Question

A weather rocket is launched straight up. Seconds into the flight, its velocity is 130m/s and it is accelerating at 19 m/s^2. At this instant, the rocket's mass is 59 kg and it is losing mass at the rate of 0.55 kg/s. What is the net force on the rocket?

Answer 1

1049,5 N up

Explanation:

Rocket's motion depends on Newton 2 and Newton 3.

At time t, rocket has mass m, moves with velocity `vecv` ejecting fuel in opposite direction with velocity `vecu` at a rate `-(dm)/dt` .

At time `t+dt`, velocity is now `vecv+dvecv`, and mass `m-dm`.

Therefore momentum at time t is `vecp(t)=mvecv`

Momentum at time t + dt is `vecp(t+dt)=vecp_(rocket)+vecp_(fuel_`
`=(m-dm)(vecv+dvecv)+dm[(vecv+dvecv)-vecu]`
`=mvecv+mdvecv-vecudm`

Therefore change in momentum is `dvecp=vecp(t+dt)-vecp(t)=mdvecv-vecudm`

`therefore vecF=(dvecp)/dt=d/dtmvecv = m(dvecv)/(dt)-vecu (dm)/(dt)`

This is known as the Rocket Equation and may be used to solve this problem.

Substituting in we get

`sumvecF=59xx19 - 130xx0,55 = 1049,5 N` upwards