## College Physics (4th Edition)

(a) $p = 6.7\times 10^9~kg~m/s$ (b) $t = 1.6\times 10^7~s$
(a) We can find the momentum: $p = \gamma~m~v$ $p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~v$ $p = \frac{1}{\sqrt{1-\frac{(0.87~c)^2}{c^2}}}~(12.6~kg)~(0.87~c)$ $p = \frac{1}{\sqrt{1-(0.87)^2}}~(12.6~kg)~(0.87)(3.0\times 10^8~m/s)$ $p = 6.7\times 10^9~kg~m/s$ (b) We can find the required time the force must be applied: $F~t = p$ $t = \frac{p}{F}$ $t = \frac{6.7\times 10^9~kg~m/s}{424.6~N}$ $t = 1.6\times 10^7~s$