A 50 g bullet moving with velocity 10 m / s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be

Options

A. `100%`
B. `95%`
C. `5%`
D. `50%`

Answer

B

Explanation

Initial K.E of system = K.E of the bullet `=(1)/(2)m_(B) v_(B)^(2)`
By the law of conservation of linear momentum
`m_(B) v_(B) + 0= m_(sys) xx v_(sys)`
`rArr v_(sys)= (m_(B) v_(B))/(m_(sys)) = (50 xx 10)/(50+ 950)= 0.5m//s`
Fractional loss in K.E. `=((1)/(2) m_(B) v_(B)^(2) -(1)/(2)m_(sys) v_(sys)^(2))/((1)/(2) m_(B)v_(B)^(2))`
By substituting `m_(B)= 50 xx 10^(-3)kg, v_(B)= 10m//s`
`m_(sys)= 1kg, v_(1)= 0.5m//s` we get
Fractional loss `=(95)/(100) therefore` Percentage loss= 95%