Question

A 2060 kg railroad flatcar, which can move with negligible friction, is motionless next to a platform. A 241 kg sumo wrestler runs at 5.3 m/s along the platform (parallel to the track) and then jumps onto the flatcar.
(a) What is the speed of the flatcar if he then stands on it?
(b) What is the speed of the flatcar if he then runs at 5.3 m/s relative to it in his original direction?
(c) What is the speed of the flatcar if he then turns and runs at 5.3 m/s relative to the flatcar opposite his original direction?

Answer 1

Answer

given,

mass of railroad = 2060 Kg

mass of sumo wrestler = 241 Kg

speed = 5.3 m/s

a) using conservation of momentum

     m v = (m + M) V

     241 x 5.3 = (241 + 2060)V

    $V = \dfrac{1277.3}{2301}$

      V = 0.556 m/s

b) Let v be the relative speed

  Again using conservation of momentum

  241 x 5.3 =241 x (5.3+v)  + 2060 v

  241 x 5.3 =241 x 5.3+241 x v + 2060 v

  2301 v = 0

       v = 0 m/s

c)  the speed of the flatcar if he then turns and run sat 5.3 m/s relative to the flat car opposite his original direction

  241 x 5.3 =241 x (v - 5.3)  + 2060 v

  241 x 5.3 = - 241 x 5.3+241 x v + 2060 v

  2301 v = 2554.6

      v = 1.11 m/s