Two boats both having a mass of 150 kg including passengers in it are at rest. A sack of mass 50 kg makes 1st boat having total mass of 200 kg. It is thrown to the second boat with a velocity whose horizontal component is $2 {ms}^{- 1}$ , relative to water. Calculate the distance (in m to nearest integer) between the boat 8.5 seconds after the throw if the sack spent 0.5 s in air. Neglect the resistance of air and water.

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Detailed Solution

First boat

$0 = 50 \times 2 + 150 v_{1}$

$\Rightarrow v_{1} = - \frac{2}{3} {ms}^{- 1}$

Second boat

$100 = 200 v_{2}$

$v_{2} = \frac{1}{2} {ms}^{- 1}$

$T = \frac{2 u_{y}}{g} = 0 .5 u_{y} = 2 .5 {ms}^{- 1}$

Initial distance $= 0.5 \times 2 = 1 m$

distance traveled by first boat in 0.5 s $= 0.5 \times \frac{2}{3} = \frac{1}{3} m$

Distance after 0.5 s : $= u_{rel} \times t = (\frac{1}{2} + \frac{2}{3}) \times 8 = \frac{28}{3} m$

Total distance = $= 1 + \frac{1}{3} + \frac{28}{3} = \frac{32}{3} = 10.67 m$

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