Hint 5

(a) By conservation of 4-momentum,

$${\sf p}_0={\sf p}_f+{\sf p}_X,$$

where the subscript $X$ refers to the exhaust. Since, for the moment, we are not interested in the final energy of the rocket, we can rewrite this as an equation for $\hbox{\ss p}_f$ and square both sides to get

$$-M_f^2c^2=-M_0^2c^2-2{\sf p}_0\cdot{\sf p}_X.\eqno(1)$$

The initial 4-momentum of the rocket and the 4-momentum of the (massless) exhaust are

$${\sf p}_0=(0,0,0,iM_0c),\qquad\qquad {\sf p}_X= (E_X/c)(-1,0,0,i),$$

and hence

$${\sf p}_0\cdot{\sf p}_X=-M_0E_X.$$

We can therefore use (1) as an equation for the energy $E_X$ of the exhaust. This gives

$$E_X={M_0^2-M_f^2\over 2M_0}c^2.$$

(b) (c) You now have everything you need to write down ${\sf p}_X$and ${\sf p}_f$.

(d) If the final speed of the rocket is $v$, then its energy can be written in the form

$$E_f=\gamma(v)M_fc^2.$$

Equate this to the final energy you found in part (c) and solve for $v$.