Two bodies of masses `11kg` and `11.5kg` are connected by a long light string passing over a smooth pulley Calculate velocity and height ascended/descended by each body at the end of `4s` .

Answer

` 0.872 m//s, 1.744m` .

Explanation

Here, `m_(1) = 11 kg, m_(2) = 11.5 kg`
`:. a = ((m_(2)-m_(1)) g)/(m_(2) + m_(1))= ((11.5 -11))/(11.5 + 11) 9.8 = (4.9)/(22.5)`
`=0.218 m//s^(2)`
From` upsilon = u + a t = 0 + 0.218 xx4 = 0.872m//s`
From ` s = ut + (1)/(2)at^(2) = 0+(1)/(2) xx0.218(4)^(2)`
`= 1.744m` .