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 ms-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 ms-2)

Question

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 ms-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 ms-2) (A) 30 m (B) 10 m (C) 40 m (D) 20 m

Tardigrade Admin · Accepted Answer

Time taken for the particles to collide,

t = \frac{d}{V _{re l}} = \frac{100}{100} = 1 sec

Speed of wood just before collision =

g t = 10 m / s

& speed of bullet just before collision

v - g t

=

100 - 10 = 90 m / s

Now, conservation of linear momentum just before and after the collision -

- (0.02) (1 v) + (0.02) (9 v) = (0.05) v

\Rightarrow 150 = 5 v

\Rightarrow v = 30 m / s

Max. height reached by body

h = \frac{v ^{2}}{2 g}

h = \frac{30 \times 30}{2 \times 10} = 45 m

∴

Height above tower =

40 m