Answer
The speed of the flatcar increases by $~~1.1~m/s$
Work Step by Step
The man's mass is $\frac{915~N}{9.8~m/s^2} = 93.37~kg$
The flatcar's mass is $\frac{2415~N}{9.8~m/s^2} = 246.43~kg$
When the man is running, let the flatcar's velocity be $v$
Then the man's velocity is $v-4.00~m/s$
We can use conservation of momentum to find $v$:
$p_f = p_i$
$(93.37~kg)(v-4.00~m/s)+(246.43~kg)(v) = (246.43~kg+93.37~kg)(18.2~m/s)$
$(93.37~kg)(v)+(246.43~kg)(v) = (339.8~kg)(18.2~m/s)+(93.37~kg)(4.00~m/s)$
$v = \frac{(339.8~kg)(18.2~m/s)+(93.37~kg)(4.00~m/s)}{339.8~kg}$
$v = 19.3~m/s$
The speed of the flatcar increases by $~~1.1~m/s$
