A bead can slide on a smooth straight wire and a particle of mass m attached to the bead by a light string of length L . The particle is held in contact with the string taut and is then let fall. If the bead has mass 2m then when the string makes an angle θ with the wire, the bead will have slipped a distance______???

Question

A bead can slide on a smooth straight wire and a particle of mass m attached to the bead by a light string of length L . The particle is held in contact with the string taut and is then let fall. If the bead has mass 2m then when the string makes an angle &theta; with the wire, the bead will have slipped a distance______???

vikas askiitian expert · Answer

ur question is not clear .... pls post the diagram then it will be easy to solve ....

Raunaq Mehta · Answer

since there is no force in the x direction conserve momentum2m(0)+m(l)=2m(x )+m(lcosq+x)solving you get c as answer.

Shubham Gupta · Answer

As no force acts in the horizontal direction, center of mass of the bead and the particle should remain at the same x-coordinate.
Initial x-coordinate of the center of mass from the original position of the bead
= (2m * 0 + m * L) / 3 = L/3
Final position of the center of mass if the bead has moved by a distance x,
= [2mx + m(x + Lcosa)] / 3m
= (3x + Lcosa) / 3
Equatting COM in both the cases,
3x + Lcosa = L
=> x = L(1 - cosa) / 3

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