# Part B) only A smooth wedge of mass M, rests on a smooth horizontal table. Initially, a particle of mass m is gently placed on its face inclined at an angle `alpha` to the horizontal. A) Show,...

**Part B) only**

A smooth wedge of mass M, rests on a smooth horizontal table. Initially, a particle of mass m is gently placed on its face inclined at an angle `alpha` to the horizontal.

A) Show, using the principle of conservation of momentum, or otherwise , that when the particle has acquired a velocity v relative to the wedge, the velocity of the wedge is `(cos (alpha))/(M+m) ` .

B)If, at this instant, the particle impinges on an inelastic obstacle fixed to the wedge and comes to rest relative to the wedge, find the velocity of the wedge and the impulse on the table.

Answer for part A) is at http://www.enotes.com/homework-help/smooth-wedge-mass-m-rests-smooth-horizontal-table-451089

### 1 Answer | Add Yours

If, at this instant, the particle impinges on an inelastic obstacle fixed to the wedge and comes to rest relative to the wedge, find the velocity of the wedge and the impulse on the table.

In an inelastic collision , both mass fixed i.e (M+m)kg. Let velocity of the new system be `V_w` m/s . So relative to wedge particle has no velocity.In this collision K.E. conserved ( In general some K.E. lost)

Thus K.E. before collision

`=(1/2)mv^2`

where v is velocty of moving particle of mass m.

The K.E. after collision

`=(1/2)(M+m)V^2_w`

Thus by law of conservation of energy

K.E. before collision= K.E. after collision

`(1/2)mv^2=(1/2)(M+m)V^2_w`

`V^2_w=(mv^2)/(M+m)`

`V_w=sqrt((mv^2)/(m+M))`

**which will be the velocity of the wedge after collision.**