Question

Bullet of mass 10 gram moving with speed 200cm/s is stoped within 5 cm of target .the average retarding force offerd by target is?

Answer 1

it is the answer.....

Answer 2

Hello

Initial velocity of the bullet = 200 cm/sec
Or in si units 2 m/sec

Final velocity of the bullet (as it stops) = 0m/sec

Distance covered while the retardation =5 cm=0.05 m

Mass of the bullet = 10 g =0.01 kg

Now,
We know that
2as = v² - u²

Since final velocity of 0
So,
2as = - u²

a = - u²/2s
So,
$a = \frac{ - 4}{2 \times 0.05} \\ a = - 40m \: per \: {sec}^{2}$
Now,

From the newton's second law of motion we know :-
F = ma

So,
F = 0.01 * (-40)
F = (-0.4) N

So,
The average regarding force is 0.4 newton in opposite direction of the motion.

Hope this will be helping you ✌️