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A bowler throws a ball horizontally along east direction with speed of 144 km/hr. The batsman hits the ball such that it deviates from its initial direction of motion by 74º north of east direction, without changing its speed. If mass of the ball is $\frac{1}{3} kg$ and time of contact between bat and ball is 0.02 s. Average force applied by batsman on ball is

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800 N, 53º East of North

800 N, 53º North of West

800 N, 53º West of North.

800 N, 53º North of East

answer is C.

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Detailed Solution

$v = 144 km / hr = 144 \times \frac{5}{18} = 40 m / s$
$∆ p = 2 mv \sin \frac{θ}{2} = 2 \times \frac{1}{3} \times 40 \times \frac{3}{5} = 16 kg m / s$

$F = \frac{∆ p}{∆ t} = \frac{16}{0.02} = 800 N$

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