## Engineering Mechanics: Statics & Dynamics (14th Edition)

$e=0.0113$
We can determine the required restitution as follows: We apply the conservation of linear momentum in the $x$-direction $m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_{x_2}}+m_Bv_{B_{x_2}}$ We plug in the known values to obtain: $0.5(-6)+(0.5)(4)(\frac{3}{5})=0.5v_{A_{x_2}}-0.5v_{B_{x_2}}$ This simplifies to: $v_{Ax}-v_{Bx}=-3.6~~~$[eq(1)] According to the conservation of linear momentum in the $y$-direction $\frac{4}{5}mvB_1=mvB_{y_2}$ $\implies v_{By}=(\frac{4}{5})(4)=3.2m/s\uparrow$ and $vB_x=3.2tan30=1.848m/s\leftarrow$ We plug in this value in eq(1) to obtain: $v_{Ax}=1.752m/s\leftarrow$ Now $e=\frac{v_{Ax_2}-v_{Bx_2}}{v_{B1}-v_{A1}}$ We plug in the known values to obtain: $e=\frac{-1.752-(-1.848)}{4(\frac{3}{5})-(-6)}$ This simplifies to: $e=0.0113$