A bullet of mass m fired at `30^(@)` to the horizontal leaves the barrel of the gun with a velocity v . The bullet hits a soft target at a height h above the ground while it is moving downward and ground while it is moving downward and emerge out with half the kinetic energy it had before hitting the target . Which of the following statements are correct in respect of bullet after it emerges out of teh target ?

Options

A. The velocity of the bullet will be reduced to half its initial value
B. The velocity of the bullet will be more than half of its earlier velocity
C. The bullet will move in a different parabolic path
D. The internal energy of the particles of the target will increase .

Answer

B, D

Explanation

9B) From law of conservation of energy ,
`U_(i)+K_(i) =U_(f)+K_(f)`
` 0+1/2 mv^(2) =U_(f) +K_(f)`
`((v)^(2))/2 = (v^(2))/2 = - gh`
`(v)^(2)=v^(2)-2gh rArr v = sqrt(v^(2)-2gh) " "...(i)`
where v is speed of the bullet just before hitting the target . Let speed after emerging form the target is v . then ,
`1/2 (mv)^(2) = 1/2 [1/2m(v)^(2)]` (given)
` :. 1/2 m(v)^(2) = 1/4 m(v)^(2) `
` :. 1/2 m(v)^(2) = 1/4 m[v^(2)-2gh]`
` (v")^(2) =1/4 m[v^(2)-2gh]`
`(v")^(2) =(v^(2)-2gh)/2 `
`= (v^(2))/2 - gh`
`v.. =sqrt((v^(2))/2-gh)" " ...(ii)`
By taking ratio (1) to (2)
`(v.)/(v..) =(sqrt(v^(2)-2gh))/((sqrt(v^(2)-2gh))/(sqrt(2)))=sqrt(2)`
`v .. = (v.)/(sqrt(2))`
` = v^(2) (v/2)`
`(v..)/(v/2)=sqrt(2)`
` = 1.414 gt 1 `
` v.. gt (v.)/2 `
(D) As the velocity of the bullet changes to v.. which is less than v hence , path followed will change and the bullet reaches at point B instead of A . as shown .
(D) As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . Therefore , thier internal energy increases .