A 23.00 k g lead sphere is hanging from a hook by a thin wire 3.80 m long, and i

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A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up V = m/s V = m/s SubmitMy AnswersGive Up A 23.00kg lead sphere is hanging from a hook by a thin wire 3.80m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 7.00kg steel dart that embeds itself in the lead sphere. Part A What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision? V = m/s SubmitMy AnswersGive Up

Explanation / Answer

Hey I solved a similar question Please apply your numbers and solve with the same procedure. Thank You. Contact kiranchegg@yahoo.in if any doubts. :)

Question: A 16.00kg lead sphere is hanging from a hook by a thin wire 3.20m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 3.50kg 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?

This where you went wrong is at the top of the loop, you assume the speed will be zero. If the speed is zero at the top of the loop the string would slack before that and not be able to complete the loop. Your method would be correct if instead of a string connecting the sphere, we have a massless rod.

The string must stay taut at all times.
So at the top of the loop, the speed minimum speed, Vt, must be fast enough to negate the force of gravity.
Centrifugal force = Gravitational force
(m + M) Vt^2/r = (m + M) g
=> Vt^2 = r g.........eq1


So, we can go on and solve the problem as follow
applying conservation of energy, where "V" is the speed of the system after the collision
(1/2) (m + M) V^2 = (m + M)g 2 r + (1/2)(m + M) Vt^2
=> V^2 = 4 g r + Vt^2........eq2

Conservation of momentum, where "v" is the speed of the dart before collision
m v = (m + M) V
=> V = m v/(m + M).........eq3

Plug eq1 and eq3 into eq2 and solve for v
[m v/(m + M)]^2 = 4 g r + g r = 5 g r
=> v = (5 g r)^(1/2) (m + M)/m = 69.76 m/s ANS

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