Answer
(a) $1.2\times 10^3N$
(b)$1.2\times 10^4N$
Work Step by Step
(a) Before we find the required average force, we have to find the change in momentum
$\Delta P=m(v_f-v_i)$
$\implies \Delta P=(0.35)(0-(-10))$
$\implies \Delta P=3.5\space Kg.m/s$
We know that
$F_{avg}=\frac{\Delta P}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{3.5}{3\times 10^{-3}}$
$F_{avg}=1.2\times 10^3 N$
(b) As $F_{avg}=\frac{\Delta P}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{3.5}{3\times 10^{-4}}$
$F_{avg}=1.2\times 10^4N$
