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Suppose a box of mass $m$ is thrown at angle $\theta$ with velocity $v$. At the topmost point,the box explodes into two identical smaller boxes. One of the boxes falls straight to the ground at that point.
I have seen that conservation of momentum is used to calculate the velocity of other box after the explosion. But gravity is always acting downwards. So how can momentum be conserved?
The horizontal component of momentum is conserved since there are no forces that act in this direction.
The same cannot be said in the vertical direction since gravity is acting on the boxes.
So there will be conservation of momentum in the x-direction. You can see this mathematically in the equation $$2mv_0\cos(\theta)=0\times m + mv_2$$ which relates the horizontal momentum before the box separates to the horizontal momentum after it separates. Note how the first term on the RHS vanishes because this corresponds to the part of the box that moves vertically downward with no component of velocity in the x-direction.
Momentum in the x direction will be conserved when looking at the two boxes alone.
Momentum in the y direction will be conserved when looking at the two boxes in the earth as one system, since it's the earth that imparts momentum to the boxes. This is easy to manage because it's well known that the earth will impart $mg\Delta t$ momentum to an object with mass $m$ that's in free fall for time $t$.
It is true that momemtnum is conserved only if external force is zero. But here as in written in photo: $$2m(v_0 \cos\theta)=m(0)+mv_2$$
This equation represents momentum conservation only in horizontal/x direction, where there is no external force (if mentioned no air resistance or wind). Gravity acts only in downward/y direction so in that direction, we can't conserve momentum.
Conservation of momentum us used here because the explosion forces are assumed to be much larger than the gravitational force, so that on the time scale of the explosion, the effects of gravity can be ignored.
