$2.2.9^\ast$. A body of mass $m_2$ moving at a velocity $v$ strikes a stationary body of mass $m_1$. The force arising from the interaction of the bodies, linearly dependent on time, grows from zero to a value $F_0$ during a time $t_0$, and then uniformly decreases to zero during the same time $t_0$. Determine the velocity of the bodies after the interaction, assuming all motion occurs along one straight line.
Solution
From the definition of Newton's second law in impulse form, it is known that $F dt = dp$. Then, by finding the area under the graph $F(t)$, we obtain the change in momentum. In our case: $$\int_0^{2t_0} F(t) dt = F_0 t_0 = \Delta p$$
Since the first body was at rest: $$m_1 v_1 = \Delta p \implies v_1 = \frac{F_0 t_0}{m_1}$$
On the second body, the force acts in the opposite direction, correspondingly decreasing its momentum, then: $$m_2 v_2 = m_2 v - F_0 t_0 \implies v_2 = v - \frac{F_0 t_0}{m_2}$$
