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A rod AB of mass M and length 8l lies on...

A rod `AB` of mass `M` and length `8l` lies on a smooth horizontal surface. A particle, of `A` mass `m` and velocity `v_(0)` strike's the rod perpendicular to its length, as shown in Fig. As a result of the collision, the centre of mass of the rod attains a speed of `v_(0)//8` and the particle rebounds back with a (not to scale) speed of `v_(0)//4`. Find the following: a. The ratio `M//m`. b. The angular velocity of the rod about `O`. c. The coefficient of restitution `'e'` for the collision. d. The velocities of the ends `'A'` and `'B'` of the rod, namely, `v_(A)` and `v_(B)` respectively.

Text Solution

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Consevation of linear momentum
`mv_(0)=M(v_(0))/8-m(v_(0))/4`
`8m=M-2mimplies10m=M`
`impliesM//m=10`
Conservation of angular momentum about the point of collision
`0=(M(81)^(2))/12omega-M (v_(0))/81`
`impliesomega=(3v_(0))/1281`
Using coefficient of restitution, equation
`[((v_(0))/8+omegal)-(-(v_(0))/4)]=I(v_(0)-0)impliese=51/1281`
Velocity at end `A`
`v_(A)=(v_(0))/8+omega.4l=28/128v_(0)`
`impliesv_(A)=7/32v_(0)`
Velocity at end `B`
`v_(B)=(v_(0))/8+1/32v_(0)`
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