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Question

A uniform rod of mass M and length 2 m lies on a smooth horizontal table. A particle of mass m moving on the table with a velocity u strikes the rod perpendicularly at an end and stops there. Find the distance travelled by the center of the rod (in metres) by the time it turns through a right angle. Given M=4m.


A
π/12
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B
π/6
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C
π/3
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D
π
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Solution

The correct option is B π/6
Let S be the distance travelled by the COM of the rod.


Since, Fext=0 on system of (rod+particle), hence linear momentum will be conserved.
Pi=Pf
mu=m(0)+Mv
v=(muM) ...(1)
Here, v is the velocity of centre of mass of the rod after the impact.

Also τext=0, hence applying angular momentum conservation about reference point as COM.
Li=Lf ...(i)
where Lf=Ltrans+LRot
r=0, as velocity vector passes through centre, hence Ltrans=Mv0r=0
Putting in Eq. (i),
mu(l2)=Iω
mu(l2)=Ml212×ω
ω=6muMl

τext=0α=0
Applying kinamatic equation for angular displacement of rod,
θ=ωt ...(ii),
Here, angle rotated by rod θ=π/2
t=π/2(6muMl)
t=πMl12mu ...(iii)
In that time interval, linear distance travelled by the rod's centre of mass is:
S=v.t
Putting v and t from Eq. (i) & (iii),
S=muM×πMl12mu
S=πl12
S=π×212=π6 m

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