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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


Answers : (1)


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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