Posted by Anonymous on .
A 41.8 kg girl is standing on a 142. kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.40 m/s to the right relative to the plank.
(a) What is her velocity relative to the surface of the ice?
b) What is the velocity of the plank relative to the surface of the ice?
Physics URGENT!!! -
Let v(g) and v(p) be the velocity of the girl and the plank relative to the ice surface. Then v(g)-v(p) is the velocity of the girl relative to the plank, so that
v(g)-v(p) = 1.4
From the law of conservation of linear momentum m(g)v(g) +m(p)v(p) =0
41.8v(g) +142v(p) = 0
v(g) = 142v(p)/41.8 =- 3.4 v(p)
1.4 = v(g) - v(p) = - 3.4 v(p) – v(p) = - 4.4 v(p)
v(p) = - 1.4/4.4 = -0.32 m/s.
v(g) = - 3.4 v(p) = -3.4 (-0.32)=1.09 m/s