Car hitting a tree - calculating the force?

  • Thread starter fawk3s
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  • #1
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So a car hits a tree.

F=ma

So lets say it was driving at 90 km/h = 25 m/s, and after the crash lets say its speed was 0 m/s.
a=v-v0/t

But how could I calculate the t?

This is not a homework question.

Thanks in advance,
fawk3s
 

Answers and Replies

  • #2
Doc Al
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There's no simple way to get the time. It depends on too many factors.
 
  • #3
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The cars crumple zones crumples. The tree bends and breaks. In fact maybe the tree is knocked down and the car keeps going with some velocity less than the initial velocity. But lets say it is a big solid tree so that the car will come to rest after the tree plows four feet (about 1.2 meters) into the crumple zone what does this get us?

I get about 0.096 sec to stop and a=260m/sec^2= about 26.6gees lets hope there is an air bag.

I used d=0.5*a*t^2 + v*t
and v=a*t
with d=1.2m and v=25m/sec
 
  • #4
In case of calculating time in a collision, it is a complex job

actually, when we are considering ideal rigid bodies, or point masses, we assume the collisions to be instantaneous


but in the practical world, whenever a collision takes place, a part of the interacting bodies compress, and the compressed part in turn exert a resistive force


the time has to be calculated experimentally by the help of high speed cameras
thus we use the quantity impluse
 
  • #5
342
1
The cars crumple zones crumples. The tree bends and breaks. In fact maybe the tree is knocked down and the car keeps going with some velocity less than the initial velocity. But lets say it is a big solid tree so that the car will come to rest after the tree plows four feet (about 1.2 meters) into the crumple zone what does this get us?

I get about 0.096 sec to stop and a=260m/sec^2= about 26.6gees lets hope there is an air bag.

I used d=0.5*a*t^2 + v*t
and v=a*t
with d=1.2m and v=25m/sec

There are so many things confusing me here.

Firstly, the d=1.2 is taken pretty randomly, but then, let it be. (Though I was actually thinking about a situation where the tree stands its ground.)
Then how do you solve the first equation and for what exactly?
And how can you tell that the tree is going to be moving at 25m/s after the time t?

Thanks in advance,
fawk3s
 
  • #6
246
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fawk3s, the tree does hold its ground the car crumples. The tree is moving with an initial velocity of 25m/s with respect to the car because that is what you stated in the problem. At the end of the event the tree and the car are standing still with respect to each other so 25m/s=t*a so a=25/t m/sec^2 substitute into the first equation and solve for t given we have d=1.2m. Yes, 1.2m is semi-random but seems a typical value from the experimental data of seeing cars after impacts.

I guess you wanted the force well we have a=26g so what is the weight of the car? 1000lbs times 26g force in lbs is 26,000 lbs.
 
  • #7
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Hi guys, i could be wrong but see if this make sense?

Assuming the tree keep standing after the impact of the car, the kinetic energy involved are changed into other form of energy by deforming the car(heat, sound etc...).
Let's say the car weight m kg, we have m and we have v, so we can get the kinetic energy, so if we know how much distance did the car move after impact before it comes to a total stop (how much the car crumbled inwards), if the OP wants the force, then we can use the work = force x distance formula to get the force applied? Does it look right to you guys?

And this lead me to ask a question myself, i hope u guys dont mind :D. If i have 2 cars (with same mass and speed), one moving from east, and one moving from west, then when they collide, there will be no net momentum (meaning both car with stop), but what about the force involved? in my mind both cars should have the same deformation inward (assuming they are exact same car), but if i go Newtons 3rd law, car A exert force on car B, car B exert same force back on car A, then the force cancels out, and no net force = no damage which doesn't make sense, what am i thinking wrong?

Thanks!
 
  • #8
Doc Al
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And this lead me to ask a question myself, i hope u guys dont mind :D. If i have 2 cars (with same mass and speed), one moving from east, and one moving from west, then when they collide, there will be no net momentum (meaning both car with stop), but what about the force involved? in my mind both cars should have the same deformation inward (assuming they are exact same car), but if i go Newtons 3rd law, car A exert force on car B, car B exert same force back on car A, then the force cancels out, and no net force = no damage which doesn't make sense, what am i thinking wrong?
Several confusions here:
(1) The forces involved in Newton's 3rd law act on different objects, so they don't directly 'cancel out'. What determines what happens to car A is the force on car A. Just because car A also exerts an equal force on car B doesn't make the force on car A disappear.
(2) Net force = 0 is not the same as no force! Imagine an elephant stepping on your foot. The net force on your foot is zero: The downward force of the elephant is balanced by the upward force from the ground. But your foot is still crushed.
 
  • #9
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thx for the answer doc, but something on the 3rd law still doesn't make sense tho, can you please clarify it for me?
if i draw the free body diagram of the whole system (car A and B), then it will look something like this (ignoring gravity): Fa=> Car A; Car B <=Fb. Even though the forces cancel out, there will still be damages due to the fact that there were forces on the individual cars.

If i only draw the FBD of car A, it will look like this: Fa=> car A <=Fb to a (Fb to a = force from car b to car a) And the FBD for car b alone will be vice versa.

But then i am not too sure what do you mean by the forces involved in newton's 3rd law acts on different objects? Because to me, it actually looks like the forces act on both cars?

Thanks!

(and is my approach in finding the F in my previous post correct? thanks again! :))
 
  • #10
Doc Al
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But then i am not too sure what do you mean by the forces involved in newton's 3rd law acts on different objects? Because to me, it actually looks like the forces act on both cars?
The force exerted by car A on car B acts on car B; The equal and opposite force exerted by car B on car A acts on car A.

(and is my approach in finding the F in my previous post correct?)
If you know the distance traveled by the car's center of mass as it comes to rest (or the time it takes for the car to come to rest) then you can calculate the average force on the car as you suggest.
 
  • #11
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Thanks Doc! :)
 
  • #12
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fawk3s, the tree does hold its ground the car crumples. The tree is moving with an initial velocity of 25m/s with respect to the car because that is what you stated in the problem. At the end of the event the tree and the car are standing still with respect to each other so 25m/s=t*a so a=25/t m/sec^2 substitute into the first equation and solve for t given we have d=1.2m. Yes, 1.2m is semi-random but seems a typical value from the experimental data of seeing cars after impacts.

I guess you wanted the force well we have a=26g so what is the weight of the car? 1000lbs times 26g force in lbs is 26,000 lbs.

Oh, right. Think I get it now. I dont know what exactly confused me but I couldnt connect these things the first time.
Thanks!
 

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