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 Online Paper (Held On 10-Jan-2019 Morning]

A) 30 m                            

B) 40 m

C) 20 m                

D)   10 m

Correct Answer: B

Solution :

  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