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A piece of wood of mass 0.03 kg is dropp...

A piece of wood of mass 0.03 kg is dropped from the top of a building 100 m high. At the same time, a bullet of mass 0.02 kg is fired vertically upward with a velocity of 100 m/s from the ground the bullet gets embedded in the wooded piece after striking. find the height to which the combination rises above the building before it starts falling take `g=10m//s^(2)`

Text Solution

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Time taken for the particles to collide,
`t=d/(v_("rel"))=100/100=1s`
Speed of wood just before collision = gt=10 m/s and
sped of bullet just before colision =v-gt=100-10
`=90` m/s
`S=100xx1-1/2xx10xx1=95m`
Now, using conservation of linear momenum just before and after the collision
`=(0.03)(10)+(0.02)(90)=(0.05)v`
`implies150=5v`
`:.v=30m//s`
Max height reached by body
`h=(v^(2))/(2g)=(30xx30)/(2xx10)=45m`

`:.` Height above tower =40 m
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