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A piece of wood of mass 0.03 kg is dropped from the top of a building 100m 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 wooden piece after striking it. Find the height to which the combination rises above the building before falling freely. ( g = 9.8 m/s2 )

```
2 years ago

Arun
25768 Points
```							Let m and m' be the mass of the wooden piece and the bullet respectively.In this way, if t is time after which bullet strikes with the wooden piece. we get, t = 1 sec. Let the velocity of striking of the Using the equation, v = u + atv = 9.8 m/sand v' = 90.2 m/sAccording to the mv - m'v' = (m +m')V  ....(i) (as the direction of both velocities are opposite)Using this equation you will find the vale of V.Let the body will rise upto height h above the building. At this point total mechanical energy of the combination is = 1/2 (m +m')V2 + (m +m')gx ...(ii)total mechanical energy of the combination at the max height = 0 + (m +m')g(100+h)  ...(iii)Equating (ii) nad (iii) and substituting the value of x. you will get the answer.
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions