Правка разделов «Statement», «Answer»
en/2.2.5.md
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| ### Statement | |||
| − | $2.2.5.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $2.2.5^\ast$. A box with sand of mass $M$ lies on a horizontal plane, the coefficient of friction with which is $\mu$. At an angle $\alpha$ to the vertical, a bullet of mass $m$ flies into the box with a velocity $v$ and almost instantly gets stuck in the sand. How long after the bullet hits the box will it start moving and then stop? At what value of $\alpha$ will it not move at all? | |||
| ### Solution | |||
| Let the stopping time of the bullet be $dt$. Then, in order to stop, the box can only move horizontally. This means that at the moment of impact, the vertical projection of the bullet's momentum is extinguished by the normal reaction force of the support (sand) $N_1$: | |||
| $$N_1 dt = mv \cos\alpha \implies N_1 = \frac{mv \cos\alpha}{dt}$$ | |||
| Let's write down the friction force at this moment: | |||
| $$F_{\text{fr}} \le \mu N = \mu((M+m)g + N_1) = \mu(M+m)g + \mu\frac{mv \cos\alpha}{dt}$$ | |||
| Since $dt$ is an infinitesimally small quantity, then: | |||
| $$N_1 \gg (M+m)g \implies F_{\text{fr}} \le \mu\frac{mv \cos\alpha}{dt}$$ | |||
| Now let's write the change in the momentum of the box with sand over $dt$: | |||
| $$\Delta p = F_{\text{fr}} dt \le \mu mv \cos\alpha$$ | |||
| ($F_{\text{fr}}$ and $\Delta p$ reach their maximum value only in the situation where the box continues to remain at rest). At the moment of impact, the box with the bullet acquires a horizontal momentum: | |||
| $$p_1 = mv \sin\alpha$$ | |||
| (the horizontal projection of the bullet's momentum). | |||
| In order to move the box, it is necessary that: | |||
| $$p_1 > \Delta p$$ | |||
| $$mv \sin\alpha > \mu mv \cos\alpha$$ | |||
| Thus, we obtained the condition for the motion of the box: $\tan\alpha > \mu$. | |||
| After the collision, during the motion of the box, the friction force will act on it: | |||
| $$F_{\text{fr}} = \mu(M+m)g$$ | |||
| This force decreases the momentum $p_1 - \Delta p$ acquired by the box, so the following is true: | |||
| $$F_{\text{fr}} t = p_1 - \Delta p$$ | |||
| $$t = \frac{p_1 - \Delta p}{F_{\text{fr}}}$$ | |||
| $$t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(M+m)g}$$ | |||
| @@ -42,7 +36,3 @@Solution | |||
| #### Answer | |||
| $t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(m+M)g}$ for $\tan\alpha > \mu$; | |||
| for $\tan\alpha \le \mu$ the box will not move. | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $2.2.5.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.2.5^\ast$. A box with sand of mass $M$ lies on a horizontal plane, the coefficient of friction with which is $\mu$. At an angle $\alpha$ to the vertical, a bullet of mass $m$ flies into the box with a velocity $v$ and almost instantly gets stuck in the sand. How long after the bullet hits the box will it start moving and then stop? At what value of $\alpha$ will it not move at all? | $2.2.5^\ast$. A box with sand of mass $M$ lies on a horizontal plane, the coefficient of friction with which is $\mu$. At an angle $\alpha$ to the vertical, a bullet of mass $m$ flies into the box with a velocity $v$ and almost instantly gets stuck in the sand. How long after the bullet hits the box will it start moving and then stop? At what value of $\alpha$ will it not move at all? | ||
| ### Solution | ### Solution | ||
| Let the stopping time of the bullet be $dt$. Then, in order to stop, the box can only move horizontally. This means that at the moment of impact, the vertical projection of the bullet's momentum is extinguished by the normal reaction force of the support (sand) $N_1$: | Let the stopping time of the bullet be $dt$. Then, in order to stop, the box can only move horizontally. This means that at the moment of impact, the vertical projection of the bullet's momentum is extinguished by the normal reaction force of the support (sand) $N_1$: | ||
| $$N_1 dt = mv \cos\alpha \implies N_1 = \frac{mv \cos\alpha}{dt}$$ | $$N_1 dt = mv \cos\alpha \implies N_1 = \frac{mv \cos\alpha}{dt}$$ | ||
| Let's write down the friction force at this moment: | Let's write down the friction force at this moment: | ||
| $$F_{\text{fr}} \le \mu N = \mu((M+m)g + N_1) = \mu(M+m)g + \mu\frac{mv \cos\alpha}{dt}$$ | $$F_{\text{fr}} \le \mu N = \mu((M+m)g + N_1) = \mu(M+m)g + \mu\frac{mv \cos\alpha}{dt}$$ | ||
| Since $dt$ is an infinitesimally small quantity, then: | Since $dt$ is an infinitesimally small quantity, then: | ||
| $$N_1 \gg (M+m)g \implies F_{\text{fr}} \le \mu\frac{mv \cos\alpha}{dt}$$ | $$N_1 \gg (M+m)g \implies F_{\text{fr}} \le \mu\frac{mv \cos\alpha}{dt}$$ | ||
| Now let's write the change in the momentum of the box with sand over $dt$: | Now let's write the change in the momentum of the box with sand over $dt$: | ||
| $$\Delta p = F_{\text{fr}} dt \le \mu mv \cos\alpha$$ | $$\Delta p = F_{\text{fr}} dt \le \mu mv \cos\alpha$$ | ||
| ($F_{\text{fr}}$ and $\Delta p$ reach their maximum value only in the situation where the box continues to remain at rest). At the moment of impact, the box with the bullet acquires a horizontal momentum: | ($F_{\text{fr}}$ and $\Delta p$ reach their maximum value only in the situation where the box continues to remain at rest). At the moment of impact, the box with the bullet acquires a horizontal momentum: | ||
| $$p_1 = mv \sin\alpha$$ | $$p_1 = mv \sin\alpha$$ | ||
| (the horizontal projection of the bullet's momentum). | (the horizontal projection of the bullet's momentum). | ||
| In order to move the box, it is necessary that: | In order to move the box, it is necessary that: | ||
| $$p_1 > \Delta p$$ | $$p_1 > \Delta p$$ | ||
| $$mv \sin\alpha > \mu mv \cos\alpha$$ | $$mv \sin\alpha > \mu mv \cos\alpha$$ | ||
| Thus, we obtained the condition for the motion of the box: $\tan\alpha > \mu$. | Thus, we obtained the condition for the motion of the box: $\tan\alpha > \mu$. | ||
| After the collision, during the motion of the box, the friction force will act on it: | After the collision, during the motion of the box, the friction force will act on it: | ||
| $$F_{\text{fr}} = \mu(M+m)g$$ | $$F_{\text{fr}} = \mu(M+m)g$$ | ||
| This force decreases the momentum $p_1 - \Delta p$ acquired by the box, so the following is true: | This force decreases the momentum $p_1 - \Delta p$ acquired by the box, so the following is true: | ||
| $$F_{\text{fr}} t = p_1 - \Delta p$$ | $$F_{\text{fr}} t = p_1 - \Delta p$$ | ||
| $$t = \frac{p_1 - \Delta p}{F_{\text{fr}}}$$ | $$t = \frac{p_1 - \Delta p}{F_{\text{fr}}}$$ | ||
| $$t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(M+m)g}$$ | $$t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(M+m)g}$$ | ||
| @@ -42,7 +36,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(m+M)g}$ for $\tan\alpha > \mu$; | $t = \frac{mv(\sin\alpha - \mu \cos\alpha)}{\mu(m+M)g}$ for $\tan\alpha > \mu$; | ||
| for $\tan\alpha \le \mu$ the box will not move. | for $\tan\alpha \le \mu$ the box will not move. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||