A cart of mass M has a pole on it from which a ball of mass mu hangs from a thin string attached at point P the cart and ball have initial velocity V the cart crashes onto another cart of mass m and sticks to it (figure) If the length of the string is R, the smallest initial velocity (in m//s) for which the ball can go in circles around point P is Neglect friction and assume M, `m gtgt mu` Given `m=1kg M=2 kg R=2m`

Answer

`0030`

Explanation

As mu lt lt m, M, momemtum conservation
`MV=(M+m)V'`
gives for the velocity of the two carts after collision,
`V'=(MV)/(M+m)`
Consider the circular motion of the ball atop the cart M if it were stationary In at the lowest and highest points the ball has speeds `V_(1)` and `V_(2)` respectively we have
`1/2muV_(1)^(2)=1/2muV_(2)^(2)+2mugR`,
`(muV_(2)^(2))/R=T+mug`
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` Hence the smallest `V_(1)` is given by
`1/2muV_(1)^(2)=1/2mugR+2mugR i.e. V_(1) = sqrt(5gR)`
With the cart moving, `V_(1)` 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
`V-V'=V-(MV)/(M+m)=sqrt(5gR) i.e. V=(M+m)/m sqrt(5gR)`