A bullet of mass 10 g moving horizontally with a velocity of 400 ms^(-1) strikes a wook 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 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 ms^(-

1.	A bullet of mass 10 g moving horizontally with a velocity of 400 ms^(-1) strikes a wook 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 found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be
Answer» `100 ms^(-1)` `80 ms^(-1)` `120 ms^(-1)` `160 ms^(-1)` Solution :Mass of bullet, `m = 10 g = 0.01 kg` Initial speed of bullet, `u = 400 ms^(-1)`. Mass of block , M = 2 kg LENGTH of string , l = 5 m Speed of the block after collision = `v_1` Speed of the bullet on emerging from block , `v = ?` USING energy conservation principle for the block, `(KE + PE)_("Reference") = (KE + PE)_h` `implies 1/2 Mv_1^2 = Mgh " or " v_1 = sqrt(2gh)` `v_1 = sqrt(2 xx 10 xx 0.1) = sqrt(2) ms^(-1)` Using momentum conservation principle for block and bullet system, `(M xx 0 + m u)_("Before collision") = (M xx v_1 xx MV)_("After collision")` `implies 0.01 xx 400 = 2sqrt(2) + 0.01 xx v` `implies v = (4 - 2sqrt(2))/(0.01) =117.15 ms^(-1) ~~ 120 ms^(-1)`