# Forces in a head on collision

#### northtreker

Hello. I'm a writer going back and doing some fact checking to make an important scene more believable. It involves a character, Lux in the backseat of a jeep which has just suffered a head on collision with a bus. But I think I am doing something wrong because the numbers I get don't seem to make sense. In particular, when I calculate the forces Lux experiences I only come up with 10,650 newtons. I am trouble putting that in context. But I have found a number of sources suggesting a professional punch averages something like ~3,000 Newtons...but that's only after cursory exploration. So...I guess I need to know if this is a good metric. ~3 times an average punch doesn't seem like enough force to cause the widespread fatalities head on collisions produce...although the reason I started dredging up my old snippets of memory from high school physics was to make sure that Lux surviving was in any way plausable.

Anyway the math:

m1 = 1600kg the jeep

v1 = 20m/s jeep's initial velocity

m2 = 18000kg the bus

v2 = -25m/s the bus's initial velocity. This is supposed to be negative because the two are approaching right?

tf = 175ms :: 0.175s I looked at the force curves on sled tests (the accelerometer used to study car accidents at the curves tended to return to 0 between 150 and 200 milliseconds

ti = 0ms

m{1+2-debris} = 1600kg + 18000kg - 100 kg :: 19,500kg ((NOTE: typical weight of windows in car is 45kg)) the mangled remains of the two entangled automobiles

(m1v1)/(tf-ti) + (m2v2)/(tf-ti) = (m{1+2-debris}v3)/(tf-ti)

solving for v3

(m{1+2-debris}v3)/(tf-ti) = (m1v1)/(tf-ti) + (m2v2)/(tf-ti)

(m{1+2-debris}v3) = [(m1v1)/(tf-ti) + (m2v2)/(tf-ti)] * (tf-ti)

v3 = ([(m1v1)/(tf-ti) + (m2v2)/(tf-ti)] * (tf-ti)) / m{1+2-debris}

v3 = ([(1,600kg * 20m/s) / 0.175s] + [(18,000kg * -25m/s) / 0.175s]) * 0.175s) / 19,500kg

v3 = [(182,857kgm/s^2 + -2,571,428kgm/2^2) * .175s] / 19,500kg

v3 = (-2,388,571kgm/s^2 * .175s) / 19,500kg

v3 = -418,000kgm/s / 19500kg

v3 = -21.4m/s

_____________________________________________

F1 = [m1*(v3-v1)]/(tf-ti)

F1 = ([1,600kg * (-21.4m/s - 20m/s)] / .175s)

F1 = [(1,600kg * -41.4m/s) / .175s]

F1 = (-66,240kgm/s / .175s)

|F1| = |378,514Newtons|

F2 = [m2*(v3-v1)]/(tf-ti)

F2 = ([18,000kg * (-21.4m/s - -25m/s)] / .175s)

F2 = [(18,000kg * 3.6m/s) / .175s]

F2 = 370,285N

F(lux) = [mlux*(v3-v1)]/(tf-ti)

F(lux) = 45kg * -41.4m/s / .175s

F(lux) = 45kg * 24.12g

F(lux) = 10,650N <----Is this way too low?

F(passanger) = [mpassanger*(v3-v2)]/(tf-ti)

F(passanger) = (100kg * 3.6m/s) / .175s

F(passenger) = 100kg * 2.1g

F(passenger) = 2,060N :: a solid but not professional punch is ~3,000N

In a head on collision with a bus, while inside a jeep, my numbers are suggesting a passenger would experience 10,000 Newtons. This is significant but not catastrophic. Head on collisions are frequently fatal aren't there? Shouldn't the newtons be much higher? What am I doing wrong here?

Sorry this is such a word wall. I wanted to write this out as completely as possible in hopes somebody might catch my error(s).

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#### Khashishi

Punches can kill people though.

#### northtreker

Granted, and thank you for responding, but I was not placing a punch and a head on collision in the same category of assumed lethality. Is this a mistake?

#### SteamKing

Staff Emeritus
Homework Helper
10000 Newtons is like having someone put a 1 ton weight on you, or a little more than half the weight of the jeep.

#### AlephZero

Homework Helper
To cut down your math a bit, the jeep decelerates from 20 m/s to a stop in 0.175 s. So the average deceleration is 20 / 0.175 = 114 m/s2 or about 12g.

So for any object, the average force = 12 times its weight. In real life the acceleration will not be constant, so the maximum deceleration will be more than that.

But that it the wrong calculation if the rear seat passenger was not wearing a seat belt. In that case, he/she would be thrown forward and would decelerate faster when he/she hit something, like the person strapped into the front seat with a seat belt. And if the rear seat passenger somehow was thrown over the empty front seat, most likely he/she would go head first through the jeep windscreen and decelerate rather quickly when he/she hit the bus.

#### AlephZero

Homework Helper
Granted, and thank you for responding, but I was not placing a punch and a head on collision in the same category of assumed lethality. Is this a mistake?
You are talking about a head on crash with a relative velocity of 45 m/s or about 160 km/hour (100 mph). People don't just walk away from accidents like that. Wearing seat belts the passengers might get away with injuries like a few cracked ribs. Without seat belts, I wouldn't bet on them surviving.

#### Nugatory

Mentor
You're assuming that lux undergoes the same deceleration as the jeep. He doesn't - the jeep crumples for 175 ms as the forces of the collision slow it to zero and then accelerate it backwards. Lux, however, keeps on moving forward until he hits the backwards-moving interior of the jeep and it accelerates him backwards. Thus, Lux's change of speed happens over a shorter time and with commensurately greater force.

#### Nugatory

Mentor
Something else you might want to consider: You are (not unreasonably) assuming that the collision between bus and jeep occurs over a period of 175 ms, but there's another approach that you could take. You can calculate the time over which the collision occurs from the amount of crushing the jeep and bus experience.

First, rewrite the speeds so that the net momentum is zero. Instead of having the 1600 kg jeep moving in one direction at 20 m/sec and the 18000 kg bus moving in the other direction at -25 m/sec, think in terms of the bus moving at -3.6 m/sec and the jeep moving at 41.4 m/sec... This is still a 45 m/sec head-on collision, all we're doing is describing it from the point of view of a bystander who happens to be driving down the road at a speed of -21.4 m/sec (which is, not coincidentally, the value of $v_3$). If this is new/confusing to you, try googling for "center of mass frame"; this is a standard and valuable technique for analyzing collisions.

The interesting thing here is that after the collision, the chunk of mangled wreckage now has speed zero from this point of view. Thus, the jeep went from a speed of 41.4 m/sec to zero during the collision, so had an average speed of 21.2 meters throughout the collision (assuming linear deceleration - you can refine this assumption from the crash test curves if you want, but it won't hugely change the result).

Next question: how far did the jeep travel during the collision? Well, once the front bumper of the jeep kissed the front bumper of the bus to start the collision, the only way the jeep could travel at all would be by crushing its nose in, so the distance traveled is the amount that the two vehicles crumpled. For this collision (as AlephNull points out, 45 m/sec is a massive crash. I once saw the aftermath of a similar collision, and the smaller vehicle's engine ended up in the back seat) that amount has to be on the order of one or two meters... So we get a collision duration of 1 or 2 meters divided by 21.2 m/sec, or about 50 to 100 msecs.

You can do a similar calculation to see what happens to Lux when he collides with the interior of the jeep. Just remember that more than a few centimeters of crumple applied to his skull or rib cage will kill him.

(Be aware that there are some oversimplifications in the analysis above, most notably the assumption that the jeep does all the crumpling. That's reasonable for a bus/jeep crash).

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#### 256bits

Gold Member
Nugatory, your oversimplification is warranted with the jeep doing all the crumpling. A jeep travelling at 41.4 m/s sec and hitting a solid brick wall would result in the same damage, as this jeep travelling at 20m/s hitting a bus approaching at 25m/s sec.

I would think that any deceleration forces calculated would be that of any part of the vehicle which has not suffered any deformation from the accident. The nose has the greatest deceleration in a time approaching instantaneous. Farther rearward parts of the jeep suffer less deceleration and subsequently less force.

#### Nugatory

Mentor
I would think that any deceleration forces calculated would be that of any part of the vehicle which has not suffered any deformation from the accident. The nose has the greatest deceleration in a time approaching instantaneous. Farther rearward parts of the jeep suffer less deceleration and subsequently less force.
That is correct. The nose stops moving the instant that it touches the bumper of the bus, while everything further back gets to move forward as the structure in front collapses - the nose doesn't have any "structure in front" to spread out the deceleration time..

This is also why seat belts work, and why crumple zones are designed into modern cars. A belted passenger experiences the same deceleration as the part of the vehicle that he's fastened to, so benefits from the time it takes for the vehicle to crumple forward of the belt mounting points. An unbelted passenger is basically in the same situation when he hits the windshield as the jeep is when it hits the bus - near instantaneous change of speed means very high deceleration.

#### sophiecentaur

Gold Member
To cut down your math a bit, the jeep decelerates from 20 m/s to a stop in 0.175 s. So the average deceleration is 20 / 0.175 = 114 m/s2 or about 12g.

So for any object, the average force = 12 times its weight. In real life the acceleration will not be constant, so the maximum deceleration will be more than that.

But that it the wrong calculation if the rear seat passenger was not wearing a seat belt. In that case, he/she would be thrown forward and would decelerate faster when he/she hit something, like the person strapped into the front seat with a seat belt. And if the rear seat passenger somehow was thrown over the empty front seat, most likely he/she would go head first through the jeep windscreen and decelerate rather quickly when he/she hit the bus.
Mr Scales (A level Physics class, 1961) would have shot me if I had said that. If you say he is thrown forwards then you have to say that, on impact with the front of the bus, he moved backwards. (i.e. reference the frame of the bus). Newton 1 applies and, when considering the Earth frame, he just keeps moving and is then brought to a halt.

@northtreker
Your calculation of the forces on the jeep itself are reasonable - because there is data about the time involved in the collision, the time for the collision between the passenger and the windscreen is not known. You could assume that the front of the jeep was (near) stationary, by the time the body actually hit it. There is no way of calculating forces without similar accelerometer readings for the brain of the passenger during the impact.
There are frequent posts on PF which talk about the 'force of collision' and mostly this is usually not knowable. What can be deduced, fairly accurately, is the change in momentum during the collision. Change in momentum is know as Impulse and an impulse (simplest form) can be produced by a force times and time. Your confusion about the relative effects of a punch and impact with a windscreen demonstrates this problem.

It strikes me that the only source of reliable information for your story would be the statistics of injuries in actual impacts. Your confusion about the effect of a punch and impact with a windscreen demonstrates this problem.

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