A 45.0 kg girl is standing on a 162 kg plank. The plank, originally at rest, isf

Question

A 45.0 kg girl is standing on a 162 kg plank. The plank, originally at rest, isfree to slide on a frozen lake, which is a flat, frictionlesssurface. The girl begins to walk along the plank at a constantvelocity of 1.44 m/s relative to theplank.
a) What is her velocity relative to the surface of the ice? b) What is the velocity of the plank relative to the surfaceof the ice? A 45.0 kg girl is standing on a 162 kg plank. The plank, originally at rest, isfree to slide on a frozen lake, which is a flat, frictionlesssurface. The girl begins to walk along the plank at a constantvelocity of 1.44 m/s relative to theplank.
a) What is her velocity relative to the surface of the ice? b) What is the velocity of the plank relative to the surfaceof the ice?

Explanation / Answer

   let us take that
   vgi = velocity of the girl relative to theice
   vpi= velocity of the planck relative to theice
   vgp = 1.44 m / s
   from the given problem we can write
   vgi = vgp + vpi
   vgi = 1.44 m / s + vpi.......... (1)
(a)
   according to the law of conservation of momentum weget
   mg vgi + mpvpi = 0
   vpi = - (mg / mp)vgi........(2)
   from (1) we get
   [1 + (mg / mp) vgi] =1.44 m / s
   vgi = ....... m / s
(b)
   using the equation (2) we get
   vpi = - (mg / mp)vgi
        = ....... m / s(directed opposite to the motion of the girl)

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