When a jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum.

`F=(rhoAV_(0))V_(0)-(rhoAV_(0))V_(0)costheta=rhoAV_(0)^(2)[1-costheta]` If surface is free and starts moving due to thrust of the liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let, at any instant, the velocity of surface be `u`. Then the above equation becomes `
F=rhoA(V_(0)-u)^(2)[1-costheta]` Based on the above concept, as shown in the following figure, if the cart is frictionless and free to move in the horizontal direction, then answer the following.

Given that cross sectional area of jet `=2xx10^(-4)m^(2)` velocity of jet `V_(0)=10m//s` density of liquid `=1000kg//m^(3)` ,mass of cart `M=10 kg`.
Velocity of cart at `t = 10 s` is equal to

Options

A. `4m//s`
B. `6m//s`
C. `8m//s`
D. `5m//s`

Answer

C

Explanation

`F=2rhoA(V_(0)-u)^(2)`
`u=` speed of the cart
`m(du)/(dt)=2rhoA(V_(0)-u)^(2)`
`int_(0)^(u)(du)/((V_(0)-u)^(2))=(2rhoA)/mint_(0)^(1)dt`
`[1/(V_(0)-u)]_(0)^(u)=(2rhoAt)/m`
`1/(V_(0)-u)=-1/(V_(0))=(2rhoAt)/m=(4t)/100` ...........i
`((2rhoA)/m=(2xx10^(3)xx2xx10^(-4))/10=4/100)`
At `t=10s`
`1/(V_(0)-h) =4/10+1/10=1/2`
`impliesV_(0)-u=2`
`u=8m/s`