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Question
A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of the pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.
Solution
Given:
m1 = 20 g = 20 × 10-3 kg; m2 = 5 kg; s = 10 × 10-2 m.
Let the speed of the bullet be v. The common velocity of bullet and pendulum bob is V. According to the law of conservation of linear momentum.
V = `(m_1v)/((m_1 + m_2)) = (20 xx 10^-3v)/(5 + 20 xx 10^-3) = 0.02/5.02 v = 0.004 v`
The bob with bullet goes up with a deceleration of g = 9.8 ms-2. Bob and the bullet come to rest at a height of 10 × 10-2 m.
from III rd equation of motion
`v^2 = u^2 + 2as` here
`v^2 - 2gs = 0`
`v^2 = 2gs`
`(0.004 v)^2 = 2 xx 9.8 xx 10 xx 10^-2`
`v^2 = (2 xx 9.8 xx 10 xx 10^-2)/(0.004)^2`
v = 350 ms-1.
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