Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - General Problems - Page 197: 86

Answer

Yes

Work Step by Step

Since the collision is elastic, the kinetic energy of the system will be conserved and the objects will fly apart separately after they collide. Therefore, the cue ball will not travel in the same direction as the red ball and fall into the same hole. However, it is possible for it to fall in the right hole. Assume the red ball makes an angle of $\theta$ and the white ball makes an angle of $\alpha$ with the vertical. $P_{ix}=P_{fx}$ $P_{iy}=P_{fy}$ $mv=mv'_1cos(\theta)+mv'_2cos(\alpha)$ $0=mv'sin(\theta)-mv'_2sin(\alpha)$ $v^2=v'^2_1cos^2(\theta)+2v'_1v'_2cos(\theta)cos(\alpha)+v'^2_2cos^2(\alpha)$ $0=v'^2_1sin^2(\theta)-2v'_1v'_2sin(\theta)sin(\alpha)+v'^2_2sin^2(\alpha)$ $v^2=v'^2_1+v'^2_2+2v'_1v'_2cos(\theta+\alpha)$ $E_{iKx}=E_{fKx}$ $mv^2=mv'^2_1cos^2(\theta)+mv'^2_2cos^2(\alpha)$ $E_{iKy}=E_{fKy}$ $0=mv'^2_1sin^2(\theta)+mv'^2_2sin^2(\alpha)$ $v^2=v'^2_1+v'^2_2$ $v'^2_1+v'^2_2+2v'_1v'_2cos(\theta+\alpha)=v'^2_1+v'^2_2$ $2v'_1v'_2cos(\theta+\alpha)=0$ $cos(\theta+\alpha)=0$ $\theta+\alpha=90^o$ Using simple trigonometry (toa), we can find that $\theta=30^o$ and $\alpha=60^o$ Then, the ball will enter the right hole.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.