Answer
When the water rocket ejects water to propel itself, this creates a force pair that can be described by Newton's 3rd Law. According to this law, the force of ejection will be equal in magnitude but opposite in direction to the force acting upon the body of the water rocket. We can write this like so:
$-F_{water} = F_{rocket}$
Because F = ma = m∆v/t, then this can be rewritten as:
$-\frac{m\Delta v}{t}_{water} = \frac{m\Delta v}{t}_{rocket}$
Given that this model works over the same time period, then $t_{water} = t_{rocket}$. We can multiply both sides by $t_{water}$ to realise the following:
$-m\Delta v_{water} = m\Delta v_{rocket}$
$-p_{water} = p_{rocket}$
Thus the rocket's motion follows the law of conservation of momentum, where the momentum of the water it propels is equal to its own momentum over a given time period.
As the rocket continues to fly, the mass of water that it propels will increase as the mass of water it has left decreases. Presuming that the velocity it ejects the water at is constant, then as it continues to fly, the magnitude of the left hand side of the equation ($p_{water}$) will increase. The right hand side of the equation ($p_{rocket}$) will increase in response. However, the mass component of $p_{rocket}$ will be decreasing, so the magnitude of the velocity will increase. This is acceleration.
I'm not sure about whether the acceleration is constant or not. I'm going to presume it isn't, but if someone knows this please let me know!
Question
A football is kicked from flat ground on a windless day. It flies through the air over 30 metres, and lands after 4 seconds. What was the maximum height it reached, and what was the angle it was launched at? [3 marks]
At least I think this one works.
