## College Physics (4th Edition)

The recoil speed of the submarine is $1.0~cm/s$
We can assume the initial momentum of the system is zero. By conservation of momentum, if the torpedo gains a certain momentum $p_t$ in one direction, the submarine will gain the same magnitude of momentum $p_s$ in the opposite direction. We can find the recoil speed $v_r$ of the submarine: $p_s = p_t$ $m_s~v_r = m_t~v_t$ $v_r = \frac{m_t~v_t}{m_s}$ $v_r = \frac{(250~kg)(100.0~m/s)}{2.5\times 10^6~kg}$ $v_r = 0.010~m/s = 1.0~cm/s$ The recoil speed of the submarine is $1.0~cm/s$.