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 string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.

Options

A. 4.9 m
B. 9.8m
C. 10.6 m
D. 2.45 m

Answer

A

Explanation

(a) Acceleration of system before breaking the string was,
`a=("Net pulling force")/("Total mass")=(3g-2g)/(5)=(g)/(5)`
After 5s velocity of system,`v=at=(g)/(5)xx5=g ms^(-2)`
Now, `h=(v^(2))/(2g)=(g^(2))/(2g)=(g)/(2)=4.6m`