Two particles of masses `m_(1)` and `m_(2)` are connected with a rigid rod of length `l`. If a force `F` acts perpendicular to the rod then (`a_(1) & a_(2)` are instantaneous of `m_(1) & m_(2)`)

Options

A. `a_(2)=0`
B. `a_(1)=F/(m_(1))`
C. `a_(CM)=F/(m_(1)+m_(2))`
D. `alpha=(F(m_(1)+m_(2)))/(m_(1)m_(2)l)`

Answer

A, B, C

Explanation

`a_(cm)=F/(m_(1)+m_(2)),Fxxl/2=Ialpha`
`a_(1)=a_(cm)+l/2alpha,a_(1)=a_(cm)-l/2 alpha`