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Valter edited Sep 3, 2026, 6:49 PM

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		+	### Statement
		+
		+	$2.2.9.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.2.9^\ast$. 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:
		+	$$m_2 v_2 = m_2 v - F_0 t_0 \implies v_2 = v - \frac{F_0 t_0}{m_2}$$
		+
		+	#### Answer
		+	$u_1 = F_0 t_0 / m_1$; $u_2 = v - F_0 t_0 / m_2$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.2.9.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.2.9^\ast$. 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:
			$$m_2 v_2 = m_2 v - F_0 t_0 \implies v_2 = v - \frac{F_0 t_0}{m_2}$$

			#### Answer
			$u_1 = F_0 t_0 / m_1$; $u_2 = v - F_0 t_0 / m_2$

			#### Answer

			[Insert a concise answer or boxed result]