A strong wind of speed 33 m s-1 blows against a wall. The density of the air is 1.2 kg m-3. The wall has an area of 12 m2 at right angles to the wind velocity.

Question 1

A strong wind of speed 33 m s^-1 blows against a wall. The density of the air is 1.2 kg m^-3. The wall has an area of 12 m² at right angles to the wind velocity. The air has its speed reduced to zero when it hits the wall.

What is the approximate force exerted by the air on the wall?

A 330 N                       B 400 N                       C 480 N                       D 16 000 N

Reference: Past Exam Paper – June 2013 Paper 12 Q9

Solution:

Answer: D.

Force F = Δp / t = mΔv / t

The air has its speed reduced to zero. So, Δv = 33 – 0 = 33 m s^-1  

We need to find the mass m of the air that hits the wall and the time t.

Available data:

density ρ = 1.2 kg m^-3             area A = 12 m²                        speed v = 33 m s^-1

Density = mass / volume         giving              mass m = Density × volume

However, we do not have the volume, BUT we know that volume = A × d

Mass m = ρ × (A × d) ------------------ (1)

Again, we do not have the value of d.

BUT, speed = distance / time = d / t

So, instead of using only d, we use d/t. To make this possible, we divide equation (1) by t on both sides.

m/t = ρ × (A × d/t) = ρ × (A × v)

m/t = 1.2 × (12×33)  

Consider the equation for force again.

Force F = Δp / t = mΔv / t = (m/t) × Δv = ρ × (A × v) × Δv

Force F = 1.2 × (12×33) × 33 = 15 681.6 ≈ 16 000 N

Notes:

1.

The quantity ‘m/t’ is the rate at which the air hits the wall. That is, how much mass of air hits the wall in 1 second. Note that ‘rate’ is a basic concept learnt in maths many years back but we tend not to apply it in physics.

At A-Level, very often we will need to manipulate equations to make up for information that is not explicitly provided.

2.

From the formula F = Δp / t = mΔv /t, it is not only the Δv that should be divided by time t. Students usually have this misconception as most of the time, we divide Δv by time. But the mass can also be divided by time giving the ‘rate of … of mass’. 

Example 1: In a rocket, the fuel gets used up with time as the rockets moves up into space. The quantity ‘m/t’ may represent the rate at which fuel is being used up.

Example 2: If sand is falling from bag at constant speed, the quantity ‘m/t’ may represent the rate of fall of sand. …

3.

If we consider the units in

Force F = Δp / t                       and

Force F= … = ρ × (A × v) × Δv

We would observe that the overall unit is still the same in both cases, and equal to the unit of force. So, the equation is still correct.