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

$e=0.75$, $\Delta T=-9.65KJ$
We can determine the required coefficient of restitution and the loss of energy as follows: We know that according to the relative velocity relation: $v_{c/t}=v_{c_2}-v_{t_2}=4.167m/s~~~$[eq(1)] We apply the conservation of linear momentum $m_tv_{t_1}+m_cv_{c_1}=m_tv_{t_2}+m_cv_{c_2}$ We plug in the known values to obtain: $(5000)(8.333)+(2000)(2.778)=(5000)v_2+(2000)v_{c_2}$ This simplifies to: $5v_{t_2}+2v_{c_2}=47.22~~~$[eq(2)] After solving eq(1) and eq(2), we obtain: $v_{t_2}=5.556m/s$ and $v_{c_2}=9.722m/s$ Now $e=\frac{v_{c_2}-v_{t_2}}{v_{t_1}-v_{c_1}}=\frac{9.722-5.556}{8.333-2.778}$ $\implies e=0.75$ The kinetic energy of the system before collision is $T_1=\frac{1}{2}m_tv_{t_1}+\frac{1}{2}m_cv_{c_1}$ We plug in the known values to obtain: $T_1=\frac{1}{2}(5000)(8.333)^2+\frac{1}{2}(2000)(2.778)^2=181.33\times 10^3J$ and the kinetic energy of the system after the collision is $T_2=\frac{1}{2}m_tv_{t_2}+\frac{1}{2}m_cv_{c_2}$ We plug in the known values to obtain: $T_2=\frac{1}{2}(5000)(5.556)^2+\frac{1}{2}(2000)(9.722)^2=171.68\times 10^3J$ Now the loss in energy can be determined as $\Delta T=T_2-T_1$ $\implies \Delta T=171.68\times 10^3-181.33\times 10^3=-9.65KJ$