Question

A bullet of mass 20gm moving with a
velocity of 600m/s penetrates a brick wall. If the
depth of penetration is 36 cm, find the average
resistive force offered by the target?


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Answer 1

Given

• Mass of bullet, m = 20 g = (20/1000) kg = 0.02 kg
• Velocity with which bullet is fired, u = 600 m/s
• Depth of penetration = distance covered by bullet, s = 36 cm = ( 36/100 ) m = 0.36 m
• Final velocity of bullet, v = 0

To find

• Average resistive force offered by the target, F =?

Formulae required

• Third equation of motion

        2 a s = v² - u²

• Newton's second law of motion

        F = m a

[ Where a is acceleration, s is distance covered, v is final velocity, u is initial velocity, F is force and m is mass ]

Solution

Using third equation of motion

→ 2 a s = v² - u²

→ 2 a ( 0.36 ) = ( 0 )² - ( 600 )²

→ 0.72 a = -360000

→ a = 500,000 m/s²

Using Newton's second equation of motion

→ F = m a

→ F = 0.02 × 500,000

→ F = 10,000 N

Therefore, the force applied by bullet on the wall is 10 kilo Newton.

Since, after penetrating for a given distance bullet come to rest, therefore magnitude of force exerted by wall on the bullet will be same., only direction will be opposite. Hence

• Magnitude of Resistive force offered by target is 10,000 Newtons.