A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the stringbreaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.

Options

A. `9.8m`
B. `19.6m`
C. `2.45m`
D. `4.9m`

Answer

d

Explanation

Acceleration of the system before the string
breaks `a = ("net pulling force")/("total mass") = (3 g -2g)/((3 + 2)) = (g)/(5)`
After 5s, velocity of the system,
`upsilon = "at" = (g)/(5) xx 5 = gm//s`
Required height `h = (upsilon^(2))/(2 g) = (g xx g)/(2 g) = (g)/(2)`
` = (9.8)/(2) = 4.9m` .