#### Answer

After the bullet is embedded in the second block, the speed of the second block is 1.96 m/s

#### Work Step by Step

We first find the initial momentum of the bullet.
$p_0 = m~v_0$
$p_0 = (0.010~kg)(400~m/s)$
$p_0 = 4.0~N~s$
By conservation of momentum, the sum of the momenta of the two blocks (after the bullet is embedded in the second block) will be equal to $p_0$. We can find the speed $v_f$ of the second block.
$p_f=p_0$
$(0.50~kg)(6.0~m/s)+(0.51~kg)~v_f = 4.0~N~s$
$(0.51~kg)~v_f = 1.0~N~s$
$v_f = \frac{1.0~N~s}{0.51~kg}$
$v_f = 1.96~m/s$
After the bullet is embedded in the second block, the speed of the second block is 1.96 m/s.