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Text Solution

`(2Mv)/(M-m)``(2mv)/(M+m)``(Mv)/(M+m)``(mv)/(M+m)`

Answer :

DSolution :

Velocity of the centre of mass of the rod `+`particle system <br> `V_(CM)=(mv)/((M+m))` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C03_E01_258_S01.png" width="80%"> <br> `vecv_(p,CM)=vecv_(p)-vecv_(c)` <br> `v_(c)=(mv)/(M+m)` and `v_(p)=v` <br> `vecv_(p,CM)=(Mv)/(M+m)` <br> `vecv_(rod,CM)=vecv_(rod)-vecv_(CM)=0-(mv)/(M+m)` <br> `v_(rod,CM)=-(mv)/(M+m)` <br> Location of the centre of mass of the system rod + particle from `A` <br> ` r_(1)=(mxx0+Ml//2)/((m+M))` <br> `=(Ml)/(2(m+M))` <br> Angular momentum of the particle just before collision about `C`,<br> `L_(p,c)=mv_(p,c)r_(1)` <br> `=(M^(2)mv)/(2(M+m)^(2))` <br> Angular momentum of the rod about the centre of mass of system `C`, <br> `L_(rod,CM)=Mv_(rod,CM)(l/2-r_(1))` <br> `=M((mv)/(M+m))((ml)/(2(M+m)))` <br> `=(Mm^(2)vl)/(2(M+m)^(2))` <br> Moment of inertia about the vertcal axis passing through `C`, <br> `I_(c,rod)=I_(0)=M(l/2-r_(1))^(2)` <br> `I_(c,rod)=(Ml^(2))/12+m((ml)/(2(M+m)))^(2)` <br> `I_(c,"particle")=mr_(1)^(2)=m((Ml)/(2(M+m)))^(2)` <br> `I_(c)=(Ml^(2))/12+M((ml)/(2(M+m)))^(2)+m((Ml)/(2(M+m)))^(2)` <br> `I_(c)=(M(M+4m))/(12(M+m))L^(2)` <br> As no external forces are acting on the system rod `+` particle, hence the velocity of the centre of mass of the system will remain constant.<br> `v_(CM)=(mv)/(M+m)` <br> For angular velocity about `C` <br> `mvr_(1)=I_(c)omega` <br> `omega=(mvr_(1))/(I_(c))=(mv((ml)/(2(M+m))))/(((M(M+4m)L^(2))/(12(m+m)))` <br> `=(6mv)/((M+3m)L)`**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**