Physics Question

[tu plang]

Nah, he's actually right. If you hit an identical car head on and you're both doing 50km/h, then it will provide exactly the right force and deformation to keep the impact point fixed (as I said before), and you'll effectively be running into a solid wall at 50km/h also. What I said about relative velocity before stands, but it's relative TO THE IMPACT FRONT.

mmm my bad, i think my two posts basically took contradictive view points. i was in a world of thermodynamic pain last night, dynamics was last week

just to vent some anger, they managed to fuck up the numbers on the exam so that our rankine-cycle power station was running at an efficiency of -0.3. i guess you could interpret it as a "trick question" but when you ask the lecturer whats going on and he says "you've done it wrong" in not so many words, you tend to think they really have fucked up. 400 odd people had the same issue.

 

[S.]

mmm my bad, i think my two posts basically took contradictive view points. i was in a world of thermodynamic pain last night, dynamics was last week

just to vent some anger, they managed to fuck up the numbers on the exam so that our rankine-cycle power station was running at an efficiency of -0.3. i guess you could interpret it as a "trick question" but when you ask the lecturer whats going on and he says "you've done it wrong" in not so many words, you tend to think they really have fucked up. 400 odd people had the same issue.

hah, you got NOTHIN on RMIT's academic fuxcellence. On Wednesday we had a Motor Vehicle Design exam, only worth 20% of the subject overall, that consisted of 13 multiple choice questions (so each question is worth about 1.5%... a fair bit!). The test was done online (though we had to be at the uni computer labs so they could check that we weren't cheating) and it was marked straight away. We were given a practice quiz that we could do from home beforehand, and being a computerised test, it too was marked straight away. Naturally, the exam was suspiciously similar to the practice quiz (read: identical but questions in a different order).

Now get this: two of the questions on the exam (that were also on the practice quiz) had DIFFERENT ANSWERS. The options you could choose from were the same, but what was considered the right answer was DIFFERENT! Not only did the two answers there contradict each other, but they also BOTH contradicted what the lecture slides/notes said!

AND... there was a THIRD question that actually had TWO correct answers (according to the notes), but obviously only one was marked correct. Biggest fuckup of a test I've ever seen in my life, and our lecturer should seriously be sacked for that. That's just utter crap.

Oh and then there was the question in our Comp Eng exam that asked about shit we learnt IN A DIFFERENT SUBJECT, LAST SEMESTER but that we didn't actually touch on THIS semester. And it was worth nearly a quarter of the entire exam's marks. Friggin ridiculous.

To anyone reading and considering studying engineering at RMIT: DON'T. There are a few really good lecturers and a metric arseload of shithouse ones.

 

[tu plang]

I sent our lecturer an email this afternoon requesting a worked solution to the question. If i dont get full marks for the question i'll definitely be paying him a visit.

By the end of the exam i had come to terms with the possibility that i just wasnt seeing it correctly (especially having put my hand up in an exam room with 750 people in it to ask the lecturer a question - him telling me i was wrong), it happens and you dont see the light until its pointed out to you, but when i found out it wasnt just me - fucking ropable. i seriously spent 20 minutes plugging the same numbers into the calculator over and over again hoping for a different number.

Would explain why he doesn't have a job in the real world, his powerplant was running off the grid as a coal burning, lake heating device.

to add insult to injury, the exam had been identical in format for the past 7 semesters, bar one question on one occasion, this semester was time for a change... after they reused last semesters mid-semester without even bothering to change the date on the cover sheet.

 

[No_Style]

This may have already been answered, but I can't be arsed to read all the responses

Basically it depends if your talking about damage to the cars or passengers. The force vectors will be very different with two cars colliding head on as they are relatively free to move (in translation and rotation etc) unlike if you were to hit a brick wall that is of infinite tensile strength and therefore cannot deform at all.

Assuming an ideal head on crash (yes an ideal head on crash ) and the above wall then the energy disipated will be twice as much if you hit the wall.

Ke = 0.5xMxV^2 will get you to that conclusion.

Now assuming you take crumple zones into account, well in the head on case you have two cars each absorbing half the total energy present in the system (assuming the perfect crash) if you hit the wall then you have 1 car absorbing twice the total energy present in the previous system = you get uber fucked up.

If you hit a stationary car however at 100km/h then during the collision both cars will move some distance in the original direction of motion of the travelling car. In this case the energy dissipation is different as the centre of mass of the entire system moves during in the collision so energy is this disipated in this process. If your gonna hit a car head on, swerve into a parked one instead (unless it's mine).

It depends if your talking about a purely elastic, non-elastic or somewhere in the middle type accident.
But ultimately in the two scenarios you outlined just the energy difference tells you that you wanna take the head on collision, let alone if you consider the impulse then the head on seems like a really good deal.

 

[Bike_Breaker]

Hey, We have been covering this in my physics class at schools recently aswell. From what i understand the force of 2 cars hitting each other head on at the same speed will resault in an eqaul force to each car. now if one car was to hit a brick wall at double the speed of the prior situation then the force applied to the car will depend on the mass of the wall(wheather the force is enough to move it) which will then determen the time it took for the car to stop. This all means that because the wall is only going to abosorb a negligable amount of force the most damge will occcur when the car hits a wall rather then head on where the force is halved!

 

[k3n!f]

This all means that because the wall is only going to abosorb a negligable amount of force the most damge will occcur when the car hits a wall rather then head on where the force is halved!

Uh...... you may want to rethink that.

 

[No_Style]

Uh...... you may want to rethink that.

No, he doesn't mate. He reached the right conclusion but through an incorrect line of reasoning.

 

[k3n!f]

No, he doesn't mate. He reached the right conclusion but through an incorrect line of reasoning.

The force isn't halved when hitting the moving car though.

The velocity of one car (at 50kmph) RELATIVE to the other car is still 100kmph. So the only difference between the two crashes (assuming in both cases everything sticks together) is that in the second case there will be two crumple zones. This means a lower rate of deceleration, which means less damage.

Or am I total off here? (Been a while since I've done any physics...)

 

[PINT of Stella. mate!]

Hey, We have been covering this in my physics class at schools recently aswell. From what i understand the force of 2 cars hitting each other head on at the same speed will resault in an eqaul force to each car. now if one car was to hit a brick wall at double the speed of the prior situation then the force applied to the car will depend on the mass of the wall(wheather the force is enough to move it) which will then determen the time it took for the car to stop. This all means that because the wall is only going to abosorb a negligable amount of force the most damge will occcur when the car hits a wall rather then head on where the force is halved!

Does anyone other than me spot the irony of your post compared with your signature?

 

[No_Style]

The force isn't halved when hitting the moving car though.

The velocity of one car (at 50kmph) RELATIVE to the other car is still 100kmph. So the only difference between the two crashes (assuming in both cases everything sticks together) is that in the second case there will be two crumple zones. This means a lower rate of deceleration, which means less damage.

Or am I total off here? (Been a while since I've done any physics...)

Yeah your off mate. Basically the key factor to look at here is the velocities.

Ignore the relative velocity crap, just work out the Kinetic energy that each vehicle has.

HEAD ON:
Kinetic Energy = 1/2 x Mass x Velocity^2

So taking mass as 1000kg for all cars, the kinetic energy of one car in the head on crash is:

(1/2)x1000x13.8888888^2 = 96450.617 J

Now double it since both cars will have the same amount of energy and you get the total energy that must be disipated in the crash:

96450.6.17 x 2 = 192901.235 J

WALL:

Right the energy that must be disipated in this crash:

(1/2)x1000x27.7777777^2 = 385.803.469 J
= 2x192901.235 J

Therefore twice the total energy must be disipated when one car hits a wall at 100km/h compared to when two cars hit each other at 50km/h.

Now take into account the fact that all the energy is directed towards a single vehicle (i.e. 1 set of crumple zones) then the impulse is going to be horrific, compared to when you have two sets of crumple zones absorbing half the energy, as is the case with the head on.

 

[bikesarefun]

OK. Start with the basics: Force = (change in momentum)/(change in time), Momentum = mass x velocity

If you look at force, the relative velocity of two objects is irrelevant. The important factor is how quickly something stops. That is going to be a question of damping to an extent. There is also the issue of energy lost due to deformation, shearing, and so on. From all of that, it ends up being a question of how strong the wall is. If the wall does not deform on impact of a 100 km/h car (which isn't going to happen with brick), the wall scenario wins. Assuming the car crumples like a car anyway. In the real world, the 100 km/h car goes straight through the wall.
If you were worrying about cars hitting cars, in one case both at 50 km/h and in another case with one stationary and one at 100, then it becomes more interesting. It becomes a question of tyre friction coefficients. And I wouldn't even start guessing...

But in the energy domain, for the 2 cars you have E=1/2mv^2 + 1/2mV^2 = mv^2, and for the wall you have E=1/2m(2v)^2 = 2v^2. Again, this only works if the wall doesn't break.

 

[S.]

OK. Start with the basics:

Force = (change in momentum)/(change in time)

Momentum = mass x velocity

In a nutshell, the relative velocity of two objects is irrelevant. The important factor is how quickly something stops. That is going to be a question of damping to an extent. There is also the issue of energy lost due to deformation, shearing, and so on.

From all of that, it ends up being a question of how strong the wall is. If the wall does not deform on impact of a 100 km/h car (which isn't going to happen with brick), the wall scenario wins. Assuming the car crumples like a car anyway. In the real world, the 100 km/h car goes straight through the wall.

If you were worrying about cars hitting cars, in one case both at 50 km/h and in another case with one stationary and one at 100, then it becomes more interesting. It becomes a question of tyre friction coefficients. And I wouldn't even start guessing...

Tyre friction coefficients have nothing to do with impact speeds - people have already stated that the impact is taking place AT THESE SPEEDS. Whether or not the car was doing 200 or 500 or 700km/h before that and then braked has nothing to do with it, you're reading well beyond what the question actually is. Relative velocity to point of impact is everything. Zero relative velocity = zero crash.

 

[No_Style]

Force = (change in momentum)/(change in time)

If you were worrying about cars hitting cars, in one case both at 50 km/h and in another case with one stationary and one at 100, then it becomes more interesting. It becomes a question of tyre friction coefficients. And I wouldn't even start guessing...

Thats not force mate, thats impulse.

The way interpret what he's saying S. is that how much resistance the parked car offers when impacted is determined by the friction of it's tyres with the road (assuming it has the handbrake or foot brakes on), not the friction of the moving car. It's easy anyway since frcitional drag is the product of the frictional coefficent for the surfaces and the normal force, from that you could work out how far the impacted car will slid when hit by another car with a determinable energy. But thats all irrelevant anyway as it hasn't been stipulated whether the collision is elastic or inelastic which will have a far bigger impact on the calulations.

 

[S.]

Thats not force mate, thats impulse.

In the hitting a stationary car scenario it's bloody easy, frcitional drag is the product of the frictional coefficent and the normal force, from that you could work out how far the impacted car will slid assuming it has it's brakes on. But thats all irrelevant anyway as it hasn't been stipulated whether the collision is elastic or inelastic which will have a far bigger impact on the calulations.

He said force = (change in m.v)/T, which by rearrangement is F.T = change in m.v which is impulse.

Whether the collision is elastic or not:
a) should be pretty obvious for a car crash. they don't tend to spring back to their original shapes, for starters.
b) isn't really relevant when determining the max force/rate of deceleration that the occupants of a car will undergo, as it only determines whether the energy converted to deformation of the structures is returned to the kinetic energy of the system.
c) profit

 

[S.]

Thats not force mate, thats impulse.

The way interpret what he's saying S. is that how much resistance the parked car offers when impacted is determined by the friction of it's tyres with the road (assuming it has the handbrake or foot brakes on), not the friction of the moving car. It's easy anyway since frcitional drag is the product of the frictional coefficent for the surfaces and the normal force, from that you could work out how far the impacted car will slid when hit by another car with a determinable energy. But thats all irrelevant anyway as it hasn't been stipulated whether the collision is elastic or inelastic which will have a far bigger impact on the calulations.

Ah yeah fair call on the parked car. However, crash accelerations can be well in excess of 20g's (into the hundreds for high speed race car crashes, and yes they can be fatal). Virtually no cars are able to accelerate (laterally or longitudinally) under grip conditions at more than 1g without aerodynamic assistance, so you're looking at <5% difference in the results. And that's assuming the handbrake works on all 4 wheels and/or the car is parked in gear or whatever... could easily be as low as 1% error.

 

[No_Style]

Yeah my screw up on the impulse thing, bikes are fun was correct.

It's not obvious though to what degree it's going to be an (in)elastic collision (what is obvious is that it's not going to be totally one or the other). The degree of elasticity will have a significant impact on the forces and time period in which the passenger experiences them.

If the cars lock together then the impulse experienced by the passenger is going to be favourable as the kinetic energy of the CoM of the system is maintained for longer period in the original direction of motion, no energy is returned to the individual vehicles and therefore to the passenger when they bounce apart.
If however they do bounce apart then velocity of any individual vehicle is not just stopped in the given time frame, it is actually reversed and in turn more energy is transfered back to the passenger.

That said, your never going to bounce off a parked car in the opposite direction, it's really only a relevant factor for an imperfect head on scenario.

 

[tu plang]

alright alright, show's over move along -> http://forums.farkin.net/showthread.php?t=69144