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College Physics

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

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BuyFindarrow_forward

College Physics

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

Gayle runs at a speed of 4.00 m/s and dives on a sled, initially at rest on the top of a frictionless, snow-covered hill. After she has descended a vertical distance of 5.00 m. her brother, who is initially at rest, hops on her back, and they continue down the hill together. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 50.0 kg, the sled has a mass of 5.00 kg, and her brother has a mass of 30.0 kg.

To determine
The speed of Gayle and brother at the bottom of the hill.

Explanation

Given info:

The vertical distance descended is 5.00m , speed of the Gayle is 4.00m/s , total vertical drop is 15.0m , mass of Gayle is 50.0kg , mass of Sled is 5.00kg , and mass of brother is 30.0kg .

The acceleration due to gravity 9.80m/s2 .

The Formula to calculate the conservation of momentum when Gayle jump on the sled is,

(mG+mS)vGS=mGuG+mSuS

  • mG is the mass of Gayle
  • mS is the mass of sled
  • vGS is the combined speed of sled and Gayle
  • uG is the initial speed of Gayle
  • uS is the initial speed of sled

Rearranging the equation gives the combined speed of Gayle and sled as,

vGS=mGuG+mSuS(mG+mS)

Substitute 50.0kg for mG , 4.00m/s for uG , 5.00kg for mS   and 0 for uS to find v .

vGS=(50.0kg)(4.00m/s)+(5.00kg)(0m/s)(50.0kg+5.00kg)=3.64m/s

The Formula to calculate the conservation of mechanical energy at the descent of 5.00m is,

12(mG+mS)v2+(mG+mS)gh=12(mG+mS)vGS2+12mBvB

  • v is the speed of sled and Gayle just before her brother hop into
  • g is the acceleration due to gravity
  • h is the descended distance
  • vB is the initial speed of brother

Rearranging the equation gives the speed of sled and Gayle just before her brother hop into

v2=12(mG+mS)vGS2+12mBvB(mG+mS)gh12(mG+mS)

Substitute 50.0kg for mG , 30.0m/s for mB ,   5.00kg for mS , 3.64m/s for vGS , 0 for vB , 5.00m for h   and 9.80m/s2 for g to find v .

v2=12(50.0kg+5.00kg)(3.64m/s)2+12(30.0kg)(0)(50.0kg+5.00kg)(9.80m/s2)(5.00m)12(50.0kg+5.00kg)

The Formula to calculate the conservation of momentum when Gayle’s brother jump onto Gayle and sled is,

(mG+mS+mB)vGSB=(mG+mS)v+mBvB

  • vGSB is the combined speed of Gayle, her brother and sled

Rearranging the equation gives the combined speed of Gayle, her brother and sled as,

vGSB=(mG+mS)v+mBvB(mG+mS+mB)

Substitute 50

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