A large block of wood of mass M=5.99 kg is hanging from two long massless cords. A bullet of mass m=10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance -9.8 cm before the (block+bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is : (take `g=9.8 ms^(-2))`

Options

A. 841.4 m/s
B. 821.4 m/s
C. 831.4 m/s
D. 811.4 m/s

Answer

C

Explanation

`m u = (m+M)v`
0.01u= (0.01+5.99)v
u= 600v ...(i)
`(1)/(2) (m+M) v^(2)= (m+M)gh`
`v= sqrt(2gh)`
`= sqrt(2 xx 10 xx 9.8 xx 10^(-2))= 1.386` ...(ii)
From equation (i) and (ii)
u= 831.43 m/s