Answer
(a) $J=3.7mN$
(b) $x=20mm$
Work Step by Step
(a) We can find the required impulse as follows:
$J=Ft$
We plug in the known values to obtain:
$J=41\times 10^{-3}\times 90\times 10^{-3}$
This simplifies to:
$J=3.7mN$
(b) We can find the required stopping length as follows:
$x=vt-\frac{at^2}{2}$
We plug in the known values to obtain:
$x=0.445\times 90\times 10^{-3}-\frac{4.94\times (90\times 10^{-3})^2}{2}$
This simplifies to:
$x=20mm$
