Question 15-58. The 15-lb suitcase A is released from rest at C. After it slides down the smooth ramp, it strikes the 10-15 suitcase B, which is originally at rest. If the coefficient of restitution between the suitcases is e = 0.3 and the coefficient of kinetic friction between the floor DE and each suitcase is uk = 0.4, determine (a) the velocity of A just before impact, (b) the velocities of A and B just after impact, and (c) the distance B slides before coming to rest. 6 ft B E D

Weight of Suitcase \omega_{A}=15 \Omega 611111 W_{B}=1006e=0.3\mu_{k}=0.4(A)K_{E_{A}}=\frac{1}{2} m(V)^{2}=0 l b . p tfinal Energy at E\begin{array}{l}P_{E}=m g z_{E}=0 \text { lb.ft } \\K E_{E}=\frac{1}{2} m\left(V_{E}\right)^{2}=\frac{1}{2} \times \frac{15}{32.2} \times\left(V_{A_{2}}\right)^{2}\end{array}Conservation of Energy\begin{array}{l}P_{A}+K E_{A}=P_{E}+K E_{E} \\90+0=0+\frac{1}{2} \times \frac{15}{32.2} \times\left(V_{A_{2}}\right)^{2} \\V_{A_{2}}=19.657 \mathrm{ft} / \mathrm{sec}\end{array}(8) Conservation of momentum\begin{array}{l}m_{A} V_{A_{1}}+m_{B V_{B_{1}}}=m_{A V_{A_{2}}}+m_{B} V_{B_{2}} \\\frac{15}{32.2} \times 19.657+0=\frac{15}{32.2} \times V A_{2}+\frac{10}{32.2} V_{B_{2}} \\\end{array}and\begin{aligned}e & =\frac{V_{B_{2}}-V_{A_{2}}}{V_{A_{1}}-V_{B_{1}}} \\0.3 & =\frac{V_{B_{2}}-V_{A_{2}}}{19.66-0} \\V_{B_{2}} & =5.898+V_{A_{2}}\end{aligned}By Eqn (4) and (2)\frac{15}{32.2} \times 19.657=\frac{15}{32.2} V_{A 2}+\frac{10}{32.2}\left(5.29 .8+V_{A}\right)9.1569=0.4658 \mathrm{VA2}+7.8316+0.3105 \mathrm{VA}_{2}\begin{array}{c}V_{A_{2}}=9.435 \mathrm{ft} / \mathrm{Sec} \leftarrow \\V_{B_{2}}=5.898+9.435=15.33 \mathrm{ft} / \mathrm{sec}\end{array}(c)\sum r_{y}=0(\pi+)N B=1096Pincipal of work and Energy\begin{array}{c}T_{1}-\sum U_{1-2}=T_{2} \\\frac{1}{2} m v_{B}^{2}+(\mu N \times \Delta x)=0 \\\frac{1}{2} \times \frac{10}{32.2} \times(15.33)^{2}+(-0.4 \times 10 \times \Delta x)=0 \\\Delta x=9.127 \rho t\end{array} ...