College Physics (4th Edition)

Published by McGraw-Hill Education

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 9

Answer

Gerald threw the ball of putty with a speed of $~30.4~m/s$

Work Step by Step

Let $m_p$ be the mass of the putty. Let $m_t$ be the mass of the target. We can use conservation of energy to find the speed $v_f$ of the target and the putty just after the collision: $KE = U_g$ $\frac{1}{2}(m_t+m_p)~v_f^2 = (m_t+m_p)~gh$ $v_f = \sqrt{2gh}$ $v_f = \sqrt{(2)(9.80~m/s^2)(1.50~m)}$ $v_f = 5.422~m/s$ We can use conservation of momentum to find the speed $v_0$ of the putty before the collision: $m_p~v_0 = (m_t+m_p)~v_f$ $v_0 = \frac{(m_t+m_p)~v_f}{m_p}$ $v_0 = \frac{(2.30~kg+0.50~kg)~(5.422~m/s)}{0.50~kg}$ $v_0 = 30.4~m/s$ Gerald threw the ball of putty with a speed of $~30.4~m/s$.

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.