According to Newton's second law:

dP/dt = F

It is convenient to rewrite this as:

dP = F dt (1)

Let's consider the change of momentum of the rocket plus fuel system and plug that into Eq. (1). We look at the rocket as it is at some time after launch. The rocket at that time contains some fuel. This is the "old state". We then look at how the momentum of that state changes:

Rocket at time t ---> Rocket at time t + dt plus some expelled burned fuel

--->

dP = (New mass of rocket) times (new velocity of rocket) + (mass of expelled fuel) times (velocity of expelled fuel) - ( old mass of rocket) times (old velocity of rocket) (2)

We define M as a function of time to be the mass of the rocket at time t. Between time
t and t + dt the mass of the rocket will increase by dM = M-dot dt. Here M-dot must, of course, be negative. The mass of the exhaust is -dM. If you insert in Eq. (2):

New mass of rocket = M + dM

new velocity of rocket = v + dv

old mass of rocket = M

old velocity of rocket = v + dv

mass of expelled fuel = -dM

velocity of expelled fuel = v - v_ex

(note that the velocity of the fuel is v_ex w.r.t. the rocket in the opposite direction in which the rocket is moving, so. w.r.t. to the ground it is v - v_ex.)

you get:

dP = M dv + v_ex dM

Here you have to ignore the higher order terms that are product of two differentials, like dv times dM. Eq. (1) then implies:

M dv + v_ex dM = F dt -->

M dv/dt = -v_ex dM/dt + F

In case of the rocket moving vertically in a gravitational field this becomes:

M dv/dt = -v_ex dM/dt - Mg (3)

To integrate this equation to find the velocity as a function of time it is not necessary to plug in the mass as a function of time, you can leave it undetermined and plug in the formula in the solution.

If you divide both sides of (3) by M, you get:

dv/dt = -v_ex 1/M dM/dt - g

And integrating this from t = 0 to t gives:

v(t) - v(0) = -v_ex Log[M(t)/M(0)] - gt --->

v(t) = v(0) - gt - v_ex Log[M(t)/M(0)] (4)

Note that the chain rule implies that the derivative of Log[M(t) ] is 1/M dM/dt . Also note that I write Log for the natural logarithm denoted by ln in the problem.

You can just plug in the figures given in problem c in Eq. (4) In problem d, what is meant is that before take off the rocket is resting on the ground so there is a normal force acting on the rocket exerted by the ground. This is not included in the equations. So, if the thrust from the exhaust is not enough to overcome gravity then the equations will seem to imply that the rocket will get negative velocity! In reality the rocket will fail to lift off.

Problem 2:

Here you insert v_0 = 0 and M(t) = M(0) - k t in (4) and integrate from t = 0 to t. You need to use that the indefinite integral of log(x) is x Log(x) - x.

The -gt term yields -1/ g t^2

and the integral of - v_ex Log[M(t)/M(0)] from t = 0 to t is:

- v_ex Int 0 to t of Log[1-kt'/M(0)] dt' =

G(t' = t) - G(t' = 0)

where

G(t') = v_ex M(0)/k [(1-kt'/M(0)) Log(1-kt'/M(0)) - (1-kt'/M(0))]

G(t) can be written as:

G(t)= v_ex M(0)/k [M(t)/M(0)Log(M(t)/M(0)) - M(t)/M(0)]

And you see that G(0) = -v_ex M(0)/k. You can thus write:

G(t) - G(0) = v_ex M(0)/k [M(t)/M(0)Log(M(t)/M(0)) - (M(t) - M(0))/M(0)] =

v_ex M(t)/k Log(M(t)/M(0)) + v_ex t =

v_ex t - v_ex M(t)/k Log(M(0)/M(t))

So, the height as a function of time is

y(t) = -1/2 g t^2 + v_ex t - v_ex M(t)/k Log(M(0)/M(t))

And you can just plug in the relevant numbers in this equation.

Problem 3

Here you should use that the center of mass can only be affected by external forces. So, the explosion won't affect the motion of the center of mass. The center of mass will move along a trajectory the shell would have moved if it hadn't exploded. In problem b, if the two parts land at the same time then they were moving at the same height at equal times. The center of mass which is in between the two parts was then also at the same height and will thus be in between the two parts when they are on the ground. So, if one part lands 100 meters beyond the target, the other part will land 100 meters short of the target.

To answer problem c, consider this case. Suppose the explosion is so violent that it causes the two shells to move at very high speeds relative to each. Suppose that the explosion shoots one part toward the ground such that it hits the target 100 meter too far. Now consider what happens as a function of the velocity of that part just after the explosion as you let it go to infinity. You choose it such that it will continue to hot the target 100 meters too far. As you increase it the time to impact will become smaller and smaller so that gravity has less and less time to act. This means that the trajectory till impact will look more and more like a straight line. So, theoretically there is no obstacle to making the velocity of that part arbitrarily high. As you let the velocity go to infinity, the other part is shot in the opposite direction at some angle toward the sky and will thus hit the ground arbitrarily far away in the opposite direction. So, it is thus certainly not true that the other part will hit the ground hundred meters short!