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Question

A rod of mass 6 m and length 6a is kept on a horizontal smooth surface. Two point masses of masses 2m and m respectively, moving perpendicular to the rod, collide with the rod and stick to the rod after collision as shown in the column-II. Point C represents centre of mass of the rod. AC=2a and CB=a. v is the linear speed of the centre of mass of the system just after collision and ω is the angular speed of the system just after collision.


A
A – (q,r), B – (s), C – (p,s), D – (p,q, r)
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B
A – (r,s), B – (p,r), C – (q,s), D – ( r,q)
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C
A – (q,r), B – (p,r), C – (q,s), D – ( p,s)
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D
A – (q,r), B – (p,q), C – (r,s), D – ( p,r)
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Solution

The correct option is A A – (q,r), B – (s), C – (p,s), D – (p,q, r)
I=112×6m×36a2+m(4a2)+2m(a2)=24ma2 for p,q,r and s

For p,
Applying conservation of linear momentum in vertical direction
2mv0mv0+6m×0=(2m+m+6m)v
v=v09^j
Now applying conservation of angular momentum we get
24ma2ω=mv02a^k+2mv0a^k
ω=v06a^k

For q,
2mv02mv0+6m×0=(2m+m+6m)v
v=024ma2ω=2mv02a^k+2mv0a^kω=v04a^k

For r,
2mv02mv0+6m×0=(2m+m+6m)v
v=024ma2ω=4mv0a^k+2mv0a^k
ω=v04a^k

For s,
mv02mv0+6m×0=(2m+m+6m)v
v=v03(^j),24ma2ω=mv02a^k+2mv0a^(k)
ω=0

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