A sphere of mass 4 kg collides with a wall, at an angle of `30^(@)` with the wall and rebounds in the direction making an angle of `60^(@)` with its original direction of motion. Find the force on the wall if the ball remains in contact with the wall for 0.1s. The initial and final velocities are the same, equal to `1ms^(-1)`

Explanation

Suppose `hatiandhatj` are the unit vector along axis X and Y respectively.

The magnitude of initial velocity `vec(v_(1))` and final velocity `vec(v_(2))` are same
`therefore |vec(v_(1))|=|vec(v_(2))|=v`
The magnitude of `p=mv`
`=4xx1`
`therefore p=4Ns....(1)`
x-component of initial momentum `vec(p_(1))`
`vec(p_(1))=mvcos60^(@)hati`
`therefore vec(p_(1))=(p)/(2)hatj`
x-component of final momentum `vec(p_(2))`
`vec(p_(2))=-mvcos60^(@)hati`
`therefore vec(p_(2))=-(p)/(2)hatj`
In above figure initial and final momentum are shown by `vec(AB) and vec(AC)`. Here `DeltaABC` is equilateral triangle
Change in momentum of sphere is obtained by substraction of two vectors
`therefore` from figure
`vec(BC)=vec(AC)-vec(AB)`
where `vec(AB)=vec(p_(1)),vec(AC)=vec(p_(2))`
`therefore Deltavecp=vec(p_(2))-vec(p_(1))`
`therefore Deltavecp=-(p)/(2)hati-(p)/(2)hatj`
`therefore Deltavecp=-phati`
`therefore Deltavecp=-4hatiNs` (from eqn. (1))
`therefore` From law of conservation of momentum change in momentum of wall `Deltavecp=4hati`
`therefore` Force exerted on wall `vecF=(Deltavecp)/(Deltat)`
`therefore vecF=(4hati)/(0.1)=40hatiN`
So force of 40 N exerted on the wall in X-direction.