Answer
$-67\ m/s$
Work Step by Step
Impulse $J=32.4\ N.s$
Time for which force applied is $t = 27\ ms = 27\times 10^{-2}\ s$
Mass of a ball $m=0.40\ kg$
Initial velocity of ball $v_i =14\ m/s$
Let the initial direction of the motion is in positive x direction.
The magnitude of the force $F_avg$ is given by
$F_{avg} =\frac{J}{\Delta t}$
$F_{avg} =\frac{32.4\ N.s}{ 27\times 10^{-2}\ s}$
$F_{avg} = 1200\ N$
Given the force is in negative direction. we use the momentum-impulse theorem:
$-F_{avg}\ t = mv_f -mv_i$
$v_f = \frac{mv_i-F_{avg}\ \Delta t }{m}$
$v_f = \frac{(0.40\ kg)(14\ m/s)-(1200\ N)(27\times 10^{-2}\ s) }{0.40\ kg}$
$v_f =-67\ m/s $
