Chapter 9 Problem 89

Problem

A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 24.0 g and a speed of 310 m/s, how high will the block rise into the air after the bullet becomes embedded in it?

Solution

The collision is totally inelastic and conserves momentum. In the second part the energy is conserved in the rising of the block and embedded bullet after the collision.

first part: $P_{i}=P_{f}$ $\rightarrow mv=(m+M)v^{\prime }$

second part: $K_{i}+U_{i}=K_{f}+U_{f}$ $\rightarrow {\frac {1}{2}}(m+M)v^{\prime 2}+0=0+(m+M)gh$

$v^{\prime }={\sqrt {2gh}}$

hence

$v={\frac {m+M}{m}}v^{\prime }={\frac {m+M}{m}}{\sqrt {2gh}}$

$\rightarrow h={\frac {1}{2g}}\left({\frac {mv}{m+M}}\right)^{2}={\frac {1}{2\left(9.80{\text{m/s}}^{2}\right)}}\left({\frac {\left(0.0240{\text{kg}}\right)\left(310{\text{m/s}}\right)}{\left(0.0240{\text{kg}}\right)+\left(1.40{\text{kg}}\right)}}\right)^{2}=1.4{\text{m}}$