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?

1 Answer
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 #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