The Fig. shows a string of equally placed beads of mass m, separated by distance d. The beads are free to slide withoutfriction on a thin wire. A constant force F acts on the first bead initially at rest till it makes collision with the second bead. The second bead then collides with the third and so on. Suppose that all collisions are elastic,

Engineering

Physics

Work Done by a Constant Force

Newtons Second Law of Motion

Collision

Question

The Fig. shows a string of equally placed beads of mass m, separated by distance d. The beads are free to slide withoutfriction on a thin wire. A constant force F acts on the first bead initially at rest till it makes collision with the second bead. The second bead then collides with the third and so on. Suppose that all collisions are elastic,

speed of the first bead immediately before and immediately after its collision with the second bead is $\sqrt{\frac{2 Fd}{m}}$ and zero respectively.

speed of the first bead immediately before and immediately after its collision with the second bead is $\sqrt{\frac{2 Fd}{m}}$ and $\frac{1}{2} \sqrt{\frac{2 Rd}{m}}$ respectively.

speed of the second bead immediately after its collision with third bead is zero.

the average speed of the first bead is .

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When a constant force F acts on the first bead (mass m) over distance d, its speed before collision is found using work-energy theorem: work done = kinetic energy. So, $F d = \frac{1}{2} m v^{2}$ , giving $v = \sqrt{\frac{2 F d}{m}}$ . For elastic collisions between identical masses, velocities swap: first bead stops, second moves at v. Thus, before collision: $\sqrt{\frac{2 F d}{m}}$ , after: 0.

Final answer: speed of the first bead immediately before and immediately after its collision with the second bead is $\sqrt{\frac{2 F d}{m}}$ and zero respectively.