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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A 20.0-kg toboggan with 70.0-kg driver is sliding down a frictionless chute directed 30.0° below the horizontal at 8.00 m/s when a 55.0-kg woman drops from a tree limb straight down behind the driver. If she drops through a vertical displacement of 2.00 m, what is the subsequent velocity of the toboggan immediately after impact?

To determine
The subsequent velocity of the toboggan after the impact.

Explanation

Given info: Mass of the toboggan is 20.0kg . The mass of the driver is 70.0kg . The chute is directed 30.0° below the horizontal. The initial velocity of the toboggan is 8.00ms-1 . Mass of the women is 55.0kg . The women drops through a vertical displacement of 2.00m .

Explanation:

The system is shown in the following figure.

The woman is falling from a vertical distance. Hence the acceleration of the women is free fall acceleration due to gravity. The velocity of the women is given by,

v=u2+2gΔy

  • Δy is the vertical displacement covered
  • v is the final velocity
  • u is the initial velocity
  • g is the free fall acceleration

Assume that the initial velocity of the women is 0 .

Substitute 0 for u , 2.00m for Δy , 9.80ms-2 for g to determine the final velocity of the women,

v=02+2(9.80ms-2)(2.00m)=6.26ms-1

Conclusion:

The system, before and after impact is shown in the following figure.

From conservation of momentum of the man, woman and the toboggan in the direction of the incline,

(mm+mw+mt)vf=(mm+mt)vi1+mwvi2sinθ

  • mm is the mass of the man
  • mw is the mass of the woman
  • <

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