A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

Asked by Pragya Singh | 1 year ago |  113

1 Answer

Solution :-

Velocity of the ball = 54 km/h

The ball is deflected back such that the total angle = 450

The initial momentum of the ball is mucosӨ 

\( \dfrac{ (0.15 \times 54 \times 1000 \times cos 22. 5)}{3600}\)

= 0.15 x 15 x 0.9239 along NO

Final momentum of the ball = mucosӨ along ON

Impulse = change in the momentum = mucosӨ – (-mucosӨ) 

= 2mucosӨ = 2 x 0.15 x 15 x 0.9239 = 4.16 kg.m/s

Answered by Abhisek | 1 year ago

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