Answer
(a) $p_0 = 290~kg\cdot m/s$ toward the east
(b) $Impulse = 290~kg\cdot m/s$ toward the west
(c) $Impulse = 290~kg\cdot m/s$ toward the east
(d) $F = 340 ~N$, which is directed toward the east
Work Step by Step
(a) $p_0 = mv_0 = (95~kg)(3.0~m/s) = 290~kg\cdot m/s$
The initial momentum is directed toward the east.
(b) Since the fullback came to a stop, the impulse is equal in magnitude to $p_0$ but in the opposite direction. Therefore,
$Impulse = 290~kg\cdot m/s$ to the west.
(c) By Newton's third law, the impulse on the tackler must be equal in magnitude to the answer in part (b), but in the opposite direction. So,
$Impulse = 290~kg\cdot m/s$ to the east
(d) $Impulse = 290~kg\cdot m/s$
$F\cdot t = 290~kg\cdot m/s$
$F = \frac{290~kg\cdot m/s}{0.85~s}$
$F = 340 ~N$, which is directed to the east