Two blocks `A` and `B` are joined by means of a slacked string passing over a massless pulley as shown in Figure. The system is released from rest and it becomes taut when `B` falls a distance `0.5 m`.
a. Find the common velocity of the two blocks just after the string becomes taut.
b. Find the magnitude of impulse on the pulley by the clamp during the small interval while string becomes taut.

Answer

`sqrt((10))/3m//s, 4/5sqrt(5)Ns`

Explanation

Velocity of `B` just before the string is taut
`v_(B)=sqrt(2g/_\h)=sqrt10m//s`
a. common velocity =v
`:. (m_(A)+m_(B))v=m_(B)v_(B)impliesv=v_(B)/3=(sqrt(10))/3m//s`
b. magnitude of impulse on `A =` magitude of impulse on
`B=1(sqrt(10)-(sqrt(10))/3)=2/3sqrt(10)m//s`
Therefore, impulse on pulley `=sqrt(2)` impulse on `A=4/3sqrt(5)Ns`