A bullet of mass 10 g moving horizontally with a velocity of 400 m s⁻¹ strikes a wood block of mass 2 kg which is suspended by 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 eme…

Question

A bullet of mass 10 g moving horizontally with a velocity of 400 m s⁻¹ strikes a wood block of mass 2 kg which is suspended by 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

Accepted Answer

(3) 120 m s⁻¹. Given: bullet m=10 g =0.01 kg at u=400 m s⁻¹; block M=2 kg; the block's centre of gravity rises h=10 cm =0.1 m. Idea: the rise tells you the block's speed just after the bullet passes through, and momentum then gives the bullet's. 1/2Mv₁²=Mgh, so v₁=√2gh=√2×10×0.1=√2 m s⁻¹. Momentum: mu=Mv₁+mv. 0.01×400=2√2+0.01v, so 4=2.83+0.01v. v=117≈120 m s⁻¹.

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