Question

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​

Answer 1

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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