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?
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
To solve this problem, we will use the concept of relative velocity.
(a) To find the girl's velocity relative to the surface of the ice, we need to consider her velocity relative to the plank and the plank's velocity relative to the ice.
The girl's velocity relative to the plank is given as 1.40 m/s to the right. Since the girl is moving in the same direction as the plank, her velocity relative to the ice will be the sum of her velocity relative to the plank and the plank's velocity relative to the ice.
Therefore, the girl's velocity relative to the ice is 1.40 m/s to the right.
(b) The velocity of the plank relative to the surface of the ice can be found by considering the forces acting on the system.
The only horizontal force acting on the system is the girl's pushing force. According to Newton's third law, the reaction force of the girl's push will be exerted on the plank in the opposite direction. These two forces cancel each other out, resulting in no net force acting on the system.
Since there is no net force, the plank will remain at rest relative to the ice. Therefore, the velocity of the plank relative to the surface of the ice is 0 m/s.