Answer:
speed of block is 0.665 m/s
Kinetic energy loss is 237.16 J
Explanation:
Here we can use momentum conservation as there is no external force on the system in horizontal direction
so here we will have
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]0.022(210) + 2(0) = 0.022(150) + 2 v[/tex]
[tex]4.62 = 3.3 = 2 v[/tex]
[tex]v = \frac{4.63 - 3.3}{2}[/tex]
[tex]v = 0.665 m/s[/tex]
Now kinetic energy loss in the system is given as
[tex]Loss = KE_i - KE_f[/tex]
[tex]Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)[/tex]
[tex]Loss = 237.16 J[/tex]
