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```				   a uniform rod of length  L and mass  2M on a smooth horizontal table .a point mass M moving horizontally at right angles to the rod with  velocity V collides with one end of rod and sticks it then
1..angular velocity of the system after collision is ?
2..the loss in k.E of the system as a whole as a result of  the collision is ?
```

6 years ago

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```										Dear akash

since there is no external force so linear momentum and angular momentum will remain conserved
MV = M(WL/2) + 2MV1
V = WL/2  + 2V1 .
V1 = V/2 - WL/4 ...........1
conservation of angular momentum
MVL/2 = MW(L/2)(L/2)  + I W
MVL/2 = MW(L/2)(L/2)  + 2ML2/12 W
V  =  WL/2  + WL/3
V   =5WL/6
W =6V/5L
from equation 1
V1 = V/5
loss in K.E = 1/2MV2   - 1/2 2M V12 - 1/2 M (WL/2)2 - 1/2 I w2

put the value and calculate

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```
6 years ago

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