Answer
The speed of the bullet-block center of mass is $~~4.95~m/s$
Work Step by Step
We can use conservation of momentum to find the velocity $v_b$ of the block:
$p_f = p_i$
$(5.20~g)(428~m/s)+(700~g)(v_b) = (5.20~g)(672~m/s)$
$(700~g)(v_b) = (5.20~g)(672~m/s)-(5.20~g)(428~m/s)$
$v_b = \frac{(5.20~g)(672~m/s)-(5.20~g)(428~m/s)}{700~g}$
$v_b = 1.81~m/s$
The resulting speed of the block is $~~1.81~m/s$
We can find the speed of the bullet-block center of mass:
$v_{com} = \frac{(5.20~g)(428~m/s)+(700~g)(1.81~m/s)}{5.20~g+700~g}$
$v_{com} = 4.95~m/s$
The speed of the bullet-block center of mass is $~~4.95~m/s$
