Answer:
[tex]p = 0.633\,\frac{kg\cdot m}{s}[/tex]
Explanation:
The combined speed after collision can be determined by means of the Principle of the Energy Conservation:
[tex]K_{A} + U_{g,A} = K_{B} + U_{g,B}[/tex]
[tex]K_{A} = K_{B} + U_{g,B} - U_{g,A}[/tex]
[tex]\frac{1}{2}\cdot (m+M)\cdot v_{A}^{2} = (m+M)\cdot g\cdot h[/tex]
[tex]v_{A} = \sqrt{2\cdot g \cdot h}[/tex]
[tex]v_{A} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(0.08\,m)}[/tex]
[tex]v_{A} \approx 1.253\,\frac{m}{s}[/tex]
The magnitud of the momentum of the combined mass immediately after the collision:
[tex]p = (0.005\,kg + 0.5\,kg)\cdot (1.253\,\frac{m}{s} )[/tex]
[tex]p = 0.633\,\frac{kg\cdot m}{s}[/tex]
