A 4700 kg block falls vertically through 6.9 m and then collides with a 380 kg pile, driving it 3.3 cm into bedrock. Assuming that the block-pile collision is completely inelastic, find the magnitude of the average force on the pile from the bedrock during the 3.3 cm descent.

Question

michele_laino · Answer

the kinetic energy of the falling block has to be equal to its potential energy, so we can write:
$$\frac{1}{2}m_B{v^2} = m_Bgh$$
where $m_B=4700$, $h=6.9$. So we can compute the magnitude of the velocity of the block when it collides with the pile:
$$v = \sqrt {2gh} $$
Now, since such collision has to be inelastic, then the work done by the resistance force $F$, has to be equal to the kinetic energy of the system block+pile, so we can write:
$$\frac{1}{2}\left( {{m_B} + {m_P}} \right){v^2} = Fd$$
where $m_P=380$, and $v=\sqrt{2gh}$, $d=0.033\,meters$. So we get:
$$F = \frac{{\left( {{m_B} + {m_P}} \right){v^2}}}{{2d}} = \frac{{\left( {{m_B} + {m_P}} \right)}}{{2d}}2gh = \frac{{\left( {{m_B} + {m_P}} \right)gh}}{d}$$

michele_laino · Answer

please substitute numeric data into my formula above

michele_laino · Answer

please I have made an error, the quantity $v$ is not $\sqrt{2gh}$. Since the collision is inelastic, total momentum is conserved, so we can write:
$${m_B}{v_0} = \left( {{m_B} + {m_P}} \right){v_1}$$
where I called $v_0=\sqrt {2gh}$. So the speed of the obkect "block+pile" is:
$${v_1} = \frac{{{m_B}{v_0}}}{{{m_B} + {m_P}}}$$
then the right formula for $F$, is:
$$F = \frac{{\left( {{m_B} + {m_P}} \right)v_1^2}}{{2d}} = \frac{{\left( {{m_B} + {m_P}} \right)}}{{2d}} \cdot {\left( {\frac{{{m_B}{v_0}}}{{{m_B} + {m_P}}}} \right)^2} = ...?$$

anonymous · Answer

Thank you , sorry I haven't responded earlier @Michele_Laino

anonymous · Answer

is it 8960156.983 N?

michele_laino · Answer

yes! correct! I got: $F=8,910,318.54$ Newtons