Solved Problem on Impulse, Linear Momentum and Collisions

Solved Problem on Impulse

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 gravitational force, using the equation of velocity as a function of displacement

\[ \begin{gather} \bbox[#99CCFF,10px] {v^2=v_0^2+2a\Delta S} \end{gather} \]

the acceleration of the motion is the acceleration due to gravity, a = g, and substituting the values

\[ \begin{gather} v^2=v_0^2+2g(S-S_{0})\\[5pt] v^2=0^2+2\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)\times(100\;\mathrm m-0)\\[5pt] v^2=0+1960\;\mathrm{\frac{m^2}{s^2}}\\[5pt] v=\sqrt{1960\;\mathrm{\frac{m^2}{s^2}}\;} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v\simeq 44.3\;\mathrm{m/s}} \end{gather} \]

b) The momentum is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {p=mv} \tag{I} \end{gather} \]

substituting the mass given to the brick and the speed, calculated in the previous item

\[ \begin{gather} p=(0.9\;\mathrm{kg})\times\left(44.3\;\mathrm{\frac{m}{s}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p=39.9\;\mathrm{kg.m/s}} \end{gather} \]

c) Applying the Impulse-Momemtum Theorem

\[ \begin{gather} \bbox[#99CCFF,10px] {I=\Delta p=p_{f}-p_{i}} \end{gather} \]

substituting the expression (I), for the initial and final values

\[ \begin{gather} I=mv_{f}-mv_{i} \end{gather} \]

the final speed, v_f = v, calculated in item (a), and the initial speed, v_i = v₀ = 0

\[ \begin{gather} I=mv-mv_{0}\\[5pt] I=(0.9\;\mathrm{kg})\times\left(44.3\;\mathrm{\frac{m}{s}}\right)-(0.9\;\mathrm{kg})\times 0\\[5pt] I=40.2\;\mathrm{\frac{kg.m}{s}}-0 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {I=39.9\;\mathrm{N.s}} \end{gather} \]

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