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

A

Explanation

AB=5

Apply conservation of linear momentum
`mu+0=mv+MV`
`(10)/(1000)xx400+0=(10)/(1000)v+2V`
`0.01v+2V=4` …..(1)
PE=KE
MgH=`1/2xxMV^2`
`2xx10xx(10)/(100)=1/2xx2xxV^2`
`implies V^2=2`
V=`sqrt2ms^(-1)`
Substituting the value of V in Eq. (1), we get
`(v)/(100)+2sqrt2=4impliesv=(4-2sqrt2)100`
=120 `ms^(-1)`