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Question

A ball of mass 500 grams is suspended with the help of a light, inextensible string of length 100 cm. The maximum value of to which the string rotates from the mean position A to reach the extreme position C is 60. Consider g=10 m/s2


ii. What will be the speed of the ball when it reaches half the value of the maximum height it can achieve?

[1 mark]

A
v = 2.5 m/s
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B
v = 3.5 m/s
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C
v = 1.5 m/s
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D
v = 5 m/s
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Solution

The correct option is D v = 5 m/s
From the previous part, we know the potential energy at F is equal to 1.25 J.
Let the velocity at point F be
v.

The potential energy at F, P.E.F = m×g×AE = 0.5×10×0.25 = 1.25 J
Kinetic energy at F, K.E.F=12mv2

Using the principle of conservation of energy,
Total energy at C = Total energy at Fm×g×AD=12mv2+(m×g×AE) 12mv2=m×g×(ADAE) v=2g(ADAE)v=2g(0.500.25) m/sv=5 m/s

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