## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can 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.