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```
a bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes and passes horizontally through a block of mass 2kg which is suspended by a string of length 5m.the center of gravity of block is found to raise a vertical distance of 0.1m . what is speed of bullet after it emerges from the block?
a bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes and passes horizontally through a block of mass 2kg which is suspended by a string of length 5m.the center of gravity of block is found to raise a vertical distance of 0.1m . what is speed of bullet after it emerges from the block?

```
4 years ago

Soham Chowdhury
26 Points
```							At first use the principle of conservation of mechanical energy...i.e..mgh=1/2mv*v....put the values...the value of v will be used in the next step...conservation of linear momentum....(m1*u1)+(m2*u2)=(m1*v1)+(m2*v2)......it wiil give the value of v1...which will be ur answer....
```
4 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