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.

mv−mu=∫F(t)dt0.45×20−0.45×0=∫0T[(α×106)t−(β×109)t2]dt 9=α×106×T22−β×109×T33 9=α×106×9×10−62−β×109×27×10−93 9=9α2−9β ⇒α−2β=2

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A ball of mass $0.45 k g$ which is initially at rest is hit by a bat. The bat remains in contact with the ball for $3 \times 10^{- 3} s .$ During this time period, the force on the ball by the bat varies with time t(in s) as $F (t) = [(α \times 10^{6}) t - (β \times 10^{9}) t^{2}] N$ 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 $α - 2 β = n,$ find n.

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answer is 2.

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Detailed Solution

$m v - m u = \int F (t) d t$

$0.45 \times 20 - 0.45 \times 0 = \int_{0}^{T} [(α \times 10^{6}) t - (β \times 10^{9}) t^{2}] d t 9 = α \times 10^{6} \times \frac{T^{2}}{2} - β \times 10^{9} \times \frac{T^{3}}{3} 9 = α \times 10^{6} \times \frac{9 \times 10^{- 6}}{2} - β \times 10^{9} \times \frac{27 \times 10^{- 9}}{3} 9 = \frac{9 α}{2} - 9 β \Rightarrow α - 2 β = 2$