2.2.9∗. 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.
For problem $2.2.9$
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: $$m_2 v_2 = m_2 v - F_0 t_0 \implies v_2 = v - \frac{F_0 t_0}{m_2}$$
