A mass of `0.98kg` attached on a spring of constant `K = 100Nm^(1)` is hit by a bules of `20gm` moving with a velocity `20ms^(1)` horizontally The bullet gas ambated and system oscilation with the mass on horizontal and surface The amplitude of oscilation will be

Options

A. `0.6 cm`
B. `6 cm`
C. `1.2 cm`
D. `12 cm`

Answer

B

Explanation

Conservation of momentum :
`(0.98 + 0.02) v = 0.02 xx 30 rArr v = 0.6ms^(-1)`
Consevation energy in oscillation we get
`(!)/(2) (0.98+ 0.02)v^(2) = (1)/(2) KA^(2)`
`:. A^(2) = (v^(2))/(K)= ((0.06)^(2))/(100) rArr A = (0.6)/(10) = 0.06m = 6cm`