An 8g bullet is fired into a 2.5 kg pendulum bob initially at rest and goes in it. The pendulum rises a vertical distance of 6cm. What is initial speed of the bullet.

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A bullet with a mass 8 g is fired into a 2.5 kg pendulum bob initially at rest. It sticks to the bob and as a result the bob rises a vertical distance of 6 cm. Use the law of conservation of energy.

Initially, the bob is at rest and its kinetic energy is 0, the potential energy of the bob is also 0. If the bullet is initially moving at v m/s, its kinetic energy is (1/2)*(8/1000)*v^2.

After collision the potential energy of the bob with the bullet stuck in it is 9.8*(6/100)*(2.5 + 8/1000). Equate this to the total energy in the system initially.

9.8*(6/100)*(2.5 + 8/1000) = (1/2)*(8/1000)*v^2

=> v^2 = 368.676

=>` v ~~ 19.2` m/s

**The initial speed of the bullet is 19.2 m/s**

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