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A batsman deflects a ball by an angle of 45^(@) without changing its initial speedwhich is equal to 54 km h^(-1) What is the impulse imparted to the ball ? Mass of the ball is0.15 g . |
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Answer» Solution :Here, ` m = 0.15 kg upsilon_(i) = 54 km//h =54 XX (5)/(18) ms^(-1) = 15 ms^(-1)` `upsilon_(f) = 54 km//h = 15ms^(-1)` Impulse imparted to the ball along X- axis ` J_(x)` = change in momentum of ball along X-axis ` = m upsilon_(fx) - (- m upsilon_(IY)) = 0.15 (15 sin 45^(@) - 0) = 1.59 kg ms^(-1)` Net impulse imparted to the ball ` J = sqrt(J_(x)^(2) + J_(y)^(2)) = sqrt(3.84^(2) + 1.59^(2)) = sqrt(17 .27) = 4.16 kg ms^(-1) (or N_(-s))` .
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