Impact

Consider a 5 kg object is traveling across a flat surface at 15 m/s, and crashes into a stationary 3 kg mass. Assuming there is no friction and the impact is perfectly elastic, what is the moving object's velocity after impact? What about the stationary object's velocity after impact?

Hint
$$$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$$$
where $$m_1$$ & $$m_2$$ are the masses of the two bodies, $$v_1$$ & $$v_2$$ are the velocities of the bodies just before impact, and $$v_1'$$ & $$v_2'$$ are the velocities of the bodies just after impact
Hint 2
For impacts, the relative velocity expression is
$$$e=\frac{v_2'-v_1'}{v_1-v_2}$$$
where $$e$$ is the restitution coefficient, $$v_i$$ is the velocity normal to the plane of impact just before impact, and $$v_i'$$ is the velocity normal to the plane of impact just after impact
During an impact, momentum is conserved while energy may or may not be conserved. For direct central impact with no external forces:
$$$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$$$
where $$m_1$$ & $$m_2$$ are the masses of the two bodies, $$v_1$$ & $$v_2$$ are the velocities of the bodies just before impact, and $$v_1'$$ & $$v_2'$$ are the velocities of the bodies just after impact
$$$(5kg)(15m/s)+(3kg)(0m/s)=(5kg)v_1'+(3kg)v_2'$$$
$$$75kg\cdot m/s=(5kg)v_1'+(3kg)v_2'$$$
For impacts, the relative velocity expression is
$$$e=\frac{v_2'-v_1'}{v_1-v_2}$$$
where $$e$$ is the restitution coefficient, $$v_i$$ is the velocity normal to the plane of impact just before impact, and $$v_i'$$ is the velocity normal to the plane of impact just after impact
The value of $$e$$ is such that:
$$0\leq e\leq 1$$ with limiting values
$$e=1$$ for perfectly elastic scenarios (energy conserved)
$$e=0$$ for perfectly plastic (no rebound)
Because our problem statement mentioned the impact is perfectly elastic:
$$$e=1=\frac{v_2'-v_1'}{15m/s-0m/s}$$$
$$$15m/s(1)=v_2'-v_1'$$$
$$$15m/s+v_1'=v_2'$$$
Thus, the moving object's velocity just after impact is:
$$$75kg\cdot m/s=(5kg)v_1'+(3kg)(15m/s+v_1')$$$
$$$75kg\cdot m/s=(5kg)v_1'+45kg\cdot m/s+(3kg)v_1'$$$
$$$30kg\cdot m/s=(8kg)v_1'$$$
$$$v_1'=3.75\:m/s$$$
Thus, the stationary object's velocity just after impact is:
$$$75kg\cdot m/s=(5kg)(3.75m/s)+(3kg)v_2'$$$
$$$75kg\cdot m/s=18.75kg\cdot m/s+(3kg)v_2'$$$
$$$v_2'=\frac{(75kg\cdot m/s-18.75kg\cdot m/s)}{3kg}=18.75\:m/s$$$
3.75 m/s, 18.75 m/s