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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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$800 N, 53 º E a s t o f N o r t h$

$800 N, 53 º N o r t h o f E a s t$

$800 N, 53 º N o r t h o f W e s t$

$800 N, 53 º W e s t o f N o r t h$

answer is C.

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Detailed Solution

$144 k m / m = 144 \times \frac{5}{18} = 40 m / s$

$\frac{| {\vec{V}}_{2} - {\vec{V}}_{1} |}{\sin 74 °} = \frac{40}{\sin (90 - 37 °)} \Rightarrow | V_{2} - V_{1} | = \frac{40 \times 2 \sin 37 ° \cos 37 °}{\cos 37 °} = 48 m / s$

change in momentum $= \frac{1}{3} \times 48 = 16 k p m / s$

$f_{a n g} = \frac{16}{0.02} = 800 N$

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