At a certain instant, the velocity of a 0.05-kg rocket is <0,2.4,0> and all significant forces on the rocket are

Question

At a certain instant, the velocity of a 0.05-kg rocket is <0,2.4,0> and all significant forces on the rocket are:

F thrust by exhaust on rocket = <0,4,0> N,

F grav by Earth on rocket = <0,-0.49,0> N

F by air on rocket = <0,-1.2,0> N

The first part asked what the net force on the rocket is which is Fnet = <0,2.31,0> N

The second part asked assume the mass of the rocket is approximately constant what will be the rocket's momentum 0.02s later: answer: <0, 0.1662, 0>

The last question is what will be the rockets displacement over this 0.02s interval

delta r = < >

The last part is the part I'm having a problem with. I don't know how to find the answer for displacemnt, the formula velocity * time = displacement isn't correct.

Answer 1

What you need here is Newton's second law. Remember the unbalanced forces F_net = ma.

where the mass is given and the net force you have calculated correctly.

If we don't have the information we need we can use the kinematic equations to determine the final position.

One thing that makes the problem easier is that the velocity and acceleration are both in the y direction, so the displacement will also be in the y-direction.

2.31 N = 0.05kg(a) so a = 2.31N/0.05kg = 46.2 m/s² in the y-direction or in vector notation (0, 46.2, 0)

Since we have now the acceleration in the y-direction and the initial velocity also in the y=direction, we can get the displacement after 0.02 seconds Δr=Δy using

Δy=v_ot + 0.5at^2Δ

Δy =(2.4)(0.02)+0.5(46.2)(0.02)² = 0.048+0.00924 =0.048924 =0.49 to the correct number of significant figures.

As a vector you should have Δr = <0, 0.49, 0>

Answer 2

IN the short time interval under consideration the net force acting on the rocket may be regarded as the constant vector Fnet = <0,2.31,0> N, this means that during the time interval the rocket follows an uniformly accelerated motion with a=Fnet/M, M=0.05 kg being the mass of the rocket.

We now that the general formula for the position under uniform acceleration starting at t₀=0 is

r(t)-r₀=at²/2+v₀t, where r₀ and v₀ are the position and velocity of the moving object at t=0.

Since the initial velocity v₀=<0,2.4,0> and the acceleration during the time interval are parallel, the motion is along a straight line starting at the initial position that we can take to be r₀=0. Accordingly, the position at the time o interest is

r(t)=Fnet/Mt²/2+v₀t=0.5<0,2.31,0> N(0.02s)²/0.05+<0,2.4,0>(0.02s)=<0,0.057,0>m

Since the motion was al along the same direcction, the distance is just the magnitude of r(t), which means that the answer is

distance=0.057m