## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 269: 97

#### Answer

The maximum height attained by the two bobs after the collision is $\frac{h}{9}$

#### Work Step by Step

Let $M$ be the mass of bob A. Then $2M$ is the mass of bob B. Let $v_A$ be bob A's speed just before the collision. We can use conservation of momentum to find the speed $v_f$ just after the collision: $m_f~v_f = m_A~v_A$ $(M+2M)~v_f = M~v_A$ $v_f = \frac{v_A}{3}$ We can use conservation of energy to find an expression for $v_A$ in terms of $h$: $KE_0=U_0$ $\frac{1}{2}Mv_A^2 = Mgh$ $v_A = \sqrt{2gh}$ We can use conservation of energy to find the maximum height $h_f$ attained by the two bobs after the collision: $U_f = KE_f$ $(3M)gh_f = \frac{1}{2}(3M)(\frac{v_A}{3})^2$ $gh_f = \frac{1}{2}(\frac{v_A}{3})^2$ $gh_f = \frac{1}{2}(\frac{\sqrt{2gh}}{3})^2$ $gh_f = \frac{1}{2}(\frac{2gh}{9})$ $h_f = \frac{h}{9}$ The maximum height attained by the two bobs after the collision is $\frac{h}{9}$.

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.