Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The minimum velocity of A is [tex]v_A= 4m/s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1m[/tex]
The initial speed of block A is [tex]u_A[/tex]
The final speed of block A is [tex]v_A = \frac{1}{2}u_A[/tex]
The initial speed of block B is [tex]u_B = 0[/tex]
The mass of block A is [tex]m_A = 7kg[/tex] gh
The mass of block B is [tex]m_B = 2 kg[/tex]
According to the principle of conservation of momentum
[tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]
Since block B at initial is at rest
[tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]
[tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]
[tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]
making [tex]v_B[/tex] the subject of the formula
[tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]
Substituting values
[tex]v_B =\frac{7 u_A}{4}[/tex]
This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A
Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of [tex]v__B'[/tex] at the top of the vertical circle
The angular centripetal acceleration would be mathematically represented
[tex]a= \frac{v^2_{B}'}{L}[/tex]
Note that this acceleration would be toward the center of the circle
Now the forces acting at the top of the circle can be represented mathematically as
[tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]
Where T is the tension on the string
According to the law of energy conservation
The energy at bottom of the vertical circle = The energy at the top of
the vertical circle
This can be mathematically represented as
[tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]
From above
[tex](T + mg) L = m v^2_{B}'[/tex]
Substitute this into above equation
[tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]
The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero
This is mathematically represented as
[tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]
making [tex]v_A[/tex] the subject
[tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]
substituting values
[tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]
[tex]v_A= 4m/s[/tex]
