Question

A bullet, having a mass of 0.04 kg moving with a velocity of 450 m/s penetrates a distance of 0.08 m into a wooden block firmly attached to the earth. Assuming a constant accelerating force, compute the following: (a) the acceleration of the bullet, (b) the accelerating force, (c) the time of acceleration, and (d) the impulse of the collision

Answer 1

(a) We can find the acceleration of the bullet from the kinematic equation:

$$v^2=v_0^2+2ad,$$ $$a=\dfrac{v^2-v_0^2}{2d},$$ $$a=\dfrac{0-(450\ \dfrac{m}{s})^2}{2\cdot0.08\ m}=-1.26\cdot10^6\ \dfrac{m}{s^2}.$$

The sign minus means that the bullet decelerates.

(b) We can find the accelerating force from the Newtons's Second Law:

$$F=ma=0.04\ kg\cdot(-1.26\cdot10^6\ \dfrac{m}{s^2})=-50625\ N.$$

(c) We can find the acceleration time from the kinematic equation:

$$v=v_0+at,$$ $$t=-\dfrac{v_0}{a}=-\dfrac{450\ \dfrac{m}{s}}{-1.26\cdot10^6\ \dfrac{m}{s^2}}=3.57\cdot10^{-4}\ s.$$

(d) By the definition, the impulse is the change in momentum, therefore we can write:

$$J=\Delta p=m(v-v_0)=0.04\ kg\cdot(0-450\ \dfrac{m}{s})=-18\ \dfrac{kgm}{s}.$$

The sign minus means that the impulse of the collision is directed in the opposite direction to the motion of the bullet.