Question

In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass of 5 x 10^-3 kg and a speed of 1110 m/s.
Required:
a. What is the average change in momentum per second for the bullets?
b. Determine the average force exerted on the wall.
c. Assuming the bullets are spread out over an area of 3.0×10^−4m^2 obtain the average pressure they exert on this region of the wall.

Answer 1

Answer:

a. ΔP/Δt =  42.6 N

b. F = 42.6 N

c. P = 142042.4 Pa = 1.42 KPa

Explanation:

a.

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

Vf = Final Speed = 1110 m/s

Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

ΔP/Δt =  42.6 N

b.

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

F = 42.6 N

c.

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

P = 142042.4 Pa = 1.42 KPa