by shamockey
 P: 453 Choose a positive direction, so that from the conservation of momentum: $$m_tv_{0t} - m_cv_{0c} = (m_t + m_c)v_f$$, where we take the direction of the truck initial velocity as positive Then, since $$v_{0t}$$ and $$v_{0c}$$ have the same magnitudes, $$(m_t - m_c)v_0 = (m_t + m_c) v_f$$ or $$v_f = \frac{m_t - m_c}{m_t + m_c}v_0$$ Because the mass of the truck is greater than that of the car, then numerator $$m_t - m_c$$ will be positive, meaning that the final velocity of the combined system (truck + car) will still be in the direction of the truck. Observe that the greater $$m_t$$ is in comparison with $$m_c$$, the closer will the final velocity be to the truck initial velocity! Therefore, the truck experiences less acceleration (same direction of velocity) than the car does (direction of velocity is reversed), and the same is true for the corresponding drivers. Hence a driver would rather prefer to be in the truck during the collision because a smaller acceleration would mean a smaller force as well (F = ma).