COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 7, Problem 47QAP
To determine

The final velocities of the resulting pieces in terms of v0.

Expert Solution & Answer
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Answer to Problem 47QAP

The final velocities of the resulting pieces in terms of v0.

  v1x=1.55v0cos45=1.10v0v1y=1.55v0sin45=1.10v0v2x=1.10v0cos30=0.953v0v2y=1.55v0sin30=0.550v0

Explanation of Solution

Given data:

Mass of object =3M

Object breaks into masses of =2M and M

  θ1=45θ2=30

Formula used:

Momentum, px=mvx

Where m= mass and vx = velocity in x- direction.

And principle of conservation of linear momentum

Calculation:

  COLLEGE PHYSICS, Chapter 7, Problem 47QAP

After the piece break into two pieces, Piece 1has a mass of M and travels off at an angle of 45 below the x-axiswith a speed of v1.

Piece 2has a mass of 2M and travels off at an angle of 30above thex-axis with a speed of v2.

We can use conservation of momentum in order to calculate thefinal velocities of the two pieces.

Since this is a two-dimensional problem, we will need to splitthe momenta into components and solve the x and y component equations.

y component:

  Piy=PfyPiy=p2fy+p1fy0=(2M)v2sin(30)(M)v1sin(45)v2=v1sin( 45)2sin( 30)=v12

x component:

  Pix=p2fx+p1fx(3M)v0=(2M)v2cos(30)+(M)v1cos(45)3v0=2v2cos(30)+v1cos(45)=2( v1 2 )( 3 2)+v12v1=323+1v0=1.55v0v2=v12=33+1v0=1.10v0

Therefore,

  v1x=1.55v0cos45=1.10v0v1y=1.55v0sin45=1.10v0v2x=1.10v0cos30=0.953v0v2y=1.55v0sin30=0.550v0

Conclusion:

Thus, for piece 1velocities in terms of v0are: -

  v1x=1.55v0cos45=1.10v0v1y=1.55v0sin45=1.10v0

Thus, for piece 2velocities in terms of v0are: -

  v2x=1.10v0cos30=0.953v0v2y=1.55v0sin30=0.550v0

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COLLEGE PHYSICS

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