Answer:
[tex]h' = 0.062\cdot h[/tex]
Explanation:
The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:
[tex]U_{g} = K[/tex]
[tex]m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = \sqrt{2\cdot g \cdot h}[/tex]
[tex]v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}[/tex]
[tex]v \approx 4.429\cdot \sqrt{h}[/tex]
Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:
[tex](0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v[/tex]
[tex]v = 1.107\cdot \sqrt{h}[/tex]
Lastly, the height reached by the two blocks is:
[tex]K = U_{g}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'[/tex]
[tex]h' = \frac{v^{2}}{2\cdot g}[/tex]
[tex]h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]h' = 0.062\cdot h[/tex]
