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by alkjsdlkjasd 1180 days ago
To quickly do the calculations:

Walking: 5 km/h * 70kg = 97 kgm/s

e-bike: 20 km/h * 90 kg = 500 kgm/s

car: 50 km/h * 2000kg = 28,000 kgm/s

4 comments

Energy is proportionate to the square of velocity. So it's:

Walking: 5 km/h * 70kg => 875 (although this is a very slow estimate for a walker) (please ignore the non-canonical units)

e-bike: 20 km/h * 90 kg => 18,000 (I think your estimate for ebike mass might be on the low side at 20kg, but whatever)

car: 50 km/h * 2000kg => 2,500,000 (and that's a fairly low speed for cars! Drunk drivers often drive faster than is wise.)

Everything else is a rounding error compared to the energy of a car.

Realistic estimates:

Walking: 6kph, 70kg => 100J (0.02% of car)

Analog bike: 18kph, 80kg => 2kJ (0.36% of car)

E-bike: 25kph, 100kg => 4.6kJ (0.82% of car)

Car: 50kph, 2000kg => 560kJ (100% of car)

Thanks. Just for the sake of fleshing out the speeding angle:

Car, 64 kph => 164% of car at 50 kph

Car, 110 kph => 480% of car at 50 kph

> Walking: 5 km/h * 70kg => 875 (although this is a very slow estimate for a walker)

In the context of this thread (a walker who’s drunk), I don’t think it’s very slow.

Also, https://en.wikipedia.org/wiki/Preferred_walking_speed:

“The preferred walking speed is the speed at which humans or animals choose to walk. Many people tend to walk at about 1.42 metres per second (5.1 km/h; 3.2 mph; 4.7 ft/s).”

https://www.sciencedirect.com/science/article/pii/S209575641...:

“The results show teenagers walk at an average speed of 1.45 m/s, young adults walk at an average speed of 1.55 m/s, middle age pedestrians walk at a speed of 1.45 m/s, older pedestrians walk at speed of 1.09 m/s, and elderly or physically disabled pedestrians walk at a speed of 1.04 m/s.”

5km/hour is about 1.4m/s; the fastest of these speeds is 5.6 km/hour.

Initial kinetic energy is not the right physical quantity to look at. Most of that kinetic energy will remain in the car/bike/…, i.e. it doesn't tell you much about how much energy will get transferred to the victim – it merely gives you a bound from above.

More details: https://news.ycombinator.com/item?id=35233887

Wouldn’t joules be more indicative of damage caused?
No. The kinetic energy of the car/bike/… doesn't tell you anything because you don't know how much energy gets transferred to the victim until you have applied momentum conservation to the elastic/inelastic problem. So, the right approach would be (in this order): Calculate momenta, calculate how much momentum gets transferred to the victim via momentum conservation, deduce the resulting change (increase) in kinetic energy of the victim. This kinetic energy will be converted into heat (= damage/injuries) in one way or another, so it's the relevant physical quantity for our considerations.

Finally, in case of an inelastic problem (likely with cars, not so likely with bikes or people), you also need to consider the energy loss during momentum transfer. Once again, this energy will become heat (= do damage), so it adds to the aforementioned increase in kinetic energy when we're interested in how much damage will be done.

Just as there is the law of conservation of energy, there is also the law of conservation of momentum. Both explain it equally well IMO.
How can mv and mv² explain the damage "equally well", when one is linear and the other is quadratic with respect to velocity?
The point is the extreme magnitude of difference between a car and a bike/person. This shows it just fine.
Momentum is always conserved. Energy is partially conserved. If it squishes, it will be more like MV, if it bounces, it's more like 1/2 mv^2.
This is awesome, thank you (and ditto for this whole chain of replies).
it's worth noting that most collisions don't happen at full speed (and combination of velocity matters a lot).