Question

A bullet of mass 0.03kg moving with the speed of 400m\s 12 cm into a fixed block of wood . calculate the average force exwerted by the wood on the bullet

Answer 1

Assuming a constant average retardation is offered to the bullet

let it be a m/s2

so using the equation of motion

v2 =u2 +2as

​Finally bullet comes to rest
so,

v= 0 

and given  u  = 75 m/s

s = 5cm  = 0.05m

substituting above we get

0 = 752 +2a(0.05)

-5625 =0.1 a

a = -56250 m/s2


this is the retardation which is offered to the motion of the bullet.

thus the average force acting on the bullet will be f = ma
m = 20 g = 0.02kg

or f = 0.02 x (-56250)

f = - 1125 N (the force is negative as it opposes the motion of the bullet and is retarding in nature.

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Answer 2

Answer:

The average force exerted by the wood on the bullet is 2*10⁴N.

Explanation:

Given:

Mass of bullet=0.03kg

Velocity(u)=400m/s

v=0m/s

Distance=12cm=0.12m

To find: Average force exerted by the wood on the bullet

Solution:

According to the equation of motions,

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

0-400²=2a0.12

-160000=0.24a

∴a=-160000/0.24

∴a=-2000000/3

We know that,

F=ma

=0.03*-2000000/3

=-20000N

∴F=2*10⁴N

Since force is a vector quantity negative sign indicates the direction.

Thus, the average force exerted by the wood on the bullet is 2*10⁴N.