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B How far will a toy car travel with a small rocket motor

  1. May 26, 2017 #1
    I want to strap an estes b4-4 rocket to a toy car and calculate how far it would travel.

    What I know:

    Mass of car: 300 grams
    Rocket nominal force: 12.8 Newtons

    I think this means the rocket would accelerate my car for a=F/m = 12.8 kg * m/s^2 / 0.300 Kg which means 12.8 m/s^2/ 0.3 = 42.67 m/s^2.

    Where do I go from here, assummig negligible friction forces?
     
  2. jcsd
  3. May 26, 2017 #2

    DrGreg

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    Assuming negligible friction forces it will travel an infinite distance. Bad assumption!
     
  4. May 26, 2017 #3

    sophiecentaur

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    How long is the rocket burn? Force times time is Impulse, which is the change in momentum.
    That will allow you to work out the speed of the car at the end of the burn.
    You can then work out how far the car will travel in a given length of time.
    If you have ever used the SUVAT equations, you could also easily work out how far the car will travel as it is accelerating.
     
  5. May 26, 2017 #4
    Total impulse it says 4.3 Ns and burn time 1s.

    So should I do 42.67 m/s^2 * 1s = 42.67 m/s? And then use s = 1/2(u+v)t = 1/2(0+42.67m/s)1s = 1/2*42.67 = 21m?

    or

    do I just do something straight with 4.3 Ns? Im still getting my head around this one.

    DrGreg
    Ok I must assume friction forces...what would be the simplest calculation, just a downwards gravity based vector due to mass?
     
  6. May 26, 2017 #5

    sophiecentaur

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    yes. The unit Ns tells you that the same result is achieved with 1n for 10s or 10N for 1s. (Like Amp Hours in a battery)
    So your momentum will be Mass X velocity and mass is 0.3kg so you can work out the final speed.
    4.3 = 0.3v
    v= 14.33'm/s
    But my point about a "given length of time" is relevant because if this is about a real experiment, the track may have to be ridiculously long for a low friction car to slow to zero. Have you enough room (starting at around 30mph)?
    PS When you write out an equation, it's not usual to include the units. Just put them in the answer. Very confusing inside an equation.
     
  7. May 26, 2017 #6
    OK wait I got that's it's final velocity will be 42m/s whereas you got 14.3 m/s.

    With 42m/s using a suvat I get 21m displacement. With your final velocity it would be about 1/4 of that, so around 5m displacement.

    Where am I wrong?
     
  8. May 27, 2017 #7

    sophiecentaur

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    Not 'wrong' - just confused by what they have told you.
    If they specify the impulse as 4.3Ns then that is the safest thing to work with. They will have measured it for several motors and it's a reliable method (standard way to measure these things, I believe). If they say that the Force it produces is 12.8N, they may mean that is the maximum force during the burn. If the burn is 1s then their 12.8N figure can't be right for the whole burn time or the impulse would be 12.8Ns and it can't be both values, can it? I suspect the 1s figure, actually.
    Stick with the 4.3Ns figure because it's the easiest to measure and, short of a typo on the spec sheet, it's likely to be more reliable as it doesn't involve time.
    Edit: It's worth while imagining what goes on in the solid fuel as it burns. It will start of as a fizz, then the burn will work its way over the total area of the cylinder, then some sort of cone shape will develop (max power here, probably) and then gradually fade out. If that's to take up the whole 1s of time, it's not surprising that the average Force is only applied, effectively, for around 1/3 of the burn time. This could easily account for the 12.8:4.3 ratio of forces that the two calculations used.
     
    Last edited: May 27, 2017
  9. May 27, 2017 #8
    Yeah, I had considered that as well, that the max thrust, stated precisely as Max, would just be a brief peak compared to the get-up and come-down parts of the burn.

    OK so using s = 1/2(u+v)t and your final velocity of 14.33m/s I get:

    1/2(0+14.33)1 = 7.15m

    That's too little. These things go 750' up with a 1 ounce rocket. According to my calcs, a 1 ounce estes rocket would weigh 28g which would put its final velocity at 153 m/s vs the 14.3 m/s of my 300g rocket. Those velocities plugged into 1/2*(u+v)*t would put their respective displacements at 7.15m and 76m. Like I said, 7.15m seems like a very short distance to travel, but I havent done it. But Im sure the estes rockets go higher than 76m, 251ft.

    So I need more tweaking.
     
  10. May 27, 2017 #9

    sophiecentaur

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    Are we talking rockets now?
    This is not the equation to use, for a start. You are assuming that the rocket flight is only 1s, which obviously makes it all a bit dodgy. To find the height achieved, you can use the 'other' method by equating the KE at the start (mv2/2) with the PE at the top (mgh). This doesn't require you to find the flight time.
    h=v2/2g
    = 42m approx
    Go over all this again and you may get more sense from it.
    When using the suvat (and any other) equations, it is essential that you can guarantee that you know the values of what you are putting in and make no assumptions. There is usually a way through to an answer but you have to choose the right one. :wink:
     
    Last edited: May 27, 2017
  11. May 27, 2017 #10
    No, im still talking cars in a straight line. But Im just comparing a 7m displacement sideways vs a 209m displacement upwards. Its a big difference so it seems its too little a horizontal distance of 7m if vertically (albeit lighter rocket vs car) its supposed to climb 209m.

    What Im still trying to figure out is how long the car should travel.
     
  12. May 27, 2017 #11

    jbriggs444

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    See the first reply in this thread.
     
  13. May 27, 2017 #12
    The negligible friction forces? Ok so could you explain to me how to calculate the distance it will travel including friction? Because if anything, those will make the calculated displacement of 7.15m even smaller, which Im pretty sure wont be the case when I light it. A small wooden car with the b44 motor strapped to it will surely travel more than 7.15m in a straight line.
     
  14. May 27, 2017 #13

    sophiecentaur

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    You are in a muddle with this, you know. If it is travelling at 14.3m/s how can it only get 7m, even in the first second of its travel? Which suvat equation suits that question? (The easiest one)
    Also, if there is no friction, how will the car ever stop on a horizontal track? Think a bit more logically and don't just take random equations to give you random answers.
    Just how do you get your 7m figure?
     
    Last edited: May 27, 2017
  15. May 27, 2017 #14
    OK well in post #8 I showed how I got it using equation s = 1/2 * (u+v) * t but I guess the 1s is too little. I took it from the rocket datasheet which states Burn time as 1s. initial velocity I used zero and final velocity the 14.3m/s calculated from your post. Why is that not the right equation?
     
  16. May 27, 2017 #15

    sophiecentaur

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    But what has the time it travels during burn time got to do with how high the quoted rocket reaches? Also, that formula assumes constant acceleration. This is why we use Impulse.
    Can you please decide what it is you actually want to find out?
     
  17. May 27, 2017 #16
    Well Ive always wanted to just find out how far a car will travel if I strap that rocket motor to it.
     
  18. May 27, 2017 #17

    jbriggs444

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    The car does not stop moving just because the rocket stops burning. Same as the rocket does not stop rising just because it stops burning.
     
  19. May 27, 2017 #18
    I understand that, so what is the proper way to calculate how far my car will travel?
     
  20. May 27, 2017 #19

    jbriggs444

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    Come up with a model that describes the friction it will encounter as it travels.
     
  21. May 27, 2017 #20
    A model? Like a vector drawing or something like that? Im not a physicist but Im guessing something like this:
     

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