In a square cut, the speed of the cricket ball changes from $30 {m s}^{- 1}$ to $40 m s^{- 1}$ during the time of its contact $△ t = 0.01 s$ with the bat. If the ball is deflected by the bat through an angle of $θ = 90^{o}$ , find the magnitude of the average acceleration $(i n \times 10^{2} {m s}^{- 2})$ of the ball during the square cut.

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Solution

As the velocity of the ball changes from $_{1}$ to $_{2}$ , the change in velocity $Δ \to v$ is given by
$| Δ \to v | = \sqrt{v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos θ}$
where $v_{1} = 30 m / s, v_{2} = 40 m / s$ and $θ = 90^{o}$ . Then,
$| Δ \to v | = 50 m / s$
$a_{a v} = \frac{| Δ \to v |}{Δ t} = \frac{50}{0.01} = 50 \times 10^{2} m / s^{2}$

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