Edits to “Statement”, “Solution”, “Answer”

astrosander edited
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revision #17732 parent #10414 ← older
en/2.7.41.md +36 −36
@@ -1,80 +1,80 @@
### Statement
−\\\\\\\$2.7.41.\\\\\\\$ A thin rod of mass \\\\\\\$m\\\\\\\$ and length \\\\\\\$l\\\\\\\$ lies on a smooth horizontal surface. A plasticine ball of mass \\\\\\\$m\\\\\\\$ with a velocity \\\\\\\$v\\\\\\\$ perpendicular to the rod hits one of its ends and sticks to it. How much heat will be released from such an impact?
+$2.7.41.$ A thin rod of mass $m$ and length $l$ lies on a smooth horizontal surface. A plasticine ball of mass $m$ with a velocity $v$ perpendicular to the rod hits one of its ends and sticks to it. How much heat will be released from such an impact?
### Solution
![ Collision of a ball with a rod |466x424, 39%](../../img/2.7.41/2.7.41_1.png)
−After collision they will move as a unit, rotating around the centre of mass, which is at a distance \\\\\\\$l/4\\\\\\\$ from the contact point of the balls
+After collision they will move as a unit, rotating around the centre of mass, which is at a distance $l/4$ from the contact point of the balls
![ Rotations around the centre of mass |403x276, 39%](../../img/2.7.41/2.7.41_2.png)
Conservation of angular momentum
−\\\\\\\$\\\\\\\$
−mv\\\\\\\\cdot\\\\\\\\frac{L}{4}=m\\\\\\\\frac{\\\\\\\\omega L}{4}\\\\\\\\cdot\\\\\\\\frac{L}{4}+I\\\\\\\\omega\\\\\\\\tag{1}
−\\\\\\\$\\\\\\\$
+$$
+mv\cdot\frac{L}{4}=m\frac{\omega L}{4}\cdot\frac{L}{4}+I\omega\tag{1}
+$$
Determening the moment of inertia, using parallel axis theorem
−\\\\\\\$\\\\\\\$
−I=\\\\\\\\frac{1}{2}mL^2+m\\\\\\\\left(\\\\\\\\frac{L}{4}\\\\\\\\right)^2=\\\\\\\\frac{7}{48}mL^2\\\\\\\\tag{2}
−\\\\\\\$\\\\\\\$
+$$
+I=\frac{1}{2}mL^2+m\left(\frac{L}{4}\right)^2=\frac{7}{48}mL^2\tag{2}
+$$
−From the \\\\\\\$(1)\\\\\\\$ and \\\\\\\$(2)\\\\\\\$
+From the $(1)$ and $(2)$
−\\\\\\\$\\\\\\\$
−v=\\\\\\\\frac{5}{6}\\\\\\\\omega L\\\\\\\\Rightarrow \\\\\\\\boxed{\\\\\\\\omega =\\\\\\\\frac{6v}{5L}}
−\\\\\\\$\\\\\\\$
+$$
+v=\frac{5}{6}\omega L\Rightarrow \boxed{\omega =\frac{6v}{5L}}
+$$
Also, in addition to rotational motion, translational motion will be observed. From the law of conservation of momentum
−\\\\\\\$\\\\\\\$
−mv=2mu\\\\\\\\Rightarrow u=v/2
−\\\\\\\$\\\\\\\$
+$$
+mv=2mu\Rightarrow u=v/2
+$$
−\\\\\\\$u\\\\\\\$ — centre of mass velocity
+$u$ — centre of mass velocity
THe initial energy was as a kinetic energy
−\\\\\\\$\\\\\\\$
−E_0=\\\\\\\\frac{mv^2}{2}
−\\\\\\\$\\\\\\\$
+$$
+E_0=\frac{mv^2}{2}
+$$
The total energy after the collision:
−\\\\\\\$\\\\\\\$
−E_f=\\\\\\\\frac{2mu^2}{2}+\\\\\\\\frac{I\\\\\\\\omega^2}{2}+\\\\\\\\frac{m(\\\\\\\\omega\\\\\\\\cdot L/u)^2}{2}
−\\\\\\\$\\\\\\\$
+$$
+E_f=\frac{2mu^2}{2}+\frac{I\omega^2}{2}+\frac{m(\omega\cdot L/u)^2}{2}
+$$
Conservation of energy
−\\\\\\\$\\\\\\\$
+$$
E_0=E_f+Q
−\\\\\\\$\\\\\\\$
+$$
−Where, \\\\\\\$Q\\\\\\\$ — the released energy
+Where, $Q$ — the released energy
−\\\\\\\$\\\\\\\$
+$$
Q=E_0-E_f
−\\\\\\\$\\\\\\\$
+$$
−\\\\\\\$\\\\\\\$
−Q=\\\\\\\\frac{mv^2}{2}-\\\\\\\\frac{mv^2}{2\\\\\\\\cdot2}-\\\\\\\\frac{7mL^2}{48}\\\\\\\\cdot \\\\\\\\left(\\\\\\\\frac{6v}{5L}\\\\\\\\right)^2\\\\\\\\cdot \\\\\\\\frac{1}{2}+\\\\\\\\frac{m\\\\\\\\cdot(6v/5L\\\\\\\\cdot L/4)^2}{2}
−\\\\\\\$\\\\\\\$
+$$
+Q=\frac{mv^2}{2}-\frac{mv^2}{2\cdot2}-\frac{7mL^2}{48}\cdot \left(\frac{6v}{5L}\right)^2\cdot \frac{1}{2}+\frac{m\cdot(6v/5L\cdot L/4)^2}{2}
+$$
After the mathematical transformations
−\\\\\\\$\\\\\\\$
−\\\\\\\\boxed{Q=\\\\\\\\frac{mv^2}{10}}
−\\\\\\\$\\\\\\\$
+$$
+\boxed{Q=\frac{mv^2}{10}}
+$$
#### Answer
−\\\\\\\$\\\\\\\$
−Q=1/10\\\\\\\\cdot mv^2
−\\\\\\\$\\\\\\\$
+$$
+Q=1/10\cdot mv^2
+$$
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