A ball of `sqrt(3)xx10^(-2)kg` hits a hard surface at `45^(@)` to normal with speed `4sqrt(2)m//s` and rebounds with `8//sqrt(3) m//s`, at `60^(@)` angle. If ball remains in contact for 0.1 sec, what force does it exert ?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Mass of the ball, \( m = \sqrt{3} \times 10^{-2} \, \text{kg} \)
- Initial speed before hitting the surface, \( u = 4\sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \) to the normal.
- Final speed after rebounding, \( v = \frac{8}{\sqrt{3}} \, \text{m/s} \) at an angle of \( 60^\circ \) to the normal.
- Time of contact, \( t = 0.1 \, \text{s} \)

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