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.