Question

a mass of 10 mg is moving with a speed of 1000 m/ s. It penetrates into a bag of sand and is brought to rest after moving 50 cm into the bag. Find the average decelerating force on the bullet. Also, find the time in which it is brought to rest.

Answer 1

If we are to obtain the average decelerating force, we may assume the motion to have a constant deceleration. So the dependence of velocity on time will be

$v(t) = v_0 - at.$

The bullet stops, so $0 = v_0 - at$ or $a = \dfrac{v_0}{t}\,.$

The distance after t seconds will be

$s(t) = v_0t - \dfrac{at^2}{2}\,.$

We may substitute $a$ from the formula above and get

$s = v_0 t - \dfrac{v_0t}{2} = \dfrac{v_0t}{2}.$

Therefore, $t = \dfrac{2s}{v_0} = \dfrac{2\cdot0.5\,\mathrm{m}}{1000\,\mathrm{m/s}} = 0.001\,\mathrm{s}.$

The deceleration will be

$a = \dfrac{1000\,\mathrm{m/s}}{0.001\,\mathrm{s}} = 10^6\,\mathrm{m/s^2}.$

Then the average force will be

$F = ma = 10\cdot10^{-6}\,\mathrm{kg}\cdot 10^6\,\mathrm{m/s^2} = 10\,\mathrm{N}.$