A bullet of mass 10 g moving horizontally with a velocity of 400 m s–1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

A bullet of mass 10 g moving horizontally with a velocity of 400 m s^–1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

160 m s^–1

100 m s^–1

80 m s^–1

120 m s^–1

Explanation:

During the collision, apply momentum conservation

(0.01)(400) + 0 = (2)V + (0.01)V

where V = $\sqrt{2gh}$

V = $\sqrt{2 × 10 × 0.1}$

V = $\sqrt{2}$

Solving V’ = 120 m / sec.