Edits to “Statement”, “Solution”, “Answer”
en/2.5.21.md
+3 −3
| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $2.5.21.$ [Insert the problem statement] | ||
| + | $2.5.21.$ A smooth "slide" of height $h$ and mass $m_1$ can slide along a horizontal plane without friction. The slide smoothly turns into a plane. What is the lowest speed of the slide, a small body of mass $m_2$, initially lying motionless on its path, will pass over the top? | ||
| ### Solution | |||
| − |  | ||
| 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 | |||
| \[u=\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}.\] | |||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}$ | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $2.5.21.$ [Insert the problem statement] | $2.5.21.$ A smooth "slide" of height $h$ and mass $m_1$ can slide along a horizontal plane without friction. The slide smoothly turns into a plane. What is the lowest speed of the slide, a small body of mass $m_2$, initially lying motionless on its path, will pass over the top? | ||
| ### Solution | ### Solution | ||
|  | ||
| 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, | 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, | ||
| \[m_1u=m_1v+m_2v\qquad\Rightarrow\qquad v=\frac{m_1u}{m_1+m_2}.\] | \[m_1u=m_1v+m_2v\qquad\Rightarrow\qquad v=\frac{m_1u}{m_1+m_2}.\] | ||
| From conservation of energy, | From conservation of energy, | ||
| \[\frac{1}{2}m_1u^2=\frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2+m_2gh\qquad\Rightarrow\qquad\frac{1}{2}m_1u^2=\frac{1}{2}(m_1+m_2)\left(\frac{m_1u}{m_1+m_2}\right)^2+m_2gh.\] | \[\frac{1}{2}m_1u^2=\frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2+m_2gh\qquad\Rightarrow\qquad\frac{1}{2}m_1u^2=\frac{1}{2}(m_1+m_2)\left(\frac{m_1u}{m_1+m_2}\right)^2+m_2gh.\] | ||
| Solving for $u$, we obtain | Solving for $u$, we obtain | ||
| \[u=\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}.\] | \[u=\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}.\] | ||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\sqrt{2gh\left(1+\frac{m_2}{m_1}\right)}$ | ||