Answer
The velocity of the 5.0-kg block immediately after the collision is $~~2.0~m/s$
Work Step by Step
We can use conservation of momentum to find the velocity $v_f$ of the 5.0-kg block immediately after the collision:
$p_f = p_i$
$(5.0~kg)~v_f+(10.0~kg)(2.5~m/s) = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)$
$(5.0~kg)~v_f = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(2.5~m/s)$
$v_f = \frac{(5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(2.5~m/s)}{5.0~kg}$
$v_f = 2.0~m/s$
The velocity of the 5.0-kg block immediately after the collision is $~~2.0~m/s~~$ in the original direction.
