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+### Statement
+
+$2.5.21.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $2.5.21$ |958x511, 31%](../../img/2.5.21/2.5.21.png)
+
+Suppose the slide has minimum initial speed $u$. In the frame of reference of the slide, the mass $m_2$ must have barely any kinetic energy left when it reaches the top of the slide. In other words, the slide and the mass must have the same speed $v$ in the original frame of reference when the mass reaches the top. From conservation of momentum,
+
+\[m_1u=m_1v+m_2v\qquad\Rightarrow\qquad v=\frac{m_1u}{m_1+m_2}.\]
+
+From conservation of energy,
+
+\[\frac{1}{2}m_1u^2=\frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2+m_2gh\qquad\Rightarrow\qquad\frac{1}{2}m_1u^2=\frac{1}{2}(m_1+m_2)\left(\frac{m_1u}{m_1+m_2}\right)^2+m_2gh.\]
+
+Solving for $u$, we obtain
+
+\[u=\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}.\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]