## College Physics (4th Edition)

The recoil speed of the car is $0.103~m/s$
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 $0.103~m/s$.