Topic Menu

Cyber-chess a chess instruction site

Questions on Newton's Laws

Q2.

The diagram shows a jet engine.

Air enters the engine at A and is heated before leaving B at a much higher speed.

 

(a) State what happens to the momentum of the air as it passes through the engine.

The momentum in the air increases.

Momentum is m x v = as speed increases the momentum will too

[1 mark]

(b) Explain, using appropriate laws of motion, why the air exerts a force on the engine in the forward direction.

There is a change of momentum of the air from position A to B, so a force is acting on the air causing that change (Newton 2: Ft = Δp).

The air exerts a reaction force on the engine of equal magnitude/size but in the opposite direction (Newton 3) therefore in the forward direction - B to A.

[3 marks]

(c) In one second a mass of 210 kg of air enters at A.

The speed of this mass of air increases by 570 m s–1 as it passes through the engine.

Calculate the force that the air exerts on the engine.

t = 1s

m = 210 kg

Δv = 570 m s–1

F = ma

F = mΔv/t

F = 210 x 570/1

F = 119,700

F = 120 kN ( for answering to 2sf)

[2 marks]

(d) When an aircraft lands, its jet engines exert a decelerating force on the aircraft by making use of deflector plates.

These cause the air leaving the engines to be deflected at an angle to the direction the aircraft is travelling as shown in the diagram below:

The speed of the air leaving B is the same as the speed of the deflected air.

Explain why the momentum of the air changes.

Momentum has two parts - mass and velocity. Velocity is a vector, therefore even though the speed does not change, the change in direction causes a change in velocity and therefore momentum.

[2 marks]

(e) The total horizontal decelerating force exerted on the deflector plates of the jet engines is 190 kN.

Calculate the deceleration of the aircraft when it has a mass of 7.0 × 104 kg.

F = ma

190 x 103 = 7.0 × 104 x a

a = - 190 x 103 / (7.0 × 104 )

a = - 2.7 m/s2

Note it is a deceleration - so should be negative.

[1 mark]

(f) The aircraft lands on the runway travelling at a speed of 68 ms–1 with the deflector plates acting.

Calculate the distance the aircraft travels along the runway until it comes to rest.

You may assume that the decelerating force acting on the jet engines remains constant.

v = 0 ms–1

u = 68 ms–1

a = - 2.7 m/s2

s = ?

v2 = u2 + 2as

0 = 682 - 2 x 2.7 x s

s = 682/5.4

s = 860 m

Again - should be 2sf

[2 marks]

(g) Suggest why in practice the decelerating force provided by the deflector plates may not remain constant.

As the plane slows the rate of intake of air will decrease. Therefore the volume/mass/amount of air passing through engine per second will decrease.

This will result in a smaller rate of change of momentum and therefore smaller force.

The decrese in air resistance as speed reduces will also play a part.

[2 marks maximum]

(13 marks Total)