# 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?

Sep 18, 2015

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 $\vec{v}$ ejecting fuel in opposite direction with velocity $\vec{u}$ at a rate $- \frac{\mathrm{dm}}{\mathrm{dt}}$ .

At time $t + \mathrm{dt}$, velocity is now $\vec{v} + \mathrm{dv} e c v$, and mass $m - \mathrm{dm}$.

Therefore momentum at time t is $\vec{p} \left(t\right) = m \vec{v}$

Momentum at time t + dt is vecp(t+dt)=vecp_(rocket)+vecp_(fuel_
$= \left(m - \mathrm{dm}\right) \left(\vec{v} + \mathrm{dv} e c v\right) + \mathrm{dm} \left[\left(\vec{v} + \mathrm{dv} e c v\right) - \vec{u}\right]$
$= m \vec{v} + m \mathrm{dv} e c v - \vec{u} \mathrm{dm}$

Therefore change in momentum is $\mathrm{dv} e c p = \vec{p} \left(t + \mathrm{dt}\right) - \vec{p} \left(t\right) = m \mathrm{dv} e c v - \vec{u} \mathrm{dm}$

$\therefore \vec{F} = \frac{\mathrm{dv} e c p}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} m \vec{v} = m \frac{\mathrm{dv} e c v}{\mathrm{dt}} - \vec{u} \frac{\mathrm{dm}}{\mathrm{dt}}$

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

Substituting in we get

$\sum \vec{F} = 59 \times 19 - 130 \times 0 , 55 = 1049 , 5 N$ upwards