A bullet of mass 20 gram strikes a pendulum of mass 5kg the centre of mass of Pendulum Rises vertical distance of 10 cm if the bullet get embedded into the pendulum calculate its initial speed
Answers
Initial speed of bullet is 280 m/s.
Explanation:
=> According to the question,
Mass of bullet, m₁ = 25 g or 0.025 kg
Mass of pendulum, m₂ = 5 kg
Vertical displacement, h = 10cm or 0.1 m
=> Suppose, a bullet strikes a pendulum with a velocity 'u'. and final velocity is suppose v.
=> According to law of conservation of linear momentum:
m₁u = (m₁ + m₂)v
v = (m₁ / m₁+m₂)*u
v = (0.025 / 0.025 + 5) * u
v = u/201
=> As per the law of conservation of energy,
1/2 (m₂ + m₁) v² = (m₂ + m₁) gh
u²/ 201² = 2 * 10 * 0.1
u² = (201)² * 2
u = 201 √2
= 280 m/s.
Thus, its initial speed is 280 m/s.
Learn more:
Q:1 A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string . The bullet crosses the block and emerges on the other side. If the center of mass of the block rises through a height of 20.0 cm , find the speed of the bullet as it emerges from the block.
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Q:2 A simple pendulum of length 1m has a wooden bob of mass 1kg. It is struck by a bullet of mass 10^-2 kg moving with a speed of 2oo m/s. The bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back. Take g=10 m/s^2
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