Answer
The magnitude of the truck's momentum increases by $~~5,900~kg~m/s$
Work Step by Step
We can convert $41~km/h$ to units of $m/s$:
$(41~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 11.39~m/s$
We can convert $51~km/h$ to units of $m/s$:
$(51~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 14.17~m/s$
We can find the initial momentum:
$p_i = mv$
$p_i = (2100~kg)(11.39~m/s)$
$p_i = 23,900~kg~m/s$ (north)
We can find the final momentum:
$p_f = mv$
$p_f = (2100~kg)(14.17~m/s)$
$p_f = 29,800~kg~m/s$ (east)
We can find the magnitude of the change in momentum:
$\Delta p = p_f-p_i$
$\Delta p = 29,800~kg~m/s-23,900~kg~m/s$
$\Delta p = 5,900~kg~m/s$
The magnitude of the truck's momentum increases by $~~5,900~kg~m/s$
