Question

10 gram Bullet moving at 200 m per second stops after penetrating 5 cm of wooden plane calculate the average force exerted on the bullet......

Answer 1

Given :
mass=10g=10/1000=0.01kg
Initial speed=u=200m/s
final velocity= o m/s
S=5cm=5/100=0.05m

From third equation of motion :V²-u²=2as

0²-(200)²=2ax0.05

-40000=ax5x2/100

-40000=a/10

a=-400000m/s²

From Newton's Second law:
F=ma
=0.01x(-400000)
=-4000N

∴ A force of 4000N acts in opposite direction.



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Answer 2

HELLO THERE!

First, let's analyze the given data!

Mass of the bullet (m) = 10g.

Convert this into kg.

Since 1g = 0.001kg, 10g = 10 x 0.001kg = 0.01kg

Initial velocity (u) of the bullet = 200 m/s

Since the bullet comes to rest, its final velocity (v) = 0.

It stops after penetrating 5 cm wooden plane. So, distance covered by it is 5 x 0.01 m = 0.05m

Now applying the third equation of motion,

v² = u² + 2aS (where a [retardation] is -a)

Hence, v² = u² - 2aS

=> 0 = (200)² - 2 x a x 0.05

$=> a = \frac{40000}{0.1}$

=> a = 4 x 10⁵ m/s²

Now, since the average retarding force is asked,

Apply Newton's Second Law of Motion,

F = ma

=> F = 0.01 x 4 x 10⁵ N

=> F = 4000N (4 x 10³ N)

So, average retarding force exerted on the bullet is 4000N.

THANKS!