Question

# A uniform bar lies on a smooth horizontal table. Two point masses collide with it and stick to the bar after the collisions, as shown in the figure. After the collisions -

A
velocity of centre of mass of the system is 0 m/s.
B
angular velocity about centre of mass of the system is 0.6 rad/s
C
angular velocity about centre of mass of the system is 0.2 rad/s
D
total energy of the system is 0.6 J

Solution

## The correct options are A velocity of centre of mass of the system is 0 m/s. C angular velocity about centre of mass of the system is 0.2 rad/s D total energy of the system is 0.6 JLet us suppose, velocity of centre of mass of the system is v and angular velocity about centre of mass of the system is ω. Figure as per question - We know that a free system of bodies rotates about its centre of mass after collision because moment of inertia is minimum about the axis passing through COM. So, position of COM of system w.r.t COM of the uniform bar, [i.e taking origin at COM of uniform bar] xcom=m1d1+m2d2+m3d3m1+m2+m3 =2×−1+1×2+8×02+1+8=0  [ from figure ] Therefore, COM of the whole system will lie at the COM of the bar. Since, there is no net external force on the system, on applying law of conservation of linear momentum. pi=pf ⇒m1v1+m2v2+m3v3=(m1+m2+m3)v ⇒2×−1+1×2+8×0=(2+1+8)v [ from figure ] ⇒v=0 As there is no net external torque on the system, on applying law of conservatuion of angular momentum about COM of the system, Li=Lf ⇒m1v1d1+m2v2d2+m3v3d3=Iω ⇒2×1×1+1×2×2+8×0×0=(2×12+1×22+8×6212)ω [ I=m1d21+m2d22+mL212 from figure ] ⇒30ω=6 ⇒ω=0.2 rad/s Now, Total energy, E=12Iω2=12×(2×12+1×22+8×6212)×0.22 ⇒E=0.6 J Therefore, option (a), (c) and (d) are correct.

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