Answer
$1.1\times 10^3 N$, $4.7\times 10^2N$
Work Step by Step
We can find the required average force as follows:
$F_{avg}=\frac{\Delta P}{\Delta t}$
$F_{avg}=\frac{m(v_f-v_i)}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{0.16(0-25)}{3.5\times 10^{-3}}$
$\implies F_{avg}=\frac{-4}{3.5\times 10^{-3}}$
(The negative sign shows that the change in impulse is in the opposite direction to that of initial momentum)
The above equation simplifies to:
$F_{avg}=-1.1\times 10^3 N$
Thus the magnitude of the average force is: $F_{avg}=1.1\times 10^3 N$
Now for $\Delta t=8.5\times 10^{-3}s$
$F_{avg}=\frac{4}{8.5\times 10^{-3}}$
This simplifies to:
$F_{avg}=4.7\times 10^2N$
