Question

A 20 gram bullet moving at 300 m/s stops after penetrating 3 cm of a bone. calculate the avg force exerted by the bullet.

Answer 1

The initial velocity of bullet is 300m/s.
Final velocity=0
Distance covered=3cm=0.03m

So,using the formula v^2=u^2-2as
We have
0=(300)^2-2a(0.03)
Or 0.06a=90000
=> a=90000/0.06
=> a= 1500000m/s^2.

So force =mass times acceleration
=(0.02kg)(1500000)
=30000N

Answer 2

Answer:

The magnitude of average force exerted by the bullet will be equal to 3×10⁴N.

Explanation:

We have given, the initial velocity of the bullet, u= 300m/s

The final velocity of the bullet will be, v = 0

The mass of the bullet, m= 20g = 20×10⁻³Kg

The distance covered by the bullet before comes to rest, S = 3cm

First, calculate the acceleration of the bullet,

From the third equation of motion:

v² - u² = 2aS

(0)² - (300)² = 2(a)(0.03)

0.06a = - 90000

a = - 15 × 10⁵m/s²

Here, negative sign shows that the retardation in the bullet.

We know the Newton's second law of motion:-

F = ma

$F = (20*10^{-3}kg)(-15*10^{5}ms^{-2})$

$F= -3*10^{4}N$

Therefore, the magnitude of average force exerted by the bullet will be 3×10⁴N.        

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