Answer:8 m/s
Explanation:
Given
[tex]m_1=1 kg[/tex]
[tex]m_2=7 kg[/tex]
kinetic Energy of [tex]m_1=32 J[/tex]
initially [tex]m_2[/tex] is at rest and let say [tex]m_1[/tex] is moving with velocity u
kinetic Energy of [tex]m_1 is =\frac{m_1u^2}{2}=32 [/tex]
[tex]u^2=64[/tex]
[tex]u=8 m/s[/tex]
In Completely inelastic collision both mass stick together and move with common velocity
Suppose v is the common velocity
[tex]m_1u+0=(m_1+m_2)v[/tex]
[tex]v=\frac{1\times 8}{1+7}=1 m/s[/tex]
therefore Final velocity with which both blocks moves is 1 m/s
