A 0.009 kg bullet fired through a door enters at 803 m/s and leaves at 617 m/s. If the door material is known to exert an average resistive force of 5620 N on bullets of this type at usual speeds, find the thickness of the door.

Answer:

The thickness of the door is 0.4230 m

Explanation:

Given;

mass of bullet, m = 0.009 kg

initial velocity of the bullet, u = 803 m/s

final velocity of the bullet, v = 617 m/s

average resistive force of the door on the bullet, F = 5620 N

Apply Newton's second law of motion;

Force exerted by the door on the bullet = Force of the moving bullet

F = ma

where;

F is applied force

m is mass

a is acceleration

Also, Force exerted by the door on the bullet = Force of the moving bullet

[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]

where;

v is the final velocity of the bullet

u is initial velocity of the bullet

t is time

We need to calculate the time spent by the bullet before it passes through the door.

[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]

Distance traveled by the bullet within this time period = thickness of the door

This distance is equivalent to the product of average velocity and time

[tex]S = (\frac{u+v}{2}) t[/tex]

where;

s is the distance traveled

[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]

Therefore, the thickness of the door is 0.4230 m

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