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  • A bullet of mass m fired at 30^° to the horizontal leaves the barrel of the gun with a velocity v.

A bullet of mass m fired at 30^{\circ} 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?
(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 continue to move along the same parabolic path.
(d) The bullet will move in a different parabolic path.
(e) The bullet will fall vertically downward after hitting the target.
(f) The internal energy of the particles of the target will increase.

Answers (1)

The answer is the option (b), (d), (f)

If we assume KE1 and KE2 to be the kinetic energy of the bullet pre and post target then,

KE_{2} = \frac{1}{2} KE_{1}

\frac{1}{2} MV_{2}^{2} =\frac{1}{4} MV_{1}^{2}

V_{2}^{2} = \left (\frac{V_{1}}{\sqrt{2}} \right )^{2} = 0.707 V_{1}

b) V_{2} = 0.707 V_{1}

Hence the velocity of the bullet post it has reached the target is greater than the previous velocity. Hence, option b is the correct choice.

d) the path of the bullet will be parabolic as it has horizontal velocity, and the new parabolic path will have both the components (horizontal and vertical). So, after the target it will be a new parabolic path and hence option (d) is correct.

f) the internal energy of the particles of the target will increase, and some parts of the kinetic energy have converted to heat. Hence option f is correct.

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infoexpert24

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