Block A is sliding across a horizontal and frictionless surface when it collides

Question

Block A is sliding across a horizontal and frictionless surface when it collides with block Bas shown in the figure. Block Bis attached to a massless string of length L 1 m and is free to rotate as a pendulum. The speed of block Aafter the collision is half its speed before the collision. Block Bwas 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?

Explanation / Answer

Given

   mass of block A , mA = 7 kg

   mass of block B , mB = 2 kg

   length of the string is L = 1 m
speed of block A after the collision with B is v = v0/2

where v0 is initial speed of the block A

in order to complete the vertical circle by the block B , it should have the kinetic energy which is equal to gravitaional potential energy at the top of the circle that is

   0.5*mB*vB^2 = mB*g*h

here h is = 2*R = 2*1 = 2 m

   equating , vB = sqrt(2*g*2R)

from conservation of momentum
  
   mA*v0+ mB*vB = mA*v0/2 + mB*vB

   mA*v0+ 0 = mA*v0/2 + mB*sqrt(2*g*2R)

   mA(v0-v0/2) = mB*sqrt(2*g*2R)

   v0 /2 = (mB/mA)(sqrt(2*g*2R))
  
   v0 = 2(mB/mA)(sqrt(2*g*2R))
substituting the values

   v0 = 2((2/7)(sqrt(2*9.8*2*1)))

   v0 = 3.57771 m/s

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