As µ << m, M, momentum conservation  
        MV = (M + m)V'
gives for the velocity of the two carts after collision,

${\mathrm{V}}^{\mathrm{\prime }}=\frac{\mathrm{MV}}{\mathrm{M}+\mathrm{m}}$

Consider the circular motion of the ball atop the cart M if it were stationary. If at the lowest and highest points the ball has speeds V₁ and V₂ respectively, we have 

$\frac{1}{2}{\mathrm{\mu V}}_1^2=\frac{1}{2}{\mathrm{\mu V}}_2^2+2\mathrm{\mu gR}$

$\frac{{\mathrm{\mu V}}_2^2}{\mathrm{R}}=\mathrm{T}+\mathrm{\mu g}$

where T is the tension in the string when the ball is at the highest point. The smallest V₂ is given by T = 0. Hence the smallest V₁ is given by 

$\frac{1}{2}{\mathrm{\mu V}}_1^2=\frac{1}{2}\mathrm{\mu gR}+2\mathrm{\mu gR}\text{ i.e. }{\mathrm{V}}_1=\sqrt{5\mathrm{gR}}$

With the cart moving, V₁ is the velocity of the ball relative to the cart. As the ball has initial velocity V and the cart has velocity V' after the collision, the velocity of the ball relative to the cart after the collision is V – V'. Hence the smallest V for the ball to go round in a circle after the collision is given by 

$\mathrm{V}-{\mathrm{V}}^{\mathrm{\prime }}=\mathrm{V}-\frac{\mathrm{MV}}{\mathrm{M}+\mathrm{m}}=\sqrt{5\mathrm{gR}}$       i.e.   $\mathrm{V}=\frac{\mathrm{M}+\mathrm{m}}{\mathrm{m}}\sqrt{5\mathrm{gR}}$