 4y agoWhy would that be? Isn't it the energy spent tearing your body apart that is relevant?
- tzs 4y agoA typical locomotive at 1 mph has more momentum than a Ford F-150 at 45 mph, but the latter is far more likely to kill you in collision so it doesn’t look like it is momentum that kills.
- pif 4y agoYou are both wrong, It's the acceleration that kills you.
- SketchySeaBeast 4y agoAcceleration, which is caused by a transfer of energy?
- ajuc 4y agoDoubling the mass of the car assuming everything else (crumple zones etc.) stays the same - doubles the acceleration experienced by the pedestrian. X_p0 = X of the pedestrian before collision X_c0 = X of the car before collision X_p1 = X of the pedestrian after the collision X_c1 = X of the car after the collision m_ = mass v_ = velocity a_ = acceleration t = time //conservation of momentum: m_p * v_p0 + m_c * v_c0 = m_p * v_p1 + m_c * v_c1 m_p * (v_p0 - v_p1) = m_c * (v_c1 - v_c0) v_p1 - v_p0 = m_c * (v_c0 - v_c1) / m_p // a = delta v / t a = (v_p1 - v_p0)/t = m_c * (v_c0 - v_c1) / (m_p * t)
- s1artibartfast 4y agoThis is the wrong way to think of it. At some point the mass is irrelevant. If an inelastic lead object the size of a car hits you, the energy transfers is the same as if a lead 747 hits you at the same speed. Both objects have an overwhelming amount of energy and momentum compared to a nearly stationary 100 kg person. They'll transfer enough energy to bring the person up to their speed, likely mangling them in the process. Trauma depends on how quickly this energy is transferred to the individual. More to your point, both a car and a truck have more than enough energy to not be meaningfully slowed down by a pedestrian. The difference is strike location, contact area, and crumple zones to slow the transfer.
- causi 4y agoDon't forget that crashes aren't just an oblivious driver smashing into a pedestrian at full speed. Many, perhaps a majority involve some degree of braking.