question_answer

A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity \[100\text{ }m{{s}^{-1}}\]from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - \[(g=10\,m{{s}^{-2}})\] [JEE Main 10-Jan-2019 Morning]

A) 30 m

B) 40 m

C) 20 m

D) 10 m

Correct Answer: B

Solution :

[b]

Bullet will collide with piece of wood at \[t=1\]second

at \[t=1\] second

Velocity of piece of wood \[=u=0+10\times 1\]

\[=\text{ }10\text{ }m\text{ }m/s\]

Velocity of bullet = v

\[=\text{ }100-10\times 1=90\text{ }m/s\uparrow \]

from momentum conservation \[\left( {{p}_{i}}={{p}_{f}} \right)\]

\[{{p}_{i}}=90\times 0.02-0.03\times 10=1.8-0.3=1.5\]

\[{{p}_{f}}=\left( 0.03+0.02 \right)\text{ }{{v}_{f}}=0.05\text{ }{{v}_{f}}\]

\[\therefore \,\,{{v}_{f}}\,=\,\frac{1.5}{0.05}\,=\,30\,m/s\]

Height from the point of collision H

\[=\,\,\frac{30\times 30}{2\times 10}\,\,\,=\,\,\,45\,m\]

At time of collision system is h distance below top of tower where \[h=\frac{1}{2}\times 10\times {{1}^{2}}=5\]

\[\therefore \] Height above tower \[=\text{ }45-5=40\] metre