Answer
The change in the cart's velocity is $~~+4.4~m/s$
Work Step by Step
We can use conservation of momentum to find the velocity of the cart after the man jumps off:
$p_f = p_i$
$(75~kg)(0)+(39~kg)(v_f) = (75~kg+39~kg)(2.3~m/s)$
$v_f = \frac{(114~kg)(2.3~m/s)}{39~kg}$
$v_f = 6.7~m/s$
We can see that the change in the cart's velocity is $~~+4.4~m/s$.
