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Rolling cart

  1. Oct 18, 2004 #1
    A 1000 kg cart is rolling to the right at 1.70m/s . A 70.0 kg man is standing on the right end of the cart.What is the speed of the cart if the man suddenly starts running to the left with a speed of 7.00 m/s relative to the cart?


    really boggles my mind, should i be using

    =(m1 +m2)Vf - m1(Vi)/m2
    or whats going on?!
     
  2. jcsd
  3. Oct 19, 2004 #2
    (m1+m2)vb=m1*v1a+m2*v2a
    we put the positive direction as right
    (m1+m2)vb/m1-m2*v2a/m1=v1a

    Im not sure how you define the masses but i think you missed (m1 +m2)Vf/!!m2!!
     
  4. Oct 19, 2004 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Conservation of momentum. Initially, the 1000kg cart is rolling to the rightare 1.7 m/s while the man is standing still, so the total momentum is (mass)(velocity) 1700 kgm/s.

    Let v be the speed of the cart after the man starts running. When the man (mass 70.0 kg) runs to the right at 7.00 m/s, relative to the cart, his speed relative to the ground is v- 7 and so his momentum, relative to the ground, is 70(v- 7)= 70v- 490 kgm/s. The momentum of the cart is 1000v and so the total momentum is now
    1070v- 490 . By conservation of momentum, that must be
    1070v- 490= 1700=> 1070v= 2190 so v= 2190/1070= 2.05 m/s to the right.
     
  5. Oct 20, 2004 #4
    Isn't the initial momentum a result of the mass of both the cart and the man, 1070 x1.7 or 1819 to the right ? If so, leads to 2.15 m/s
     
  6. Oct 20, 2004 #5

    Doc Al

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    Staff: Mentor

    Yes, the initial momentum must include the man moving with the cart.
     
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