A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it

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A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it

is struck horizontally by a 5.00-kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-04-02T11:26:19+03:00

Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = $\sqrt{rg}$ = $\sqrt{2.80 \times 9.8}$ = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

$P.E_{highest} = mgh$

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as $P.E_{highest} = mg(2r)$

where u is the velocity at the lowest point. Consequently, the modified equation is

= $\sqrt{v^{2} + 4gr}$

= $\sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}$

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = $\frac{(m_{1} + m_{2})u}{m_{2}}$

= $\frac{(20 + 5) \times 11.71}{5}$

= $\frac{292.75}{5}$

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.