Question

Q:

Answer 1

@Vincent-Lyon.Fr

Answer 2

The trolley is an open system.
In order to use Newton's second law, you need to define a closed system* and write down its momentum at instant t and its momentum at instant t + dt.
Since no net horizontal force acts on the system, those two momenta will be equal.

I will get back to you later.

*closed system will be trolley + small mass dm of water that will fall in the trolley between instants t and t+dt

Answer 3

okay ..let at any instant t....mass of the water in the trolley is say m'
now
linear momentum at t : (M+m')v + (dm)u
and at time t+dt : (M+m'+dm)(v-dv)
as the velocity of the trolley decrease with time
right?

Answer 4

@Vincent-Lyon.Fr if you won't mind i can answer it

Answer 5

he neglected friction so mv = mv.
the volume of water getting out of the nozzle = area* hieght
\[V= \frac{ \pi d ^{2} u t }{ 4 }\]
then the mass will be \[m= \frac{ \pi d ^{2} u t \rho }{ 4 }.\]
now we will get the 2 components of the velocity
x direction = u
y direction = g(10) t
Now use conservation of momentum
mv = mv
\[\frac{ \pi d ^{2} u t \rho }{ 4 } \sqrt{u ^{2}+a ^{2}t ^{2}}= (M+\frac{ \pi d ^{2} u t \rho }{ 4 } )V\]
Thus just divide by the term next to V and it is solved ( you can simplify it as you want).
\[V = \frac{ \pi d ^{2} u t \rho }{ 4(M +\frac{ \pi d ^{2} u t \rho }{ 4 } ) } \times \sqrt{u ^{2}+a ^{2}t ^{2}}.\]
I assumed that there's no distance at the first between the trolly and nozzle (if their was you will just add that time as the water won't reach it until it covers this distance between them and then will be continually reach trolly in no time).