A 50.0-kg student stands on a 150-kg wooden plank. Both start at rest on a frozen pond (assume the interface between pond and plank is frictionless). The student then walks with a constant velocity of 1.10 im/s relative to the plank. (a) What is the velocity of the plank relative to the frozen pond? (b) What is the velocity of the student relative to the frozen pond? (Hint: use velocities that are in the reference frame of the pond.)

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14-11-2019
Physics

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A 50.0-kg student stands on a 150-kg wooden plank. Both start at rest on a frozen pond (assume the interface between pond and plank is frictionless). The student then walks with a constant velocity of 1.10 im/s relative to the plank. (a) What is the velocity of the plank relative to the frozen pond? (b) What is the velocity of the student relative to the frozen pond? (Hint: use velocities that are in the reference frame of the pond.)

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sitiadriani sitiadriani · Answer 1 · 2019-11-14T12:20:01+01:00

Answer:

(a) the velocity of the plank relative to the frozen pond is 0.825 m/s

(b) the velocity of the plank relative to the frozen pond is 0.275 m/s

Explanation:

Given information:

student's mass, [tex]m_{s}[/tex] = 50kg

wooden plank's mass, [tex]m_{p}[/tex] = 150 kg

Note : all the velocity is in vector

student's velocity relative to the plank,[tex]v_{sp}[/tex] = 1.10i m/s

student's velocity relative to frozen pond, [tex]v_{sf}[/tex]

[tex]v_{sf}[/tex] = [tex]v_{sp}[/tex] + [tex]v_{p}[/tex]

(a) the velocity of the plank relative to the frozen pond, [tex]v_{pf}[/tex]

to calculate the velocity of the plank relative to the frozen pond, we can start from the momentum, p.

The total momentum is

p = [tex]m_{s}[/tex] [tex]v_{s}[/tex] + [tex]m_{p}[/tex] [tex]v_{p}[/tex]

[tex]v_{s}[/tex] is the velocity of student relative to the frozen pond

[tex]v_{p}[/tex] is the velocity of the plank relative to the frozen pond

at rest on a frozen pond, the momentum is 0. so,

p = 0

[tex]m_{s}[/tex] [tex]v_{s}[/tex] + [tex]m_{p}[/tex] [tex]v_{p}[/tex] = 0

[tex]m_{s}[/tex] [tex]v_{s}[/tex] = - [tex]m_{p}[/tex] [tex]v_{p}[/tex]

*the negative indicates that the student and the plank are moving in opposite direction

now subtitute student's velocity relative to frozen pond, [tex]v_{sf}[/tex] to the equation.

[tex]m_{s} ({v_{sp} + v_{p} )[/tex] = - [tex]m_{p} v_{p}[/tex]

[tex]m_{s}v_{sp} + m_{s} v_{p}[/tex] = - [tex]m_{p} v_{p}[/tex]

- [tex](m_{s} v_{p} + m_{p} v_{p3}[/tex]) = - [tex]m_{s} v_{p}[/tex]

-[tex](m_{s} + m_{p} )v_{p}[/tex] = - [tex]m_{s} v_{p}[/tex]

[tex]v_{p}[/tex] = -[tex](\frac{m_{s} }{m_{s} +m_{p} })[/tex]

[tex]v_{p}[/tex] = - [tex](\frac{50}{150+50} )(1.10i)[/tex]

[tex]v_{p}[/tex] = -0.825 i m/s

the velocity of the plank relative to the frozen pond is 0.825 m/s in the opposite direction

(b) the velocity of the student relative to the frozen pond

[tex]v_{sf}[/tex] = [tex]v_{sp}[/tex] + [tex]v_{p}[/tex]

[tex]v_{sf}[/tex] = (1.10i m/s) + (- 0.825i m/s)

[tex]v_{sf}[/tex] = (1.10i m/s - 0.825i m/s)

[tex]v_{sf}[/tex] = 0.275i m/s

the velocity of the plank relative to the frozen pond is 0.275 m/s.