A triangular block `ABC` of mass `m` and side `2a` lies on a smooth horizontal plane is shown. There point masses of the `m` each strikes the block at `A, B` and `C` with speed as shown. After the collision the particle come to rest. Then

Options

A. the centre of mass of `DeltaABC` remains stationary after collision
B. the centre of mass of `DeltaABC` moves with a velocity`v` along `x`-axis after collision
C. the triangular block rotates with an angular velocity `omega=(2sqrt(3)mva)/I` about its centroid axis perpendicular to its plane
D. a point lying at a distance of `(1/(2sqrt(3)ma))` from centroid `G` on perpendicular bisector of `BC` is at rest just after collision

Answer

B, C

Explanation

By law conservation of linear movementum we `mv hati+mv hatj+mv(-hatj)+0`
have `=vec(P)_("triangular wedge")+0`
`rArrvec(P)=mv hati` since the net linear momentum imparted to the tangular which is along `x`-axis end in non zero, so the centre of mass of the wedge `ABC` will move along `x-`axis.