Question

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 emerges 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 the target?

1. The velocity of the bullet will be reduced to half its initial value.
2. The velocity of the bullet will be more than half of its earlier velocity.
3. The bullet will continue to move along the same parabolic path.
4. The bullet will move in a different parabolic path.
5. The bullet will fall vertically downward after hitting the target.
6. The internal energy of the particles of the target will increase.

Answer 1

b,  d and f

Explanation:

Consider the adjacent diagram for the given situation in the question.

b. Conserving energy between "O" and "A"

`U_i + K_i = U_f + K_f`

⇒ `0 + 1/2 mv^2 = mgh + 1/2 mv^'`

⇒ `(v^')^2/2 = v^2/2 = - gh`

⇒ `(v^')^2 = v^2 - 2 gh`

⇒ `v^' = sqrt(v^2 - 2 gh)` ......(i)

Where v' is the speed of the bullet just before hitting the target. Let speed after emerging from the target is v" then,

By question, = `1/2 (mv^")^2 = 1/2[1/2 m(v^')^2]`

= `1/2 m(v^")^2`

= `1/4 m(v^')^2`

= `1/4 m[v^2 - 2 gh]`

⇒ `(v^")^2 = (v^2 - 2 gh)/2 = v^2/2 - gh`

⇒ `v^" = sqrt(v^2/2 - gh)`  .....(ii)

From equations (i) and (ii)

`v^'/v^" = (sqrt(v^2 - 2 gh))/(sqrt(v^2 - 2 gh)/(sqrt(2))) = sqrt(2)`

⇒ `v^" = v^'/sqrt(2) = v^2 (v^'/2)`

⇒ `v^"/(v^'/2) = sqrt(2)` = 1.414 > 1

⇒ `v^" > v^'/2`

Hence, after emerging from the target velocity of the bullet (v") is more than half of its earlier velocity v’ (velocity before emerging into the target).

d. As the velocity of the bullet changes to v’ which is less than v¹ hence, the path, followed will change and the bullet reaches point B instead of A', as shown in the figure.

f. As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target. Therefore, their internal energy increases.