Answer
$v_{truck}=6.25m/s;v_{car}=21.7m/s$
Work Step by Step
We can find the required speed of the car and the truck as follows:
According to law of conservation of momentum
$P_i=P_f$
$\implies m_cv_c+m_tv_t=m_cv_c^{\prime}+m_tv_t^{\prime}$
$\implies m_c(0)+m_tv_t=m_cv_c^{\prime}+m_tv_t^{\prime}$
This simplifies to:
$v_t-v_t^{\prime}=\frac{m_c}{m_t}v_c^{\prime}$.....eq(1)
According to law of conservation of energy
$\frac{1}{2}m_cv_c^2+\frac{1}{2}m_tv_t^2=\frac{1}{2}m_cv_c^{\prime 2}+\frac{1}{2}m_tv_t^{\prime2}$
This simplifies to:
$v_t^2-v_t^{\prime2}=\frac{m_c}{m_t}v_c^{\prime 2}$.......eq(2)
Dividing eq(2) by eq(1), we obtain:
$v_t+v_t^{\prime}=v_c^{\prime}$......eq(3)
Adding eq(1) and eq(3), we obtain:
$2v_t=(\frac{m_c}{m_t}+1)v_{c^{\prime}}$
$\implies v_c^{\prime}=\frac{2v_tm_t}{m_c+m_t}$
We plug in the known values to obtain:
$v_c^{\prime}=\frac{2(15.5m/s)(1720Kg)}{732Kg+1720Kg}=21.7m/s$
From eq(3) $v_t^{\prime}=v_c^{\prime}-v_t$
We plug in the known values to obtain:
$v_t^{\prime}=21.75m/s-15.5m/s=6.25m/s$
