Q.

A ball of mass  0.45kg which is initially at rest is hit by a bat. The bat remains in contact with the ball for 3×103s. During this time period, the force on the ball by the bat varies with time t(in s) as  F(t)=[(α×106)t(β×109)t2]N where α and β are constants. The ball's speed, immediately as it loses contact with the bat is 20 m/s. If the relation between  α and  β is  α2β=n, find  n.

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Detailed Solution

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mvmu=F(t)dt

0.45×200.45×0=0T[(α×106)t(β×109)t2]dt         9=α×106×T22β×109×T33       9=α×106×9×1062β×109×27×1093        9=9α29β  α2β=2

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A ball of mass  0.45 kg which is initially at rest is hit by a bat. The bat remains in contact with the ball for 3×10−3s. During this time period, the force on the ball by the bat varies with time t(in s) as  F(t)=[(α×106)t−(β×109)t2]N where α and β are constants. The ball's speed, immediately as it loses contact with the bat is 20 m/s. If the relation between  α and  β is  α−2β=n, find  n.