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A block of mass 2 kg is moving with a speed of $2 {ms}^{- 1}$ at this instant say t = 0 it is struck by a jet releasing water at a rate of $1 {kgs}^{- 2}$ with a speed of $6 {ms}^{- 1}$ . Assume that water comes to rest just after striking the block the acceleration of block just after t = o s is

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6 {ms}^{- 2}

4 {ms}^{- 2}

2 {ms}^{- 2}

none of these

answer is C.

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Detailed Solution

Velocity of water $u = 6 {ms}^{- 1}$

Mass $(\frac{Δ m}{Δ t}) = 1 kg/s$

Force exerted by water on block is

$F = (\frac{m}{t}) u = 1 \times 6 = 6 N$

Time taken to come to rest to water is v = u + at

$\Rightarrow 0 = u - at \Rightarrow u = \frac{F}{m} t \Rightarrow t = \frac{6 \times 1}{6} = 1 s$

In this 1s (just after the t = 0s) ‘a’ attain by block is $a = \frac{v - u}{t} = \frac{0 - 2}{t} = - 2 {ms}^{- 2}$

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