Answer
(a) $I=200\space Kg.m/s$
(b) $F_{avg}=800 N$
Work Step by Step
(a) The required impulse can be determined as follows:
As $\Delta P=mv$.......eq(1)
We know that
$v_f=v_i+at$
$\implies v_f=0+at$
$\implies v_f=(16)(0.25)=4m/s$
Putting values in equation (1), we obtain:
$\Delta P=(50)(4)=200\space Kg.m/s$
Impulse is given as
$I=\Delta P$
$\implies I=200 \space Kg.m/s$
(b) We can find the average force as
$F_{avg}=\frac{\Delta P}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{200}{0.25}$
This simplifies to:
$F_{avg}=800 N$
