Answer
$0.5m/s$
$16875J$
Work Step by Step
We apply the law of conservation of momentum
$m_r(v_r)_1+m_c(v_c)_1=(m_r+m_c) v$
We plug in the known values to obtain:
$(15000)(1.5)+12000(-0.75)=(15000+12000)v$
This simplifies to:
$v=0.5m/s$
The kinetic energy of the system before coupling is given as
$T_1=\frac{1}{2}m_r(v_r)^2_1+\frac{1}{2}m_c(v_c)^2_1$
$\implies T_1=\frac{1}{2}(15000)(1.5)^2+\frac{1}{2}(12000)(-0.75)^2=20250J$
and the kinetic energy after coupling is given as
$T_2=\frac{1}{2}(m_r+v_c)(v)^2$
$\implies T_2=\frac{1}{2}(15000+12000)(0.5)^2=3375J$
Now the difference in kinetic energy is
$\Delta T=T_1-T-2=20250-3375=16875J$
