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Question

A piece of wood of mass 0.03kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02kg is fired vertically upward with a velocity 100ms−1 from the ground. The bullet after collision 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 Take, g=10ms−2)

A
20m
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B
30m
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C
10m
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D
40m
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Solution

The correct option is D 40m
Velocity of bullet is very high compared to velocity of wooden block so, in order to calculate time for collision, we take relative velocity nearly equal to velocity of bullet.
So, time taken for particles to collide is
t=dvrel=100100=1s
Speed of block just before collision is;
v1=gt=10×1=10ms1
Speed of bullet just before collision is
v2=ugt
=10010×1=90ms1
Let v = velocity of bullet + block system, then by conservation of linear momentum, we get
(0.03×10)+(0.02×90)=(0.05)v
v=30ms1
Now, maximum height reached by bullet and block is
h=v22gh=30×302×10
h=45m
Height covered by the system from point of collision = 45 m
Now, distance covered by bullet before collision in 1 s.
=100×112×10×12=95m
Distance of point of collision from the top of the building
=10095=5m
Maximum height to which the combined system reaches above the top of the building before falling below =455=40m

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