Edits to “Statement”, “Answer”
en/2.2.12.md
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| @@ -1,11 +1,5 @@ | |||
| ### Statement | |||
| − | $2.2.12.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $2.2.12.$ A projectile explodes at the highest point of its trajectory at a horizontal distance $L$ from the cannon into two identical fragments. One of them returned to the cannon along the original trajectory of the projectile. Where did the second fragment fall? | |||
| ### Solution | |||
| The projectile explodes at the highest point of the trajectory, which means it has no vertical velocity component. Let its velocity be $\vec{v}$. Since one of the fragments returned to the cannon, its velocity will be $-\vec{v}$. | |||
| Let's write the law of conservation of momentum on the horizontal axis: | |||
| $$m_0 \vec{v} = -m \vec{v} + m \vec{v}_2$$ | |||
| $$\vec{v}_2 = \frac{m_0 + m}{m} \vec{v} = 3\vec{v}$$ | |||
| The velocity of the second fragment is $3$ times greater than $v$, which means it will cover a horizontal distance to the ground of $3L$ from the point of the explosion, or $4L$ from the cannon. | |||
| @@ -20,7 +14,3 @@Solution | |||
| #### Answer | |||
| At a horizontal distance of $4L$ from the cannon. | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $2.2.12.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.2.12.$ A projectile explodes at the highest point of its trajectory at a horizontal distance $L$ from the cannon into two identical fragments. One of them returned to the cannon along the original trajectory of the projectile. Where did the second fragment fall? | $2.2.12.$ A projectile explodes at the highest point of its trajectory at a horizontal distance $L$ from the cannon into two identical fragments. One of them returned to the cannon along the original trajectory of the projectile. Where did the second fragment fall? | ||
| ### Solution | ### Solution | ||
| The projectile explodes at the highest point of the trajectory, which means it has no vertical velocity component. Let its velocity be $\vec{v}$. Since one of the fragments returned to the cannon, its velocity will be $-\vec{v}$. | The projectile explodes at the highest point of the trajectory, which means it has no vertical velocity component. Let its velocity be $\vec{v}$. Since one of the fragments returned to the cannon, its velocity will be $-\vec{v}$. | ||
| Let's write the law of conservation of momentum on the horizontal axis: | Let's write the law of conservation of momentum on the horizontal axis: | ||
| $$m_0 \vec{v} = -m \vec{v} + m \vec{v}_2$$ | $$m_0 \vec{v} = -m \vec{v} + m \vec{v}_2$$ | ||
| $$\vec{v}_2 = \frac{m_0 + m}{m} \vec{v} = 3\vec{v}$$ | $$\vec{v}_2 = \frac{m_0 + m}{m} \vec{v} = 3\vec{v}$$ | ||
| The velocity of the second fragment is $3$ times greater than $v$, which means it will cover a horizontal distance to the ground of $3L$ from the point of the explosion, or $4L$ from the cannon. | The velocity of the second fragment is $3$ times greater than $v$, which means it will cover a horizontal distance to the ground of $3L$ from the point of the explosion, or $4L$ from the cannon. | ||
| @@ -20,7 +14,3 @@Solution | |||
| #### Answer | #### Answer | ||
| At a horizontal distance of $4L$ from the cannon. | At a horizontal distance of $4L$ from the cannon. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||