Answer
$9.1$ kg.m/s
Work Step by Step
Initial speed of the hand $v_i = 13\ m/s$
Final speed of the hand $v_f = 0\ m/s$
Mass of the hand $m=0.7\ kg$
Duration of the collision $\Delta t=5\times 10^{-3}\ s$
Since the change in momentum is equal to the impulse applied;
$J= m(\Delta v)$
$J=m(v_f - v_i)$
$J=0.7\ kg (0\ m/s - 13\ m/s)$
$J=-9.1\ kg. m/s$
Therefore, the magnitude of impulse is $J=9.1\ kg.m/s$.
