Question

A bullet hits a sand box with a velocity of 20m/s and penetrates it up to a distance of 6cm find the deceleration of the bullet in the sand box

Answer 1

dist = 6cm = 0.06 m
v= 20m/s
u=0m/s
v^2=u^2+2as
here u = 0
then v^2=2as
20×20=2×a×0.06
a=-3333.3333m/s^2

therefore retardation = 3333.333m/s^2

Answer 2

Answer:  

The deceleration is $\bold{3333.33\ \mathrm{m} / \mathrm{s}^{2}.}$

Solution:

The deceleration is defined as the negative acceleration exerted by the body to come to rest from motion. As the bullet hits a sand box with velocity 20 m/s, it becomes the initial velocity exerted by the bullet and thus it covers a distance of 6 cm in the sand box with this velocity.

After reaching 6 cm distance the bullet stops so the final velocity will be zero. The question is to find the negative acceleration exerted by the bullet when it stops after covering 6 cm distance.

So the given quantities are

Initial Velocity u = 20 m/s

Final Velocity v = 0

Distance S = 6 cm = 0.06 m

Using Newton’s third equation of motion, we can easily determine the deceleration:

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

Thus,

$a=\frac{v^{2}-u^{2}}{2 s}=\frac{0-(20)^{2}}{2 \times 0.06}$

$\bold{a=\frac{-400}{0.12}=-3333.33\ \mathrm{m} / \mathrm{s}^{2}}$