## College Physics (4th Edition)

The total mass of matter and antimatter required is $1.35\times 10^4~kg$
We can find the kinetic energy of the starship: $K = (\gamma-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{(0.3500~c)^2}{c^2}}}-1)(2.0\times 10^5~kg)(3.0\times 10^8~m/s)^2$ $K = (1.0675-1)(2.0\times 10^5~kg)(3.0\times 10^8~m/s)^2$ $K = 1.215\times 10^{21}~J$ We can find the total mass $M$ of matter and antimatter required to produce this amount of energy: $E = Mc^2$ $M = \frac{E}{c^2}$ $M = \frac{1.215\times 10^{21}~J}{(3.0\times 10^8~m/s)^2}$ $M = 1.35\times 10^4~kg$ The total mass of matter and antimatter required is $1.35\times 10^4~kg$.