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Force on the car in a collision

  1. May 31, 2015 #1

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    1. The problem statement, all variables and given/known data
    Car A with mass 1000 kg is accelerating at 1ms-2 at car B.
    Car B with mass 1000 kg is accelerating at 1ms-2 at car A.
    What is the force on car B at the collision?

    2. Relevant equations
    F = ma
    Newton's third law

    3. The attempt at a solution
    Car A force:
    F = ma
    F = 1000 x 1
    F = 1000 N

    Car B force:
    F = ma
    F = 1000 x 1
    F = 1000 N

    Then using newton's third law, since every action has an equal opposite reaction force, if car B was colliding with a stationary object then the force on car B after colliding is 1000 N. But since car A is travelling at car B with 1000 N of force then that also has to be added into car B's total collision force. So therefore car B has a total collision force of 2000 N.

    Is the answer right? Also is my explanation correct?
     
  2. jcsd
  3. May 31, 2015 #2
    From Newton's third law, each force involves two objects, e.g. weight is exerted on an _apple_ by the _Earth_. What are the objects involved in each of the two forces you've identified?
     
  4. May 31, 2015 #3

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    I'm not really sure what you are asking but the objects involved is the 2 cars, car A and car B. Is this what you are asking for?
     
  5. May 31, 2015 #4
    Not really, but maybe I'm misinterpreting the problem. If car A is accelerating at 1 m/s2 "at car B", what is causing that acceleration? Also, what kind of force is it (e.g. gravitational, normal, friction, tension, ...)?
     
  6. May 31, 2015 #5

    haruspex

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    The question makes no sense.
    We are not told the velocities at collision, so cannot assess the momenta. Neither are we told the duration of the collision, so there is no way to determine the forces during impact (and they would not be constant during the impact anyway).
    Where does this question come from?
     
  7. May 31, 2015 #6

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    It was a question my friend had. Can't you explain this through newton's third law of motion?
     
  8. May 31, 2015 #7
    You can, but you need a lot more detail first. That's why I was asking about the objects and types of forces involved, and even that's just the beginning.
     
  9. May 31, 2015 #8

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    If it was a collision between 2 cars that are travelling at each other in a straight line and they have a "perfect collision". Is this enough information?
     
  10. May 31, 2015 #9

    haruspex

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    In your attempt to do so, you used an acceleration before impact and related it to a force after impact. That does not constitute a valid application of the law. It has to be the net force producing the acceleration. The accelerations here are produced, I assume, by the engines.
     
  11. May 31, 2015 #10

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    How about if I reworded the question a bit: Two cars are travelling at each other with 1000 N of force each. What will be the impact force of one of the cars?
     
  12. May 31, 2015 #11
    What does it mean to travel with 1000N of force? Sorry for all the questions, but these are things you need to think through.
     
  13. May 31, 2015 #12

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    No problems, its my bad that I did not explain the question properly. What I mean by 1000N of force is that the car has a force of 1000N. Eg. Car A has a force of 1000N, car B has a force of also 1000N but in a opposite direction to car A so they will collide together. Is this enough information?
     
  14. May 31, 2015 #13
    Okay, but what does "a force of 1000N" mean in this question? Usually we speak about cars having a velocity (speed and direction) when talking about collisions (e.g. when the police take measurements to determine each car's speed before a collision and decide whether to give a ticket for dangerous driving). Sometimes we'll talk about a car's weight in Newtons (e.g. a 9810N car, which would have a mass of 1000 kg) but that's not so common.

    It's a fantastic idea to create your own questions to help understand how the physics ideas fit together. You still have a lot to review, but you'll come out of it much farther ahead!
     
  15. May 31, 2015 #14

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    Force of 1000N means the car has a force of 1000N applied to it. So the car has a net force of 1000N.
     
  16. May 31, 2015 #15
    But how does that relate to the second car?
    Since 2 objects are involved in every force, what other object is exerting that 1000N force on the car?
    What direction is the 1000N force in?
    Is that before the collision, during the collision, or after the collision?
     
  17. May 31, 2015 #16

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    Car A and car B both have 1000N of force applied to it. The force are in opposite direction.
    The engine is exerting the force on the car. The force from the engine on the car is greater than 1000N. The 1000N is from the force from the engine subtracted by the friction.
    The 2 cars are travelling in opposite directions.
    Eg. Car A has 1000N applied to it to the left. Car B has 1000N applied to it to the right. So the 2 cars are travelling at each other.
    This is before the collision. Disregard air resistance so that before the collision, right up to it the net force of each car is 1000N. I am trying to find what is the force after the collision.
     
  18. May 31, 2015 #17

    haruspex

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    This is no different from your original statement, that the cars have mass 1000kg and are being accelerated at 1m/s2. The longer that state lasts before collision, the greater the speed at impact, so the greater the forces during the impact.
    Suppose the time from rest to impact is t1, that during that time each car is propelled towards the other with a net force F, and the duration of the impact is time t2. The average force during impact will be ##F\frac{t_1}{t_2}##.
     
    Last edited: May 31, 2015
  19. May 31, 2015 #18

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    How about if the force of each car right before the collision is 1000N? Could you then explain it with Newton's third law? My understanding is that if a car has a force of 1000N right before a collision and it hits a brick wall then the force on the car immediately after the collision would be 1000N because of the third law. The question is to help me understand what happens if the car collided with a object that is moving at it? Will the force on the car immediately after the collision be the force of the object (it collided with) + the reaction force of the car?

    I apologise for my questions, I have only started physics recently so I am not completely sure as to how to explain a question.
     
  20. May 31, 2015 #19

    haruspex

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    As Scott has tried to explain to you, this doesn't mean anything.
    An object has momentum, an object has KE, but an object does not "have" force. Force is something one object exerts on another.
    The way you are using the term 'force' here sounds like what you are intuitively thinking of is momentum. The cars will have momentum just before impact. The units of momentum are kg m/s, or N s (Newton-seconds).
     
  21. May 31, 2015 #20

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    Ok I think I get where you are coming from now. The purpose of the question is for me to understand Newton's third law, that every action has an equal opposite reaction.

    How I understand is if a car hits a stationary object with 1000N of force it will have 1000N of reaction force. Is this the correct way to think about it?

    So if my understanding above is correct then what happens if the object that the car hits was not stationary? So if the car hits the object with 1000N of force and the object hits the car with also 1000N of force, will the reaction force on the car be 2000N? I got this from:
    Total force = reaction force of the car + the action force of the object.

    The equation is something I though of and is not an equation I learnt from school so it might be wrong.

    Also thanks for helping me even though my question is incorrect.
     
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