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B Rear Car Impact

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  1. May 19, 2017 at 10:30 PM #1
    Hi all,
    I am currently self teaching physics and have come across something i don't know how to do.

    There are 2 cars, Cars 1 and 2, travelling at a velocity of 60km/h (16.67m/s) in the eastern direction with a gap of 5 metres between each other. Car 2 is in front of Car 1. Car 2 decelerates to 50km/h (13.89m/s) over 1 second (deceleration of roughly 2.78m/s^2). Car 1 knows that since it is going 2.78m/s faster it will close the 5m gap in roughly 1.8 seconds, so it steers north to avoid collision (assume it does not/cannot brake). Hence, Car 1 makes a partial impact with the rear of Car 2 as it is steering north. Out of the 1.8m rear car width of Car 2, Car 1 makes impact with 40cm of it.

    My question is how does this 40cm impact affect the velocity in the eastern direction of Car 1?

    Here are some more values I have.

    Both the cars weigh 1820kg (same car).
    Rolling friction of 0.43m/s^2, (779.27 N)
    Have not come up with a steering force yet, if it is relevant just assume a value.

    Here is a diagram.
    Basic Map.png Basic map 2.png


    Furthermore, if anymore values are required to solve for this, just assume them. (Most of these numbers are made up by myself as the only stimulus I was given is a police report based on the incident.)

    And ignore the other features of the images i have uploaded, they are for the rest of the scenario.

    Thanks all :)
     
  2. jcsd
  3. May 19, 2017 at 11:22 PM #2

    haruspex

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    It is not possible to use that known impact width without knowing the mechanical properties of the car bodies.
    If they are perfectly rigid and unbreakable then the only difference from a full width impact will be the extent to which each car rotates in the impact. For that, you need to know the moments of inertia about a vertical axis.
    More realistically, the less the impact width the more the bodies will crumple locally, reducing the overall shock.
     
  4. May 19, 2017 at 11:45 PM #3
    Thanks for the quick reply haruspex,

    I'll elaborate my scenario a bit more. After this impact, Car 1 crashes into a tree coming to a stop. For this i have derived a function where my I am required to input the velocity in the eastern direction to determine the velocity, and displacement in the resultant direction of the two vectors. So i am required to know the eastern velocity after this impact to determine the force Car 1 hits the tree with.

    Yes that does make sense, but of course cars aren't perfectly rigid and unbreakable. Lets say i state a crumple distance one of the cars have after impact, would this be considered a measure of the mechanical properties of the cars? Furthermore, how would i be able to use this to determine Car 1's velocity in the eastern direction?

    If this is not possible (which I don't doubt) may you suggest another suitable approach to this situation?

    Thanks
    Paiwand
     
  5. May 19, 2017 at 11:57 PM #4

    haruspex

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    I've thought of a way we might be able to avoid that.
    Suppose that the impulse exerted over the duration of the impact is J. If we estimate the moments of inertia, and use the mid point of the collision width as the point of impact, we can calculate the resulting changes in velocity and rotation rates. These must combine such that the cars remain in contact at first. That should give us enough to eliminate J and calculate those velocities.
     
  6. May 20, 2017 at 12:26 AM #5
    Since i'm self teaching physics I haven't looked into moments of inertia much yet so could you further explain what it is and its application?

    From my quick google search I've gathered the formula for a moment of inertia to be I=mr^2 , would the impact between the cars not decrease the radius from the centre of the car(s) due to crumpling? How would that be factored into calculating the moments of inertia?
     
  7. May 20, 2017 at 12:33 AM #6

    haruspex

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    The impulse J is along the line of relative motion of the cars. The moment of this impulse is found by multiplying it by the distance from the car's mass centre to the line of action of the impulse. I.e. it is only the lateral distance that matters. This means it will not change as the crumpling progresses.
    It will change as the cars rotate, though. We will have to assume the crumpling completes before the cars have rotated much. We can then treat them as a single body, rotating together.
    At that point, we have to consider what the tyres are doing on the road. Even though the cars are not braking, the rotation will mean there is sideways skidding.
     
  8. May 20, 2017 at 12:46 AM #7
    Since Car 1 (blue) is already steering north (left) would this skidding not be very minimal? The tires of the car have already rotated due to the steering meaning it is (after impact) roughly facing the resultant direction.

    Would the line of action of the impulse start at the car's mass centre and end at the centre of impact width?

    Furthermore, may you please provide some formulas for these, it would greatly benefit me with this scenario.

    Thank you very much for the help so far :)
     
  9. May 20, 2017 at 1:06 AM #8

    haruspex

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    The front tyres of the rear car won't skid, true, but the rear ones will. The rear of the car will swing out to the left.
    I think the front car's tyres won't skid much, but the front right tyre might a bit.
    No.
    As I wrote, the impulse is along the line of travel of the vehicles, so its line of action is parallel to the front car. Its moment arm for the front car will be half the car's width minus half the impact width.
    The rear car is a bit more complicated because it is at a slight angle, but I think we can neglect that.
    (You have posed a complicated problem. Several simplifications will be needed.)
    If each car has a moment of inertia I about a vertical axis through its centre, and the moment arms are x, then the cars will each rotate at rate ω=Jx/I.
    At the same time, if the velocities just before impact are u1 and u2, and just after impact the contact point is moving at speed v, and the cars have the same mass, we have, just after impact:
    mass centre of front car is moving at speed v-xω
    mass centre of rear car is moving at speed v+xω
    mu1 - J = m(v+xω)
    mu2 + J = m(v-xω)

    Because of the sideways skid of the rear wheels of the rear car, it will tend to lose its rotation. The front car won't at first. So the front car will turn through a greater angle, maybe ending sideways across the road, or even pointing back the wrong way. (I believe I have observed this in movie crashes.) Of course, by that time the front car will be skidding on all four wheels. With a violent enough crash it will roll.
    The driver of the rear car may have a better chance of steering out of the skid.
     
  10. May 20, 2017 at 1:50 AM #9
    Thank you for the help so far haruspex but,

    I am still struggling to understand these concepts, and am finding it hard to visualize and apply what you've have told me (cons of self teaching physics). If you don't mind could you use some arbitrary numbers and carry out these calculations? Even a diagram will help if you cannot do the calculations. I think it will help me understand it a bit better.

    Regardless of my understanding, I really appreciate the help.
     
  11. May 20, 2017 at 2:25 AM #10

    haruspex

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    Go 55 seconds into .
    (There may be later ones in this compilation that are also relevant.)
     
  12. May 20, 2017 at 3:40 AM #11
    After attempting how you told me to approach this scenario, I've come up with:

    Assuming no friction of any form

    The velocity of Car 1 being 14.53 m/s after the moment of impact
    and
    The velocity of Car 2 being 16.02 m/s after the moment of impact

    This is using the values i stated at the start of the thread.

    Does this sound reasonable to you?

    Thank you so much for everything :)
     
    Last edited: May 20, 2017 at 4:17 AM
  13. May 20, 2017 at 4:28 AM #12

    haruspex

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    How do you get that? What I wrote would imply car 2 ends up with a lower mass centre speed than car 1.
     
  14. May 20, 2017 at 4:46 AM #13
    How would the the rear car be faster than front car after impact?
     
  15. May 20, 2017 at 5:02 AM #14

    haruspex

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    Look at the equations I wrote in post #8. Where the cars are in contact, they have the same velocity. But both cars are rotating clockwise. That means the mass centre of the rear car is going a little faster, while that if the front car is going a little more slowly.
     
  16. May 20, 2017 at 5:34 AM #15
    The rear car is applying a steering force North (left, anti clockwise), so how could it rotate in a clockwise direction?
    This is getting confusing for me.
     
  17. May 20, 2017 at 5:48 AM #16

    haruspex

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    The impact is against the right hand side of its front. The car may be pointing left at impact but it will not be significantly rotating left. Immediately after impact it will be rotating right substantially. Even while it is still pointing a bit left that will mean its centre is moving forward faster than the impact area.
     
  18. May 20, 2017 at 5:57 AM #17
    If the rear car was to have a significantly larger mass or velocity, would there still be a clockwise rotation on the rear car?
     
  19. May 20, 2017 at 6:40 AM #18

    haruspex

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    The larger the velocity the greater the impulse and the greater the rotation.
    If its mass is concentrated around its periphery that will increase its moment of inertia and reduce its rotation.

    Edit: looking at the video I posted, the sudden twist to the left of the front car (the collision is mirror image of your set-up) is very clear. The slight twist to the left of the rear car is more subtle.
     
    Last edited: May 20, 2017 at 6:54 AM
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