From the top of a 100 m tall building is relesead at rest, a 900 g mass brick under the action of
gravitational force. Find:
a) The speed of the brick when touching the floor;
b) The momentum of brick by touching the floor;
c) The impulse of force acting on brick during fall.
Problem data:
- Fall height: S = 100 m;
- Mass of brick: m = 900 g;
- Initial speed of brick: v0 = 0;
- Acceleration due to gravity: g = 9.8 m/s2.
Problem diagram:
We choose a frame of reference pointing downward at the top of the building. As the brick is at rest, its
initial speed is zero, v0 = 0, its initial position is also zero,
S0 = 0, and the acceleration due to gravity is in the same direction as the frame
of reference (Figure 1).
Solution:
First, we must convert the mass of brick given in grams (g) to kilograms (kg) used in the
International System of Units (SI)
\[
\begin{gather}
m=900\;\mathrm{\cancel g}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel g}}=0.9\;\mathrm{kg}
\end{gather}
\]
a) The brick is in free fall under the action of gravitational force, using the equation of velocity as a
function of displacement
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{v^2=v_0^2+2a\Delta S}
\end{gather}
\]
the acceleration of the motion is the acceleration due to gravity, a = g, and substituting the values
\[
\begin{gather}
v^2=v_0^2+2g(S-S_{0}) \\[5pt]
v^2=0^2+2\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)\times(100\;\mathrm m-0) \\[5pt]
v^2=0+1960\;\mathrm{\frac{m^2}{s^2}} \\[5pt]
v=\sqrt{1960\;\mathrm{\frac{m^2}{s^2}}\;}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{v\simeq 44.3\;\mathrm{m/s}}
\end{gather}
\]
b) The momentum is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{p=mv} \tag{I}
\end{gather}
\]
substituting the mass given to the brick and the speed, calculated in the previous item
\[
\begin{gather}
p=(0.9\;\mathrm{kg})\times\left(44.3\;\mathrm{\frac{m}{s}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{p=39.9\;\mathrm{kg.m/s}}
\end{gather}
\]
c) Applying the Impulse-Momemtum Theorem
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{I=\Delta p=p_{f}-p_{i}}
\end{gather}
\]
substituting the expression (I), for the initial and final values
\[
\begin{gather}
I=mv_{f}-mv_{i}
\end{gather}
\]
the final speed, vf = v, calculated in item (a), and the initial speed,
vi = v0 = 0
\[
\begin{gather}
I=mv-mv_{0} \\[5pt]
I=(0.9\;\mathrm{kg})\times\left(44.3\;\mathrm{\frac{m}{s}}\right)-(0.9\;\mathrm{kg})\times 0\\[5pt]
I=40.2\;\mathrm{\frac{kg.m}{s}}-0
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{I=39.9\;\mathrm{N.s}}
\end{gather}
\]
DE
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FR
PT-BR
