A block of mass m = 1 kg slides with velocity `upsilon = 6 m//s` on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle `theta` before momentarily coming to rest. If the rod has mass M = 2 kg, and length `l = 1m`, the value of `theta` is approximately: (take `g = 10 m//s^2`)

Options

A. `49^@`
B. `63^@`
C. `69^@`
D. `55^@`

Answer

B

Explanation

`mv l = I Omega `
` Omega = (mv l)/( I )`
` 1/2 I omega^2 =(m+M) g l_("com") ( 1- cos theta)`
` 1/2 ((mvl)^2)/(I ) = (m+M) gl_("com") (1- cos theta)`
`I=((Ml^2)/(3)+ml^2)`
` I= ((2)/(3) +1) =5/3`
` ( 36 xx 3 )/( 2xx5 ) =( 3xx10xx2 )/( 3 ) = ( 1 - cos theta )`
`(27 )/( 50 ) = ( 1- cos theta)`
`cos theta =1 - (27 )/(50 )`
` cos theta =1-(23)/(50) `
` theta = 60^@`