Question

A bullet of mass of 20g travelling with a velocity of my 20m/s penetrates a sand bag and comes to rest in 0.05sec find the distance through which it penetrates in the sand bag

Answer 1

given -

t = 0.05 s

v = final speed = 0 (bullet comes to stop)

u =initial speed= 20(speed of bullet)

s = distance

from equation of motions we've-

v-u/a = t which can be rewritten as v-u/t = a

hence , a = v-u/t...........(i)

from another equation of motion we've-

$v^{2}$ = $u^{2}$ + 2as

$v^{2}$-$u^{2}$/2a= s

$\frac{(v+u)(v-u)}{1/\frac{2(v-u)}{t}}=\frac{v+u(t)}{2}$

= $\frac{20(0.05)}{2}$= 0.5m

ans = 0.5m