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Momentum conservation gives: $Mv=(M+m)v_{′}$

$⇒$ velocity of the two carts after collision $v_{′}=M+mMv $

Consider the circular motion of the ball atop the cart if it were stationery. If at the lowest and highest points the ball has speeds $v_{1}$ and $v_{2}$, respectively, then

$21 μv_{1}=21 μv_{2}+2μgR$

and $Rμv_{2} =T+μg$,

where $T$ is the tension in the string when the ball is at the highest point.

The smallest $v_{2}$ is given by $T=0$.

$⇒$ The smallest $v_{1}$ is given by

$21 μv_{1}=21 μgR+2μgR$

$⇒v_{1}=5gR $

When the cart is moving, $v_{1}$ is the velocity of the ball relative to the cart. As the ball has initial velocity, velocity of the ball relative to the cart after the collision is $(v−v_{′})$. Hence, the smallest velocity for the ball to go round in a circle after the collision is given by:

$v_{1}=(v−v_{′})=v−(M+mMv )=5gR $

$⇒v=(mM+m )5gR $

$⇒$ velocity of the two carts after collision $v_{′}=M+mMv $

Consider the circular motion of the ball atop the cart if it were stationery. If at the lowest and highest points the ball has speeds $v_{1}$ and $v_{2}$, respectively, then

$21 μv_{1}=21 μv_{2}+2μgR$

and $Rμv_{2} =T+μg$,

where $T$ is the tension in the string when the ball is at the highest point.

The smallest $v_{2}$ is given by $T=0$.

$⇒$ The smallest $v_{1}$ is given by

$21 μv_{1}=21 μgR+2μgR$

$⇒v_{1}=5gR $

When the cart is moving, $v_{1}$ is the velocity of the ball relative to the cart. As the ball has initial velocity, velocity of the ball relative to the cart after the collision is $(v−v_{′})$. Hence, the smallest velocity for the ball to go round in a circle after the collision is given by:

$v_{1}=(v−v_{′})=v−(M+mMv )=5gR $

$⇒v=(mM+m )5gR $

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