Question

A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy? If the bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?

Answer 1

Mass of bullet, (m) = 0.015 kg
Initial velocity of bullet, (v₁) = 400 m/s
Distance, (S) = 0.02 m
So, initial kinetic energy of bullet,
$K.E=\frac{1}{2}mv$²
Therefore, initial kinetic energy,
$$\begin{gathered} {\left(K.E\right)}_1=\frac{1}{2}(0.015)(400{)}^2\mathrm{J} \\ =\boxed{1200\mathrm{J}} \end{gathered}$$

Similarly, final kinetic energy as the final velocity is zero,
(K.E)₂ = 0
So,
Work done = Change in kinetic energy
Therefore work done,
Work done = (K.E)₂ – (K.E)₁
(Force) (Distance) = (0 –1200) J
So,
$$\begin{gathered} \mathrm{Force}=-\frac{1200}{0.02} \\ =-60000 \\ =\boxed{-\left(6\right)(10{)}^4\mathrm{J}} \end{gathered}$$

Negative sign shows that the force applied is opposite to the direction of motion of the bullet.