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Answer:
[tex]5\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of clay = m
[tex]m_2[/tex] = Mass of block = 3m
[tex]u[/tex] = Velocity of clay
[tex]v[/tex] = Velocity of block = -3 m/s
[tex]v_c[/tex] = Velocity of combined mass = -1 m/s
As mometum in the system is conserved we get
[tex]m_1u+m_2v=(m_1+m_2)v_c\\\Rightarrow u=\dfrac{(m_1+m_2)v_c-m_2v}{m_1}\\\Rightarrow u=\dfrac{(m+3m)\times -1-3m\times -3}{m}\\\Rightarrow u=\dfrac{5m}{5}\\\Rightarrow u=5\ \text{m/s}[/tex]
The original velocity of the clay mass was [tex]5\ \text{m/s}[/tex].