A particle A of mass 2m is held on a smooth horizontal table and is attached to one end of an inelastic string which runs over a smooth light pulley at the edge of the table. At the other end of the string there hangs another particle B of mass m, the distance from A to the pulley is l. The particle A is then projected towards the pulley with velocity u.

(a) Find the time before the string becomes taut, and shown that after the string becomes taut, the initial velocity of A and B is `4u//3`.
(b) Find the common velocity when A reaches the pulley (assume that B has not yet reached the ground).

Answer

(a) `2(u)/(g)`, (b) `(sqrt(4u^(2)+6gl))/(3)`

Explanation

(a) When A is projected to the right, B falls freely under gravity. The string is tensioned when A and B travel the same distance. Let t be the time in which they cover the same distance. Then u(distance covered by A) `=(1)/(2)g t^(2)` (distance covered by B) or `t=2(u)/(g)`
Let v be the initial common velocity. when the string is tensioned, the two bodies experience teh same impulsive force but in opposite directions.
Change in momentum of `A=2mv_(f)-2m2mv-2m u`
But `u-("initial")=g t=g(2u)/(g)=2u`
Change in momentum of `B=mv-m xx 2u`
Therefore, by Newton's law of motion
`(2mv-2m u)=-(mv-2m u)`
or `3mv=4m u` or `v=(4u)/(3)`
(b) Let a be the common acceleration,
`mg-T=ma`,
`T=2ma :. a=(g)/(3)`
`:. v_(2)-v_(0)^(2)=(2g)/(3)l'`
where `l'` is the distance covered by A or B with acceleration.
Now `l = l' +u t = l+u(2u)/(g) rArr l' = l-(2u^(2))/(g)`,
Here `v_(0)=(4u)/(3)`
`v_(2)=(16u^2)/(9)+(2g)/(3)(l=(2u^2)/(g))`
`=(16u^(2))/(9)+(2gl)/(3)-(4u^(2))/(3)=(4u^(2)+16gl)/(9)`
`:. v=(sqrt(4u^(2)+6gl))/(3)`.