Review & Synthesis: Chapters 6-8 - Review Exercises - Page 324: 19

After the collision, bob A rises to a height of $0.57~m$ and bob B rises to a height of $2.3~m$

Let the masses be $m_A$ and $m_B$. Note that $m_B = 2m_A$. Let $v_A$ be the initial velocity of bob A. Let $v_B$ be the initial velocity of bob B. Let $v_A'$ be the final velocity of bob A. Let $v_B'$ be the final velocity of bob B. We can use conservation of energy to find the velocity of bob A just before the collision: $KE = U_g$ $\frac{1}{2}m_Av_A^2 = m_Agh$ $v_A = \sqrt{2gh}$ $v_A = \sqrt{(2)(9.80~m/s^2)(5.1~m)}$ $v_A = 10.0~m/s$ We can use conservation of momentum to set up an equation. $m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$ Since the collision is elastic, we can set up another equation. $v_A - v_B = v_B' - v_A'$ $v_A' = v_B' - v_A + v_B$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$ $v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$ $v_B' = \frac{2m_A~v_A+2m_A~v_B -m_Av_B}{m_A+2m_A}$ $v_B' = \frac{2~v_A+2~v_B -v_B}{3}$ $v_B’ = \frac{(2)(10.0~m/s)+(2)(0)- (0)}{3}$ $v_B' = 6.67~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A + v_B$ $v_A' = 6.67~m/s - 10.0~m/s + 0$ $v_A' = -3.33~m/s$ After the collision, bob A is moving in the opposite direction at a speed of 3.33 m/s After the collision, bob B is moving at a speed of 6.67 m/s We can use conservation of energy to find the height that bob A reaches: $U_g = KE$ $m_Agh = \frac{1}{2}m_A(v_A')^2$ $h = \frac{(v_A')^2}{2g}$ $h = \frac{(3.33~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 0.57~m$ We can use conservation of energy to find the height that bob B reaches: $U_g = KE$ $m_Bgh = \frac{1}{2}m_B(v_B')^2$ $h = \frac{(v_B')^2}{2g}$ $h = \frac{(6.67~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 2.3~m$ After the collision, bob A rises to a height of $0.57~m$ and bob B rises to a height of $2.3~m$