# Conservation of $y$ component of momentum

Rain with mass $$mr$$, falling vertically downwards at speed $$v$$, into a truck of mass $$mt$$, moving on a horizontal surface at speed $$u$$ inital, ignoring friction and air resistance. Taking the system to be the truck and the rain, momentum should be conserved.

x-direction momentum conservation: (truck mass x truck velocity) + (rain mass x rain velocity) before collision = (truck mass x truck velocity) + (rain mass x rain velocity) after collision

$$m_t \times u_{initial} = (m_t + m_r) \times u_{final}$$ so $$u_{final} = (m_t \times u_{initial})/(m_r + m_t)$$

So the rain + truck system slows down to conserve momentum - all good so far

y-direction momentum conservation:(truck mass x truck velocity) + (rain mass x rain velocity) before collision = (truck mass x truck velocity) + (rain mass x rain velocity) after collision

$$m_r \times v = 0$$ so momentum is clearly not conserved in the y direction but my question is why? I know physically that the rain exerts a force on the truck because of this momentum change but if I take the system to be the rain + truck shouldn't momentum conservation hold?

• I think you forget that the truck is not floating in space but that you have to take into account the force the truck excerts on the road/floor the truck is driving on. Commented May 10 at 16:04

If only the rain and the truck constitute the system, gravity is an external force and therefore $$y$$ momentum is not conserved. But if the Earth is included in the system then the system is isolated and $$y$$ momentum is conserved. That’s because the $$y$$ momentum of the center of mass of the rain + Earth is zero both before and after the rain moves with the truck. Think of the rain falling down and the Earth moving up before the vertical velocity of the rain is zero.