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

Options

A. `100 ms^(-1)`
B. `80 ms^(-1)`
C. `120 ms^(-1)`
D. `160 ms^(-1)`

Answer

C

Explanation

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)`