Answer :
As the girl walks along the plank at a constant speed of 1.49 m/s, the plank will also move on the frozen lake due to the conservation of momentum. Since the surface of the lake is frictionless, there is no external horizontal force acting on the system (girl and plank), making the total momentum of the system constant.
Let's denote the girl's mass as m1 (44.4 kg) and the plank's mass as m2 (159 kg). The girl's speed relative to the plank is v1 (1.49 m/s), and the speed of the plank relative to the frozen lake is v2.
Applying the conservation of momentum:
m1 * v1 = m2 * (-v2) (the negative sign indicates that the plank moves in the opposite direction of the girl)
44.4 kg * 1.49 m/s = 159 kg * (-v2)
Now, solving for v2:
v2 = (44.4 kg * 1.49 m/s) / 159 kg = -0.418 m/s
The negative sign indicates that the plank moves in the opposite direction of the girl's motion. The speed of the plank relative to the frozen lake is 0.418 m/s.
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The girl's velocity relative to the ice is 1.165 m/s. The plank's velocity relative to the ice is -0.325 m/s. The solution uses the conservation of momentum principle.
The problem is a classic example of the conservation of momentum on a frictionless surface. Let's solve for the velocities step-by-step:
Part (a): Velocity of the girl relative to the surface of the ice
Given:
- Mass of the girl ([tex]m_g[/tex]) = 44.4 kg
- Mass of the plank ([tex]m_p[/tex]) = 159 kg
- Speed of the girl relative to the plank ([tex]v_{gp[/tex]) = 1.49 m/s
Using the conservation of momentum, the total initial momentum is zero (as both are initially at rest). When the girl starts walking, her velocity relative to the ice ([tex]v_{gi[/tex]) and the plank's velocity relative to the ice ([tex]v_{pi[/tex]) must satisfy the momentum conservation equation.
Initial momentum: 0
Final momentum: [tex](m_g \times v_{gi}) + (m_p \times v_{pi}) = 0[/tex]
Since [tex]v_{gi} = v_{pi} + v_{gp}[/tex], the equation can be written as:
- [tex](m_g \times (v_{pi} + 1.49)) + (m_p \times v_{pi}) = 0[/tex]
- [tex]44.4 \times (v_{pi} + 1.49) + 159 \times v_{pi} = 0[/tex]
- [tex]44.4 \times v_{pi} + 66.156 + 159 \times v_{pi} = 0[/tex]
- [tex]203.4 \times v_{pi} = -66.156[/tex]
- [tex]v_{pi} = -0.325 m/s[/tex]
Now, [tex]v_{gi} = v_{pi} + 1.49 = -0.325 + 1.49 = 1.165\left m/s[/tex]
Therefore, the girl's velocity relative to the ice is 1.165 m/s.
Part (b): Velocity of the plank relative to the surface of the ice:
The velocity of the plank relative to the ice is -0.325 m/s.
Complete question:
A 44.4 kg girl is standing on a plank that has a mass of 159 kg. the plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. the girl begins to walk along the plank at a constant speed of 1.49 m/s relative to the plank. (a) What is her velocity relative to the surface of ice? (b) What is the velocity of the plank relative to the surface of ice?
