Solved Problem on Impulse, Momentum and Collisions
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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).
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 acceleration due to gravity
\[ \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \]
the acceleration of the motion is the acceleration due to gravity, a = g, and substituting the values
\[ \begin{gather} S=S_{0}+v_{0}t+\frac{g}{2}t^2\\[5pt] 100=0+0\times t+\frac{9.8}{2}t^2\\[5pt] 100=4.9t^2\\[5pt] t^2=\frac{100}{4.9}\\[5pt] t=\sqrt{20.4\;}\\[5pt] t\simeq 4.5\;\mathrm{s} \end{gather} \]
This is the interval of time for the brick to reach the ground. Velocity as a function of time is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_{0}+at} \end{gather} \]
substituting the interval of time found, a = g, and the initial speed equal to zero
\[ \begin{gather} v=v_{0}+gt\\[5pt] v=0+9.8\times 4.5 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v\simeq 44.1\;\mathrm{m/s}} \end{gather} \]

b) The momentum is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {p=mv} \tag{I} \end{gather} \]
substituting the mass and the speed, calculated in the previous item
\[ \begin{gather} p=0.9\times 44.1 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {p=39.7\;\mathrm{kg.m/s}} \end{gather} \]

c) The impulse of a force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {I=F\Delta t} \end{gather} \]
the weight is the only force that acts in the brick, given by
\[ \begin{gather} \bbox[#99CCFF,10px] {W=mg} \end{gather} \]
using the interval of time calculated in item (a)
\[ \begin{gather} I=W\Delta t\\[5pt] I=mg\Delta t\\[5pt] I=0.9\times 9.8\times 4.5 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {I=39.7\;\mathrm{N.s}} \end {gather} \]
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