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 6m/s, relative to water. Calculate the distance (in m) between the boat 8.5 sec. after the throw if the sack spent 0.5 sec. in air. Neglect resistance of air and water.

Engineering

Physics

Straight Line Motion

Relative Motion

Momentum and Energy Conservation for System of Particles

Question

Two boats both having a mass of 150 kg including passengers in it are at rest. A sack of mass 50 kg makes 1^st boat having total mass of 200 kg. It is thrown to the second boat with a velocity whose horizontal component is 6m/s, relative to water. Calculate the distance (in m) between the boat 8.5 sec. after the throw if the sack spent 0.5 sec. in air. Neglect resistance of air and water.

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1^st boat
0 = 50 × 6 + 150 v
v₁ = –2m/s $S_{1} = - 2 \times \frac{1}{2} = - 1 m$

2^nd boat 300 = 200 v₂

$v_{2} = \frac{3}{2} m / s$

$T = \frac{2 u_{y}}{g} = 0.5$ u_y = 2.5 m/s

Initial distance = $\frac{2 \times 2.5 \times 6}{10} = 3 m$

dist. = u_rel × t = 3.5 × 8 = 28
Total distance = 32 m