Answer
$32559.022J$
Work Step by Step
We apply the law of conservation of momentum to determine the required velocity
$m_t(v_t)_1+m_c(v_c)_1=(m_t+m_c) v$
We plug in the known values to obtain:
$(2500)(\frac{30\times 1000}{3600})+1500(0)=(2500+1500)v$
This simplifies to:
$v=5.208m/s$
The kinetic energy of the system before coupling is given as
$T_1=\frac{1}{2}m_t(v_t)^2_1+\frac{1}{2}m_c(v_c)^2_1$
$\implies T_1=\frac{1}{2}(2500)(\frac{30\times 1000}{3600})^2+0=86805.55J$
and the kinetic energy after coupling is given as
$T_2=\frac{1}{2}(m_t+m_c)(v)^2$
$\implies T_2=\frac{1}{2}(2500+1500)(5.208)^2=54246.528J$
Now the difference in kinetic energy is
$\Delta T=T_1-T_2$
$\implies \Delta T=86805.55-54246.528=32559.022J$
