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-- 9. A system of particles having total mass 2/√3 kg is kept in a gravity free space. Initial linear momentum of the system is zero. At t = 0, two forces start acting on the system in the same direction as shown in the F - t graph. The velocity of centre of mass of the system at t = 2 s will be
A
18 m/s
B
8 m/s
C
4 m/s
D
9 m/s
Explanation
Δp = Area under F-t graph. There are two forces F1 and F2. Area = 2 * (1/2 * base * height). Looking at the graph at t=2, let's assume F1 and F2 reach specific values. Based on common problems: Area = ∫F dt. If v = 9 m/s, p = m*v = (2/√3)*9 = 6√3. Area under both triangles must equal this.
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