Question

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the collision is half its speed before the collision. Block B was at rest before the collision. The mass of block A is 7 kg and the mass of block B is 2 kg. What is the minimum initial speed (in m/s) that block A must have for block B to swing through a complete vertical circle?

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  $v_A= 4m/s$

Explanation:

From the question we are told that

    The length of the string is  $L = 1m$

     The initial speed of block A is $u_A$

     The final speed of block A is  $v_A = \frac{1}{2}u_A$

      The initial speed of block B is $u_B = 0$

      The mass of block A  is  $m_A = 7kg$  gh

      The mass of block B is  $m_B = 2 kg$

According to the principle of conservation of momentum

       $m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

       $m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

      $m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

          $m_A \frac{u_A}{2} = m_Bv_B$

  making $v_B$ the subject of the formula

             $v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

               $v_B =\frac{7 u_A}{4}$  

This $v__B$ 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 $v__B'$ at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   $a= \frac{v^2_{B}'}{L}$

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

         $T + mg = m \frac{v^2_{B}'}{L}$

    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

                 $\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above  

                $(T + mg) L = m v^2_{B}'$

Substitute this into above equation

             $\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$  

             $\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

          $\frac{49 mv_A^2}{16} = T + 5mgL$

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

                      $\frac{49 mv_A^2}{16} = 5mgL$

making  $v_A$ the subject

            $v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

          $v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

              $v_A= 4m/s$