A `0.5 kg` ball moving with a speed of `12 m//s` strikes a hard wall at an angle of `30^(@)` with the wall. It is reflected with the same speed and at the same angle . If the ball is in contact with the wall for `0.25 s`, the average force acting on the wall is

Options

A. `48 N`
B. `24 N`
C. `12 N`
D. `96 N`

Explanation

The vector `O vecA` represents the momentum of the objecy before the collision and the vector `vecA B` represents the change in momentum of the object `Delta vecP`. As the magnitudes of `O vecA` and `O vecB` are equal the component of `O vecA` and `O vecB` along the wall are equal and in the same direction while those perpendicular to the wall are equal and opposite . Thus, the change in direction of the perpendicular component.
Hence, `Delta p = OB sin 30^(@) - (- OA sin 30^(@))`
`= mv sin 30^(@) = - (-mv sin 30^(@))`
`= 2mv sin 30^(@)`

Its time rate will appear in the from of everage force acting on the wall .
`F xx t = 2 mv sin 30^(@)`
Or `F = (2 mv sin 30^(@))/(t)`
given `m = 0.5 kg, v = 12 m//s, t = 0.25 s`
`theta = 30^(@)`
Hence, `F = (2 xx 0.5 xx 12 sin 30^(@))/(0.25) = 24 N`.