Answer
The speed of the bullet as it leaves block 1 is $~~721.4~m/s$
Work Step by Step
The speed of the bullet as it leaves block 1 is the same as its speed just before it enters block 2.
We can use conservation of momentum to find the speed of the bullet as it leaves block 1:
$p_i = p_f$
$(3.50~g)~v = (1803.5~g)~(1.40~m/s)$
$v = \frac{(1803.5~g)~(1.40~m/s)}{3.50~g}$
$v = 721.4~m/s$
The speed of the bullet as it leaves block 1 is $~~721.4~m/s$
