M is `a` fixed wedge. Masses `m_(1)` and `m_(2)` are connected by `a` light string. The wedge is smooth and the pulley is smooth and fixed `m_(1)=10kg` and `m_(1)=7.5kg` . When `m_(2)` is just released, the distance it will travel in 2 second is.

Options

A. `2.8m`
B. `7.5m`
C. `4.0m`
D. `6.0m`

Answer

A

Explanation

F.B.D of `m_(1)` and `m_(2)`
Equation of motion
For `m_(1): T-m_(1)g((1)/(2))=m_(1)a` (i)
For `m_(2): m_(2)g-T=m_(2)a` (ii)
From (i) and (ii), `a=((m_(2)-m_(1)/(2))g)/(m_(1)+m_(2))=(10)/(7)m//s^(2)`
Hence distance travelled by block `m_(2)` in 2 sec, using
`s=ut+(1)/(2)at^(2)`
`s=0xx2+(1)/(2)(10/(7))xx(2)^(2)=(20)/(7)m~~2.8m