Question

A bullet having a mass of 0.005 kg is moving with speed of 100 m/s .it penetrates into a bag of sand and brought to rest after moving 25 cm into the bag . Find the decelerating force on the bullet . Also calculate the time in which it is brought to rest.

Answer 1

Question 19160

It is given, that $S = 25 \, \text{cm} = 0.25 \, \text{m} = v_0 t_s - \frac{a t_s^2}{2}$ , where $t_s$ denotes the time for which the object moved until stop. Also, for velocity, $v = v_0 - a t$ , and for stop time $0 = v_0 - a t_s \Rightarrow t_s = \frac{v_0}{a}$ .

Plugging this formula into formula for S, obtain: $t_s = 2\frac{S}{v_0} = 0.005s$ . Hence, the acceleration is $|a| = \frac{v_0}{t_s} = \frac{100m / s}{0.005s} = 20000m / s^2$ (actually this acceleration is negative, but one needs the absolute value only). Hence, $|F| = m|a| = 0.005kg \cdot 20000m / s = 100N$ .