Question

A stone weighing 3 kg falls from the top of a tower 100 metres high and buries itself 2 metres deep in the sand . the time of penetration is???

Answer 1

A stone weighing $3\mathrm{kg}$ falls from the top of a tower 100 meters high and buries itself 2 meters deep in the sand. The time of penetration is?

Solution.

$m = 3kg,H = 100m,h = 2m,g = 9.8\frac{m}{s^2};$

The change of the potential energy of the stone is equal the work of the resistance force of sand:

$$\Delta W _ {p} = F h;$$ $$\Delta W _ {p} = m g (H + h);$$ $$m g (H + h) = F h.$$

The change of the momentum of the stone is equal product of the resistance force at the time:

$$\Delta p = F \Delta t;$$ $$\Delta p = p _ {2} - p _ {1};$$ $$p _ {1} = m v _ {1};$$

$v_{1} = 0$ - the stone is at rest.

$$p _ {1} = 0.$$ $$p _ {2} = m v _ {2};$$ $$\Delta p = m v _ {2} - 0 = m v _ {2}.$$ $$\left\{ \begin{array}{c} m v _ {2} = F \Delta t; \\ m g (H + h) = F h. \end{array} \right.$$

Divide first equation by second equation:

$$\frac {v _ {2}}{H + h} = \frac {\Delta t}{h};$$

The time of penetration:

$$\Delta t = \frac {h v _ {2}}{H + h}.$$

$v_{2}$ we will find from the law of conservation of energy.

The kinetic energy of the stone at the point $\mathbf{O}$ is equal the potential energy of it at the point $\mathbf{A}$ :

$$W _ {k} = W _ {p};$$ $$\frac {m v _ {2} ^ {2}}{2} = m g H;$$ $$v _ {2} = \sqrt {2 g H}.$$ $$\Delta t = \frac {h \sqrt {2 g H}}{H + h}.$$ $$\Delta t = \frac {2 \cdot \sqrt {2 \cdot 9 . 8 \cdot 1 0 0}}{1 0 0 + 2} = 0. 8 6 8 (s).$$

Answer: The time of penetration is $\Delta t = 0.868s$ .