Rotation 4 Question 5

6. A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar [as shown in the figure] and stick to the bar after collision. Denoting

angular velocity (about the centre of mass), total energy and centre of mass velocity by ω,E and vc respectively, we have after collision

(1991)

(a) vc=0

(b) ω=3v5a

(c) ω=v5a

(d) E=35mv2

Analytical Answer Type Questions

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Answer:

Correct Answer: 6. (a, c, d)

Solution:

Pf=0 or vc=0Li=Lf or (2mv)a+(2mv)(2a)=Iω

Here, I=(8m)(6a)212+m(2a)2+(2m)(a2)=30ma2

Substituting in Eq. (i), we get

ω=v5a

Further, E=12Iω2=12×(30ma2)v5a2=3mv25



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