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Impulse and Acceleration for a Recoiling Car

  1. Jan 14, 2015 #1
    1. The problem statement, all variables and given/known data
    A car with a mass of 1000 kg crashes into a rubber wall with an initial velocity of 30 m/s and recoils at 15 m/s. The change in momentum occurs over 1 meter. Find the force that the car experiences and time for which this force acts.

    2. Relevant equations
    v22 - v12=2ad
    F=ma
    j=Ft

    3. The attempt at a solution
    I started by plugging into the first equation and got:

    (-15)2 - 302= 2(a)(1)
    Solving gives an acceleration of -387.5 m/s^2, but I was thinking about this and there is a problem. As the recoil velocity approaches the magnitude of the initial velocity (e.g., 30 -> -30), the left-hand side of the first equation approaches zero. How can I account for the fact that the velocities are in different directions, meaning that the acceleration should be even larger?
     
  2. jcsd
  3. Jan 14, 2015 #2

    BiGyElLoWhAt

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    I think you want to look at magnitudes with the first equation. Velocity is a vector, and if you have a car going to the right (positive velocity), it then accelerates and then has a velocity to the left (negative velocity), the acceleration is obviously non zero, correct?

    I would start with your impulse equation. See where that gets you.
     
  4. Jan 14, 2015 #3
    mv2-mv1=mat => 45=at

    I know I could use a system of equations, but I was told that this is not necessary for this question. Obviously I am aware that the acceleration would be non-zero in a recoil situation, so that is where my confusion begins. Does that kinematics equation not apply to situations where the direction of velocity occurs?
     
  5. Jan 14, 2015 #4

    BiGyElLoWhAt

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    It does, but read my first sentence again. V1 and V2 are magnitudes, not the actual velocity.
     
  6. Jan 14, 2015 #5
    OK, so they are magnitudes of 15 and 30, but recoil situations require direction. The change in velocity is 45, but somehow I don't think that is the way to do it (to just do 45^2). I am not sure how to use that equation if they are supposed to be magnitudes but they are vectors.
     
  7. Jan 14, 2015 #6

    BiGyElLoWhAt

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    So you need to add the sign in, or use your thinking power to put the sign on the acceleration. Let me give you a couple of examples. The initial velocity is 10m/s in the +x and the final velocity is 12 m/s in the +x. This happens over 1m. What's the acceleration? Now the initial velocity is 10 m/s in the +x and the final is 8 m/s in +x, over 1 m. What's the acceleration?
     
  8. Jan 14, 2015 #7

    BiGyElLoWhAt

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    You don't do (45m/s)^2. You use the equation as is, but you need to understand which quantities (if any) in that equation are vectors.
     
  9. Jan 14, 2015 #8
    Right the acceleration is bigger in the 12/10 case because the difference is v^2 is 44 instead of 36. So is it just -(15^2) -(30)^2? That just seems like cheating.
     
  10. Jan 14, 2015 #9

    BiGyElLoWhAt

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  11. Jan 14, 2015 #10

    BiGyElLoWhAt

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    And also, that's right, but not the point I was trying to make. Which direction is the acceleration?
     
  12. Jan 14, 2015 #11

    haruspex

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    This is a seriously flawed question. It implies that the force is constant over the whole process, which is clearly not possible. The recoil distance must equal the impact distance, so if the force were constant KE would be fully restored.
    Also, it's not clear what "The change in momentum occurs over 1 meter" means. The change in velocity is from 30 m/s in one direction to 15m/s in the other direction. Does it mean that the car was brought to a stop (before recoiling) in a meter or in half a meter?
    That equation won't apply here. It is only for constant acceleration.
    You might think that you can apply it if you do it in two stages: deceleration and recoil. It is possible that this is what the question poser intends you to do. On that basis you might say impact force * distance = KE absorbed, recoil force * distance = KE restored. This will give you two forces differing by a factor of 2.
    But it does say it's a rubber wall. This implies a spring-like process, the force increasing linearly with extent of compression. Here you can apply SHM equations from the moment that impact starts to the point of maximum compression. You can do likewise on the recoil stage, but the modulus will be lower (since work is not conserved).
    This will lead to a correct answer, but the force will be a function of time, not a constant.
     
  13. Jan 14, 2015 #12

    BiGyElLoWhAt

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    I think that it is. I see no place where it implies that energy should be conserved (clearly it's not) I also see no place where it implies momentum should be conserved (clearly it's not). Also, based on the fact that op wrote under relevent equations ##I=Ft##, not ##i =\int Fdt## as well as some kinematics, I would say this is meant to be treated as a constant acceleration collision. My 2 cents.
     
  14. Jan 14, 2015 #13

    haruspex

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    Suppose it is constant acceleration a. Let the compression distance be x (probably intended to be 1m). Speeds are u to start with, 0 at max compression, v at finish.
    Compression phase: u2 - 02 = 2ax.
    Recoil phase: v2 - 02 = 2ax
    Conclusion: |u| = |v|.

    Yes, there is a way to get answer by assuming constant acceleration, but since that same assumption leads to a contradiction the laws of logic say that you can get any answer you like from it! If p is false then ##p \implies q## is true for any q.
     
  15. Jan 14, 2015 #14

    BiGyElLoWhAt

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    I never said the problem wasn't broken. I myself have had numerous broken problems in class. There's really not a whole lot you can do about it in my experience, especially if the professor made the assignment him/herself.

    What you are asking op to do is to find a funcion of time for the force, then find the limits of the integral that fits the problem set. This is not something you do in a class that uses kinematic equations as a basis.
     
  16. Jan 14, 2015 #15

    haruspex

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    I'm not sure you've completely grasped what I'm saying. The problem statement contains a contradiction. In consequence, it will be possible to obtain conflicting answers by entirely logical means.
    But let's start with this... how do you interpret this statement:
    ?
    Edit: interpreting the question as asking for a single constant force, it contains a contradiction.
     
    Last edited: Jan 14, 2015
  17. Jan 14, 2015 #16
    Thanks for all the replies. Even though the problem has flaws, I did get something from thinking about it. Without knowing how it was really intended, if I were to assume that the car acts as a spring and compresses a half of a meter and then springs back over a half a meter, then is it right to say that I could calculate acceleration (and thus the first half of the time) over that half of a meter with v1=30 and v2=0, and then repeat to calculate the second half of time with the respective equations? Or further, if it recoiled at 30 m/s like an ideal spring, I could approach it similarly in two pieces?
     
  18. Jan 14, 2015 #17

    haruspex

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    Yes, but I'm not at all sure whether it means that or 1m each way.
    Yes, but in either of two ways. You can pretend it's constant acceleration or assume SHM.
    Either way, you will not get a single constant force for the whole process. Maybe that's ok, and the questioner will be happy with the force expressed as a function of time.
     
  19. Jan 14, 2015 #18
    Great, thanks. Honestly I was just more curious than anything. Appreciate the insight.
     
  20. Jan 15, 2015 #19

    BiGyElLoWhAt

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    I interpreted that to mean the entire distance traveled to be 1m throughout the collision. So i guess 1/2 meter in and 1/2 out with energy loss. I however didn't pay attention to the details, as i also interpreted it as a kinematic question. Just out of curiosity head, is your class trig based or calculus based?
     
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