Правка разделов «Statement», «Solution», «Answer»
en/2.4.4.md
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| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $2.4.4.$ [Insert the problem statement] | ||
| + | $2.4.4.$ The conveyor belt moves horizontally at a speed of $u$. On the belt at a tangent to it flies a body whose speed is perpendicular to the direction of movement of the belt and at the time of hitting the belt is equal to $v$. The body slides along the belt and then stops. Find the work of the friction force applied to the body from the belt and to the belt from the body. Why is the performance different in these cases? | ||
| ### Solution | |||
| − |  | ||
| It is easy to see that the change in kinetic energy of the body is $m(u^2-v^2)/2$, which is equal to the work done by friction from the belt on the body. For the work done by friction from the body on the belt, more analysis is required. We treat the situation as an inelastic collision of the body and the belt with mass $M\gg m$. From conservation of momentum, the components of the velocity of the body and the belt moving together are $u'=Mu/(M+m)$ in the same direction as $u$ and $v'=mv/(M+m)$ in the same direction as $v$. Thus, the change in kinetic energy of the belt is | |||
| \[\frac{1}{2}M(u'^2+v'^2-u^2)=\frac{1}{2}M\left(\frac{m^2(v^2-u^2)-2Mmu^2}{(M+m)^2}\right).\] | |||
| As $m^2(v^2-u^2)\ll2Mmu^2$, we drop the first term and have an approximation $-mu^2$ for the work done by friction from the body on the belt. To check the validity of this analysis, we perform it to find the change in kinetic energy of the body | |||
| \[\frac{1}{2}m(u'^2+v'^2-v^2)=\frac{1}{2}m\left(\frac{M^2(u^2-v^2)-2Mmv^2}{(M+m)^2}\right).\] | |||
| As $M^2(u^2-v^2)\gg2Mmv^2$, we drop the second term and have an approximation $m(u^2-v^2)/2$ for the work done by friction from the belt on the body in agreement with the result from an earlier approach. | |||
| Note that while frictions between the body and the belt have equal magnitude, the trajectories of the body and the belt are different, so frictions affect the energies of the body and the belt differently. | |||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $m(u^2-v^2)/2$ and $-mu^2$ | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $2.4.4.$ [Insert the problem statement] | $2.4.4.$ The conveyor belt moves horizontally at a speed of $u$. On the belt at a tangent to it flies a body whose speed is perpendicular to the direction of movement of the belt and at the time of hitting the belt is equal to $v$. The body slides along the belt and then stops. Find the work of the friction force applied to the body from the belt and to the belt from the body. Why is the performance different in these cases? | ||
| ### Solution | ### Solution | ||
|  | ||
| It is easy to see that the change in kinetic energy of the body is $m(u^2-v^2)/2$, which is equal to the work done by friction from the belt on the body. For the work done by friction from the body on the belt, more analysis is required. We treat the situation as an inelastic collision of the body and the belt with mass $M\gg m$. From conservation of momentum, the components of the velocity of the body and the belt moving together are $u'=Mu/(M+m)$ in the same direction as $u$ and $v'=mv/(M+m)$ in the same direction as $v$. Thus, the change in kinetic energy of the belt is | It is easy to see that the change in kinetic energy of the body is $m(u^2-v^2)/2$, which is equal to the work done by friction from the belt on the body. For the work done by friction from the body on the belt, more analysis is required. We treat the situation as an inelastic collision of the body and the belt with mass $M\gg m$. From conservation of momentum, the components of the velocity of the body and the belt moving together are $u'=Mu/(M+m)$ in the same direction as $u$ and $v'=mv/(M+m)$ in the same direction as $v$. Thus, the change in kinetic energy of the belt is | ||
| \[\frac{1}{2}M(u'^2+v'^2-u^2)=\frac{1}{2}M\left(\frac{m^2(v^2-u^2)-2Mmu^2}{(M+m)^2}\right).\] | \[\frac{1}{2}M(u'^2+v'^2-u^2)=\frac{1}{2}M\left(\frac{m^2(v^2-u^2)-2Mmu^2}{(M+m)^2}\right).\] | ||
| As $m^2(v^2-u^2)\ll2Mmu^2$, we drop the first term and have an approximation $-mu^2$ for the work done by friction from the body on the belt. To check the validity of this analysis, we perform it to find the change in kinetic energy of the body | As $m^2(v^2-u^2)\ll2Mmu^2$, we drop the first term and have an approximation $-mu^2$ for the work done by friction from the body on the belt. To check the validity of this analysis, we perform it to find the change in kinetic energy of the body | ||
| \[\frac{1}{2}m(u'^2+v'^2-v^2)=\frac{1}{2}m\left(\frac{M^2(u^2-v^2)-2Mmv^2}{(M+m)^2}\right).\] | \[\frac{1}{2}m(u'^2+v'^2-v^2)=\frac{1}{2}m\left(\frac{M^2(u^2-v^2)-2Mmv^2}{(M+m)^2}\right).\] | ||
| As $M^2(u^2-v^2)\gg2Mmv^2$, we drop the second term and have an approximation $m(u^2-v^2)/2$ for the work done by friction from the belt on the body in agreement with the result from an earlier approach. | As $M^2(u^2-v^2)\gg2Mmv^2$, we drop the second term and have an approximation $m(u^2-v^2)/2$ for the work done by friction from the belt on the body in agreement with the result from an earlier approach. | ||
| Note that while frictions between the body and the belt have equal magnitude, the trajectories of the body and the belt are different, so frictions affect the energies of the body and the belt differently. | Note that while frictions between the body and the belt have equal magnitude, the trajectories of the body and the belt are different, so frictions affect the energies of the body and the belt differently. | ||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $m(u^2-v^2)/2$ and $-mu^2$ | ||