The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run ballistics tests on the pistol at the firing range. If a 0.050-kg bullet were fired from the handgun with a speed of 400 m/s and it traveled 0.080 m into the target before coming to rest, what force did the bullet exert on the target?

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

e/c, where e is its total energy

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach