Question

A bullet of mass 0.05 kg is fired from a rifle with a velocity of 800 M s-1 and penetrates to a depth of 0.2 M calculate the average force of the wood to the penentration of the bullet​

Answer 1

Solution :-

As per the given data ,

• Mass of the bullet (m)= 0.05 kg
• Velocity of the bullet(u) = 800 m /s
• Distance traveled (s)= 0.2 m
• Final velocity (v)= 0 m/s ( as the bullet comes to rest )

As the bullet moves with uniform acceleration we can use third equation of motion in order to find acceleration ,

As per third equation of motion ,

⇒ v²=u²+ 2as

Now let's substitute the given values in the above equation ,

⇒ 0 = 640000 + 2 x a x 0.2

⇒ a = - 640000 / 0.4

⇒ a = - 1600000 m /s

Here , negative sign denotes retardation

By applying newton's second law of motion ,

⇒ F = ma

⇒ F = 0.05 x - 1600000

⇒ F = - 80,000 N

⇒ | F| = 80,000 N

The magnitude of force acting on the wood is 80,000 N