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Solution :

Figure shows the situation of system just before and just after collision. <br> Initially the centre of mass of the rod is at point O. After collision when the particle sticks to the rod. Centre of mass is shifted from point O to C as shown in figure. Now the system is rotated about axis passing through C <br> Now from linear momentum conservation <br> ` mv = (M + m)v' rArr v' = (mv)/(M + m) ` <br> From Angular momentum conservation about A <br> `L_(i) = L_(f)` <br> ` 0 + 0= I omega - (m + M) v' l_(1)` <br> `rArr I omega = (m + M) v' l_(1)` <br> Put the value of `v',& l_(1)` we get <br> `omega = ( 6mv)/((M + 4m)l)` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOT_CON_JEE_PHY_C10_SLV_047_S01.png" width="80%">**Basic Property Of Rigid Body**

**Rigid Body Motion**

**Velocity And Acceleration Of A Point In Rotating Rigid Body In Pure Rotation**

**Relative Angular Velocity In Case Of A Rigid Body**

**Rotational And Translationa Motion Together**

**Rolling Motion**

**Moment Of Inertia Of Rigid Body**

**Uniform Rectangular Sheet**

**Triangular Lamina About Its Base**

**INERTIA OF SQUARE SHEET ABOUT IT'S DIAGONAL**