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Speed of car related to momentum

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    A 1000 kg car with 4 people on it, each 75 kg, is at rest. The people jump off the car with speed 2 ms-1 relative to the car. Find the speed of the car if the people jump off :

    a. simultaneously
    b. one by one


    2. Relevant equations
    conservation of momentum


    3. The attempt at a solution

    a.
    m1u1 + m2u2 = m1V1 + m2V2

    0 = 300 * 2 - 1000 * v2

    v2 = 0.6 ms-1


    b.
    (i) when person 1 jumps off :
    m1u1 + m2u2 = m1V1 + m2V2

    0 = 75 * 2 - 1225 * v2

    v2 = 6/49 ms-1

    (ii) when person 2 jumps off :
    m1u1 + m2u2 = m1V1 + m2V2

    0 = 75 * (2 + 6/49) - 1150 * v2

    v2 = 156/1127 ms-1

    (iii) when person 3 jumps off :
    m1u1 + m2u2 = m1V1 + m2V2

    0 = 75 * (2 + 156/1127) - 1075 * v2

    v2 = 7230/48461 ms-1

    (iv) when person 4 jumps off :
    m1u1 + m2u2 = m1V1 + m2V2

    0 = 75 * (2 + 7230/48461) - 1000*v2

    v2 = 0.161 ms-1

    I calculated this with respect to the car. Do I get it right? For this kind of question, what should be taken as the reference, the car or the ground?

    Thanks


    EDIT : I just realized my mistake. The reference should be ground and when person 2, 3 and 4 jump off one by one, there should be initial momentum relative to the ground. I'll amend my calculation in the next post
     
    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 7, 2010 #2
    a. the answer is still the same

    b. assume the car moves to the left and the person jumps off to the right
    (i) when person 1 jumps off :
    0 = 75 * 2 - 1225 * v2

    v2 = 6/49 ms-1

    (ii) when person 2 jumps off :
    -(75 * 6/49) - 1225 * 6/49 = 75 * (2 + 6/49) - 1150 * v2

    v2 = 312/1127 ms-1

    (iii)when person 3 jumps off :
    - 75*312/1127 - 1150 * 312/1127 = 75*(2 + 312/1127) - 1075 * v2

    v2 = 0.474 ms-1

    (iv)when person 4 jumps off :
    - 75 * 0.474 - 1075 * 0.474 = 75 * (2 + 0.474) - 1000 * v2

    v2 = 0.73 ms-1

    So the final speed of the car = 0.73 ms-1

    Do I get it right? Thanks
     
    Last edited: Feb 7, 2010
  4. Feb 8, 2010 #3
    Sorry fo bumping...

    Can someone please check my work? Thanks :smile:
     
  5. Feb 8, 2010 #4
    I agree with this part. Here's my picture (jumping off to the left):
    http://img695.imageshack.us/img695/9197/mv1.jpg [Broken]




    I don't agree with these. My picture, for part (ii) is,
    http://img246.imageshack.us/img246/6475/mv2q.jpg [Broken]

    when I use a frame at rest, although, it would be easier to make a new inertial frame for each jump, that moves at the constant speed of the previous jump. Basically, you can use a frame at rest to get [itex] \rm V_1[/itex], then a frame moving at [itex] \rm V_1[/itex] to get [itex] \rm V_2[/itex],...etc.

    The jumpers each give [itex]\rm \frac{mv}{3m+M}[/itex], [itex]\rm \frac{mv}{2m+M}[/itex], [itex]\rm \frac{mv}{m+M}[/itex], and [itex]\rm \frac{mv}{M}[/itex] respectiviely, which can be summed to find the final speed.
     
    Last edited by a moderator: May 4, 2017
  6. Feb 8, 2010 #5
    Hi dr_k

    So for the case when person 2 jumps off, the equation should be (based on your picture):

    (3*75 + 1000) * 6/49 = - (2 - 6/49) * 75 + 1150 * V2 ??


    And I don't understand about this part "it would be easier to make a new inertial frame for each jump, that moves at the constant speed of the previous jump. Basically, you can use a frame at rest to get V1 , then a frame moving at V1 to get V2, ... etc"

    Can you please explain more? Thanks
     
  7. Feb 8, 2010 #6
    Yep. Drawing the picture is 1/2 the problem.

    Well, solving for [itex] \rm V_2[/itex], in the second picture, gives [itex] \rm V_2 = V_1 + \frac{m v}{2 m + M}[/itex], where [itex] \rm V_1 = \frac{m v}{3 m + M}[/itex]. Basically the objects in the the (i) initial and (f) final picture are all traveling at [itex] \rm V_1[/itex]. Instead of carrying these terms, you could just make an inertial frame moving at constant speed [itex] \rm V_1 [/itex], and then remember to sum up the speed of the frames at the end. For example, the second picture above now looks like

    http://img199.imageshack.us/img199/4534/frames1.jpg [Broken]

    This frame says [itex] \rm V' = \frac{m v}{2 m + M} [/itex]. To get the speed for an observer at rest, just add [itex] \rm V_1 + V' [/itex] . Now do this as many times as required to get all four masses off.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 12, 2010 #7
    Hi dr_k

    Maybe I get your point. I'll try it first. Thanks a lot !! :smile:
     
  9. Feb 13, 2010 #8
    Good Morning Songoku,

    I was just looking at this problem, and I noticed that I didn't read the problem carefully enough. There is a subtle point that I neglected. The problem states "The people jump off the car with speed 2 ms-1 relative to the car". Relative to the car implies that I missed something. In part "(a) a. simultaneously", the picture should look like this:

    http://img18.imageshack.us/img18/4534/frames1.jpg [Broken]

    This picture correctly shows both 4m and M moving with [itex] \rm V_1[/itex], and therefore v is "relative" to the car.

    This implies

    [itex] \rm 0 = M V_1 - 4m (v-V_1)[/itex]

    so

    [itex] \rm V_1 = \frac{4mv}{M+4m}[/itex] .

    Putting the numbers in gives [itex] \rm V_1 = .4615 m/s[/itex] for part (a).

    Give me a few minutes, and I'll write something up for part (b).
     
    Last edited by a moderator: May 4, 2017
  10. Feb 13, 2010 #9
    For part (b):

    http://img688.imageshack.us/img688/7726/frames2.jpg [Broken]

    Here v is "relative" to the car, and we get

    [itex] \rm V_1 = \frac{mv}{4m+M}[/itex]

    Now I make a new inertial frame, moving in a straight line at constant speed [itex] \rm V_1[/itex] ,

    http://img708.imageshack.us/img708/8674/frames3.jpg [Broken]

    where

    [itex] \rm V_2 = \frac{mv}{3m+M}[/itex]

    Do this two more times, with new inertial frames, to get the last two masses off. The final speed of the car is therefore

    [itex] \rm V_{car} = V_1+V_2+V_3+V_4 = \frac{mv}{4m+M}+\frac{mv}{3m+M}+\frac{mv}{2m+M}+\frac{mv}{m+M} = .5078 m/s[/itex]
     
    Last edited by a moderator: May 4, 2017
  11. Feb 22, 2010 #10
    Hi dr k

    Ah I see my mistake for neglecting "relative" . Thanks !! :smile:
     
  12. Feb 22, 2010 #11
    I missed it as well! But I think we've got it covered now.
     
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