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Question

A 7.5 g bullet moving at 120 m/s strikes a thick wall. Assume that the bullet

undergoes a uniform deceleration and stops after penetrating 6 cm. Find

(a) The time taken by the bullet to stop.

(b) The impulse on the wall.

(c) The magnitude of the average force experienced by the wall.

Answer 1

The bullet is initially traveling with a certain momentum calculated as (m)(v) or (7.5/1000)(120) = 0.9 kg m/s. After the bullet comes to a stop, the momentum is zero (velocity = 0). So the change in momentum is 0.9 kg m/s.

The Momentum-Impulse Theorem says the change in momentum equals the impulse. The Impulse is calculated as (F)(Δt) but we can't calculate Δt yet because we don't know the force.

But we are told that the bullet undergoes a uniform deceleration. And we know that F = ma so if we can determine the deceleration, we can calculate the force. One of the kinematic equation of motion says:

v_f² = v_i² + 2a(Δx) so a = (v_f² - v_i²)/(2•Δx)

and we know v_f = 0, v_i = 120, Δx = 6/100 so:

a = (0 - 120²)/(2•0.06) = -14400/0.12 = -120,000 m/s²

That means the force is calculated as F = ma = (7.5/1000)(-120,000) = -900 N (Note that the negative sign means the force is applied in the direction opposite the velocity or direction of bullet travel)

Going back to the Impulse-Momentum Theorem, 0.9 = (900)(Δt) so Δt = 0.9/900 = 0.001 s or 1 millisecond

The Impulse equals the change in momentum so it is 0.9 Ns (units represent Force•time)

The magnitude of the force was 900 N (from above)