Question 9.50: A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden blo......

(a) We choose +x along the initial direction of motion and apply momentum
conservation:

m_{\mathrm{bullet}}{\vec{\nu}}_{i}=m_{\mathrm{bullet}}{\vec{\nu}}_{1}\ +\ m_{\mathrm{block}}{\vec{\nu}}_{2}

(5.2\ s)(672\mathrm{~m/s})=(5.2\ g)(428\mathrm{~m/s})\ +(700\mathrm{~g}){\vec{\nu}}_{2}

which yields \nu_{2}\, = 1.81 m/s.

(b) It is a consequence of momentum conservation that the velocity of the center of mass is unchanged by the collision. We choose to evaluate it before the collision:

\vec{\nu}_{\mathrm{com}}\;=\;{\frac{m_{\mathrm{bullet}}\vec{\nu}_{i}}{m_{\mathrm{bullet}}+m_{\mathrm{block}}}}={\frac{(5.2\;\mathrm{g})(672\;\mathrm{m}/s)}{5.2\;\mathrm{g}+700\;\mathrm{g}}}=4.96\;\mathrm{m/s}.