A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω. The ratio of v and ω is :

Solution

Physics sir · 4 steps

The centre of mass sits at the bar's centre, so conserve angular momentum about that point.

1. Find the pivot point. Take x from the bar's centre. The 2m mass strikes at x = -2 cm and the m mass at x = +4 cm. The centre of mass is at (20m(0) + 2m(-2) + m(4))/23m = 0, which is the bar's centre. No external torque acts on the table, so angular momentum about this point is conserved.
2. Angular momentum before. The 2m mass moves down at 2 cm left of centre, giving 2m × v × 2 = 4mv. The m mass moves up at 4 cm right of centre, giving m × v × 4 = 4mv. Both turn the system the same way (clockwise), so the total is 8mv.
3. Moment of inertia after. Bar: (20m × 12²)/12 = 240m. Stuck masses: 2m × 2² = 8m and m × 4² = 16m. Total I = 264m (in cm² units).

   8mv = 264m ω ⇒ v/ω = 264/8 = 33

4. So. v/ω = 33 (in cm), option (a). The whole system also drifts at v/23 because the net linear momentum is mv, but that does not change the spin about the centre of mass.

The idea to keep. For particles sticking to a free body, conserve angular momentum about the combined centre of mass. Check the sense of each r × p: masses moving opposite ways on opposite sides add up.

Answer from the NTA final answer key. Solution by PrepWiser.