Question

A bullet of mass 4kg when fired with a velocity of 50m/s can enter a wall up to a depth of 10cm. how much will be the average resistances offered by the wall?

Answer 1

Mass = 4 gm = 0.004 kg.

Velocity = 50 m/s.

Depth of penetration in the wall = 10 cm = 0.1 m

Kinetic energy = (1/2) x m x v^2
                        = (1/2)×0.004×2500
                        = 5 J

We know that kinetic energy = Resistive force × Distance

=>  5 N = 0.1 x F
=>  5 N = 1/10 x F
=> F = 50 N.

Hence, the average resistances offered by the wall = 50 N.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Given :-

Mass of the bullet, u = 4 g = 4 × 10⁻³ kg

Initial velocity, u = 50 m/s

Distance, S = 10 cm = 10 × 10⁻² m

Final velocity, v = 0 (since bullet comes to rest finally)

To Calculate :-

Force, F = ??

Formula to be used :-

F = ma

or F = m × (v² - u²/2S)

v² - u² = 2aS

a = v² - u²/2S

F = 4 × 10⁻³ × [(0)² - (50)²/2 × 10 × 10²]

F = 4 × 10⁻³ [- 2500 × 10²/2 × 10]

F = - 50 N.

Hence, the average resistances offered by the wall - 50 N.