Question

A bullet moving horizontally to the right (+x direction) with a speed of 500 m/s strikes a sandbag and penetrates a distance of 10.0 cm. What is the average acceleration, in m/s2, of the bullet?

Answer 1

Answer:$a=1.25\times 10^6 m/s^2$

Explanation:

Given

initial velocity of bullet$\left ( u\right )=500 m/s$

Penetration distance$\left ( x\right )=10cm$

i.e. bullet will finally stop after moving 10cm in the sandbag

Final velocity of Bullet$\left ( v\right )=0$

Using motion of equation to get acceleration

$v^2-u^2=2ax$

$0-\left ( 500\right )^2=2\left ( -a\right )\left ( 0.1\right )------\left ( - because\ it\ is\ deceleration\right )$

$a=\frac{\left ( 500\right )^2}{2\times 0.1}$

$a=1.25\times 10^6 m/s^2$