Answer:
[tex]E_k = 0.55v_1^2 J[/tex]
[tex]h = 0.0128v_1^2[/tex]
where [tex]v_1[/tex] is the initial velocity of the moving block
Explanation:
We can apply the law of momentum conservation to calculate the compund velocity post-collision:
[tex]m_1v_1 +m_2v_2 = (m_1 + m_2)v[/tex]
where [tex]m_1 = m_2 = 2.2 kg[/tex] are the masses of block 1 (the moving block) and block 2. [tex]v_1 , v_2[/tex] are the initial velocities before collision of block 1 and 2, and [tex]v_2 = 0m/s[/tex] as it's at rest initially.
[tex]2.2v_1 + 0 = (2.2 + 2.2)v[/tex]
[tex]v = 2.2v_1/4.4 = 0.5v_1[/tex]
So the kinetic energy right after the collision is
[tex]E_k = \frac{(m_1 + m_2)v^2}{2} = \frac{(2.2 + 2.2)(0.5v_1)^2}{2} = \frac{4.4*0.25v_1^2}{2} = 0.55v_1^2 J[/tex]
To calculate the horizontal level where the compound rises to, we need to apply the law energy conservation. As the compound rises its kinetic energy is converted to potential energy:
[tex]E_p = E_k[/tex]
[tex]mgh = mv^2/2[/tex]
where m is the mass and h is the vertical distance traveled, v is the velocity at the bottom, and g = 9.8 m/s2 is the gravitational acceleration.We can divide both sides by m:
[tex]gh = v^2/2[/tex]
[tex]h = v^2/(2g) = (0.5v_1)^2/(2g) = v_1^2/(8*9.8) = 0.0128v_1^2[/tex]
