Answer
The recoil speed of the car is $0.103~m/s$
Work Step by Step
The initial momentum of the system is zero. By conservation of momentum, if the shell gains a horizontal component of momentum $p_s$ in one direction, the car will gain the same magnitude of horizontal momentum $p_c$ in the opposite direction. Since the ground exerts a normal force on the car, the car does not gain an equal and opposite vertical component of momentum.
We can find the recoil speed $v_r$ of the car:
$p_c = p_{s,x}$
$m_c~v_r = m_s~v_{s,x}$
$v_r = \frac{m_s~v_{s,x}}{m_c}$
$v_r = \frac{(98~kg)(105~m/s)~cos~60.0^{\circ}}{5.0\times 10^4~kg}$
$v_r = 0.103~m/s$
The recoil speed of the car is $0.103~m/s$.
