Question

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?

Answer 1

Answer:

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.