A bullet of mass 10 g moving horizontally with a velocity of `400ms^(-1)` strikes a wooden block of mass 2kg 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

Options

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

Answer

`(1)`

Explanation

Applying momentum conservation
`(10)/(1000) xx 400 + 0 = 2 xx v_(1) + (10)/(1000) xx v_(2)`
`Rightarrow 4 = 2V_(1) + 0.01 v_(2) ....(1)`
Applying work energy theorem for block
`W = Delta KE `
`Rightarrow 2 xx10 xx0.1 = 1/2 xx 2 xx v_(1)^(2)`
`v_(1) = sqrt 2 = 1.4 m/s`
Putting the value of `v_(1)` in equation(1) `
`4 = 2xx 1.4 + 0.01 v_(2) Rightarrow v_(2) = 120 m/s`