A wedge of mass m and triangular cross- section (AB = BC = CA = 2R) is moving with a constant velocity `-v hat`I towards a sphere of radius R fixed on a smooth horizontal table as shown in figure. The wadge makes an elastic collision with the fixed sphere and returns along the same path without any rotaion. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very shot time. `Deltat,` during which the sphere exerts a constant force F on the wedge.

(a) Find the force F and also the normal force N exerted by the table on the wedge during the time `Deltat.`
(b) Ler h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge, during the interval `Deltat.`

Explanation

(i) `vecF =(2mv)/(Deltat)hati -(2mv)/(sqrt(3)Deltat)hatk, vecN =(mg + (2mv)/(sqrt(3)Deltat))hatk`, (ii) `(4mv)/(sqrt(3)Deltat)h`.
(i) The collision between sphere and wedge is elastic. As the sphere is fixed, hence the wedge will return with velocity `vhati`.
Now, linear impulse in x-direction = change in momentum in x-direction.
`therefore (F cos 30^(@))Deltat =mv-(-mv) = 2mv`
`therefore F=(2mv)/(Deltat cos 30^(@)) = (4mv)/(sqrt(Deltat))`

`vecF =((2mv)/(Deltat))hati -(2mv)/(sqrt(3)Deltat)hatk`
Now consider the equilibrium of wedge in vertical direction. During collision.
`N=mg + (2mv)/(sqrt(3)Deltat)`

or in vector form `vecN =mg + (2mv)/(sqrt(3)Deltat)hatk`
(ii) For rotational equilibrium of wedge about
its (CM), anticlockwise torque of F= Clockwise torque due to N
`therefore` Magnitude of torque of N about centre of mass = magnitude of torque of F about centre of mass =F.h.
`|vectau_(N)| =((4mv)/(sqrt(3)Deltat))h`