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A uniform rod of mass M and length a lies on a smooth table. A particle of mass m strikes it at distance a/4 from the centre and stops after the collision. Find the a) velocity of the centre of the rod and the b)angular velocity of the rod about its centre just after the collision. a) They have used linear momentum formula: mv=MV. where V is the velocity of the centre of the rod. How could they use it when the rod was undergoing rotational and not translational motion after collision? Shouldn't velocity of the centre of rod be 0 because its stationary? b) How do we find part b? Please explain the concept.

```
10 years ago

509 Points
```							here we have to apply to concepts
first is conservation of angular momentam and second is conservation of linear momentam...
initial angular momentam of particle about the center of rod is mva/4 ....
finally particle is rest but rod is rotating with angular velocity W..
so final angular momentam=Iw=initial angular momentam              (by conservation of angular momentam)
IW=mva/4
I for rod is ma2/12 about its axis passing through center perpendicular to rod...
WMa2/12 =mva/4
W= 3mv/Ma ]   .......................1
now applying conservation of linear momentam
let Vcom is the linear velocity of rod finally then
MVcom = mv
Vcom  =mv/M .......................2
```
10 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