Two identical rods each of mass m and length l are placed on a smooth horizontal surface. In case (1) a particle of mass m moving with V0 strikes the rod perpendicularly at its centre. In case (ii), another particle of mass m moving with V0 strikes the rod perpendicularly at l / 4 from the centre. In both the cases, particles come to rest just after collision. The ratio of velocity of center of mass of rod in case (i) to that in case (ii) is

In both casesapplying com,mv0=MV ⇒V=mV0M ∴V1V2=1:1

Two identical rods each of mass m and length l are placed on a smooth horizontal surface. In case (1) a particle of mass m moving with V0 strikes the rod perpendicularly at its centre. In case (ii), another particle of mass m moving with V0 strikes the rod perpendicularly at l / 4 from the centre. In both the cases, particles come to rest just after collision. The ratio of velocity of center of mass of rod in case (i) to that in case (ii) is

In both casesapplying com,mv0=MV ⇒V=mV0M ∴V1V2=1:1

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Two identical rods each of mass m and length $l$ are placed on a smooth horizontal surface. In case (1) a particle of mass m moving with V₀ strikes the rod perpendicularly at its centre. In case (ii), another particle of mass m moving with V₀ strikes the rod perpendicularly at $l$ / 4 from the centre. In both the cases, particles come to rest just after collision. The ratio of velocity of center of mass of rod in case (i) to that in case (ii) is

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1:2

1:1

4:1

2:1

answer is A.

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Detailed Solution

In both cases
applying com,
${mv}_{0} = MV \Rightarrow V = \frac{{mV}_{0}}{M} ∴ \frac{V_{1}}{V_{2}} = 1 : 1$

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