1.

A uniform disc of mass m and diameter 2R moves forward towards another uniform disc of mass 2m & diameter 2R on a frictionless surface as shown in figure When the first disc contacts the second, they stick to each other and move as a single object.

Answer»

the velocity of combined disc after the COLLISION is `(V_(0))/3`
the angular velocity of combined disc after the collision is `(omega_(2))/3`
If `omega_(0)=8/3 (V_(0))/R`, the combined disc will not rotate
If combined disc does not rotate, the energy LOSS is `19/9MV_(0)^(2)`

Solution :Momentum conservation
`MV_(0)=2MV`
`V=(V_(0))/3`
Angular momentum conservation about centre of mass
`1/2MR^(2).omega_(0)-4/3MV_(0)R=[(MR^(2))/2+M(4/3R)^(2)+1/2(2M)R^(2)+2M(2/3R^(2))]OMEGA`
`omega=3/25 omega_(0)-8/25(V_(0))/R`
For `omega=0`
`omega_(0)=(8V_(0))/(3R)`
The energy loss,
`/_\E=E_(i)-E_(f)`
`=1/2MV_(0)^(2)+1/2((MR^(2))/2).omega^(2)-1/2 (3M)V^(2)=19/9MV_(0)^(2)`


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