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Given Bob is standing in the middle of a road when hit by a truck, and making some basic assumptions here, I'm trying to figure out how far Bob would fly. I'm making the assumption that all of the momentum and kinetic energy of the truck is transferred to Bob, sending Bob flying uncrushed off into the air, and Bob travels in a perfect arc with no air resistance until he hits the ground at the end of his fall.

The horizontal distance traveled by Bob should be able to be calculated as follows:

$$d_x = \frac{p^2\sin{\theta}\cos{\theta}}{m_{Bob}^2 \ g}$$

Given the truck is loaded to 25,000 kg (~55,000 lbs) and traveling at 100 km/h (~62 mph), Bob is roughly 80 kg (~176 lbs), and the surface hitting Bob is about 10° off vertical, we can plug in our information as follows:

$$d_x = \frac{\left(25000\ \textrm{kg} * 100\ \textrm{km/h} \right)^2\sin{10°}\cos{10°}}{\left(80\ \textrm{kg}\right)^2 * 9.8\ \textrm{m/s^2}}$$

$$d_x = 1,314.89 \ \textrm{km}$$

This gives us the incredible travel distance of 1,314.89 kilometers, or 817 miles. So if Bob was hit by a truck in London he might land around the Vatican in Rome.

London to Rome (Courtesy of www.distancecalculator.net/from-rome-to-london)

Clearly, something is wrong here. Where is my mistake?

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closed as off-topic by Kyle Kanos, Jon Custer, Yashas, Bill N, John Rennie Mar 7 '17 at 16:03

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    $\begingroup$ Why would the truck's momentum be transferred fully? $\endgroup$ – Nephente Mar 5 '17 at 20:37
  • $\begingroup$ Is that 25,000 kg the mass or weight of the truck ? $\endgroup$ – StephenG Mar 5 '17 at 20:49
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    $\begingroup$ Thank goodness - I'm not the only one annoyed at the price of airline tickets. $\endgroup$ – JMLCarter Mar 5 '17 at 21:11
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    $\begingroup$ Whether or not you do the calculations correctly, the physics assumptions are ridiculous. A 250 ton truck stops instantaneously after hitting a human? $\endgroup$ – Bill N Mar 6 '17 at 19:33
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    $\begingroup$ This reminds me of the time I tried to swat a fly in mid-air and my hand came to a complete stop while the fly broke the sound barrier (and the window) in response. $\endgroup$ – Devsman Mar 6 '17 at 21:24
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Initially just consider Bob bouncing off the windscreen in a horizontal direction and see what might happen, then you can look at Bob being thrown up at an angle.

I'm making the assumption that all of the momentum and kinetic energy of the truck is transferred to Bob

is not possible.

Truck before collision mass $M$ speed $V$ and Bob after collision mass $m$ and speed $v$.

All momentum transferred $\Rightarrow v = \frac M m V$

All kinetic energy transferred $\Rightarrow v = \sqrt{\frac M m} V$


Although it will not be one, assume an elastic (kinetic energy conserved) collision and see what you get when also apply the conservation of momentum.

You will find that because the truck is so much more massive than Bob, Bob rebounds off at approximately twice the speed of the truck.

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  • $\begingroup$ It's also worth pointing out that if the collision is inelastic, Bob's kinetic energy after the collision will always be lower than it would be in the elastic case. $\endgroup$ – Michael Seifert Mar 6 '17 at 21:15
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Physics isn't my forte, but consider that very rarely does a truck hit a man and stop outright. And you can confirm this by checking how fast you're assuming the person is moving after energy exchange, and then checking how long this would give Bob in contact with the truck.

The truck here weighs around 1000 times what Bob does, so whereas the truck was going 100km/h, here you'd have Bob going 100,000km/h at peak velocity, so if his acceleration is linear he'd have half that velocity on average during acceleration. Assuming Bob is around 50cm radius from surface to center of mass, it would take him 50cm / 50,000km/h to fly out of range, or 36 microseconds. Be reasonable now, does stopping a truck in 36 microseconds sound reasonable? Plus, stopping distance for a large truck is usually around 100 meters, so if brakes designed to stop a truck as fast as possible take 100 meters to do it, how can an 80kg fleshy water balloon expect to stop it in 50cm, 0.5% of that stopping distance?

Don't bother with the math here to see the problems, logic topples all.

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  • $\begingroup$ But this doesn't explain WHY $\endgroup$ – TheEnvironmentalist Mar 5 '17 at 23:28
  • $\begingroup$ @TheEnvironmentalist You are not asking "why" anywhere in your question... Your question is: "Clearly, something is wrong here. Where is my mistake?". And Tim is directly explaining what is wrong (unrealistic) in your procedure. $\endgroup$ – Steeven Mar 6 '17 at 20:48
  • $\begingroup$ @Steeven He pointed out that there is a mistake, but not what the mistake is $\endgroup$ – TheEnvironmentalist Mar 7 '17 at 7:45

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