Question 9.52: In Fig. 9-59, a 10 g bullet moving directly upward at 1000 m......
In Fig. 9-59, a 10 g bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 5.0 kg block initially at rest. The bullet emerges from the block moving directly upward at 400 m/s. To what maximum height does the block then rise above its initial position?
We think of this as having two parts: the first is the collision itself – where the bullet passes through the block so quickly that the block has not had time to move through any distance yet – and then the subsequent “leap” of the block into the air (up to height h measured from its initial position). The first part involves momentum conservation (with +y upward):
{\bigl(}0.01\,\mathrm{kg}{\bigr)}{\bigl(}1000\,\mathrm{m/s}{\bigr)}={\bigl(}5.0\,\mathrm{kg}{\bigr)}{\vec{\nu}}+{\bigl(}0.01\,\mathrm{kg}{\bigr)}{\bigl(}400\,\mathrm{m/s}{\bigr)}
which yields \vec{{\nu}} = 12 m/s. The second part involves either the free-fall equations from Ch. 2 (since we are ignoring air friction) or simple energy conservation from Ch. 8. Choosing the latter approach, we have
{\frac{1}{2}}(5.0\,\mathrm{kg})(1.2\,\mathrm{m/s})^{2}=(5.0\,\mathrm{kg})\bigl(9.8\,\mathrm{m/s^{2}}\bigr)h
which gives the result h = 0.073 m.