Calculating Force Based On Newton's Third Law Of Motion (1)

Question 1.

A rocket burns 0.01 kg of fuel each second and ejects it as a gas with a velocity of 5, 000 m/s. What force does the gas exert on the rocket? 

Solution:

The momentum of the burning fuel provides the force that moves the rocket forward. 

Momentum of the burning fuel = mv

                                     = 0.01 x 5, 000

                                     = 50 kgm/s

 Therefore, the force the gas exerts on the rocket each second is 50 Newtons.

 Question 2.

A jet engine develops a thrust of 270 Ns when the velocity of the exhaust gases relative to the engine is 300 ms^-1. What is the mass of the material ejected per second?

Solution:

The thrust on the jet engine is equal to the momentum of the burnt gases.

 Therefore,

 Thrust = mv

 270 = m x 300

m   = 270/300

m  = 0.9 kg

The mass of the material ejected per second is 0.9 kg

 Question 3.

A rocket burns fuel at the rate of 10 kgs^-1 and ejects it with a velocity of 5 x 10³ ms^-1. The Thrust exerted by the gas on the rocket is?

Solution:

The momentum of the burning gas equals the thrust exerted on the rocket.

Momentum = mv

 Momentum of burning fuel = 10 x 5 x 10³

                             = 5 x 10⁴ Ns

Therefore, Thrust exerted by the gas on the rocket is        5.0 x 10⁴ Ns          

Question 4.

A machine gun with a mass of 5 kg fires a 50 g bullet at a speed of 100 ms^-1, the recoil speed of the machine gun is? 

Solution:

The momentum of the bullet is equal to the momentum of the gun.

MV = mv

Where M = mass of the gun, V = recoil velocity (or speed) of the gun, m= mass of the bullet, v = velocity (or speed) of the bullet.

 Therefore,

5 x V = 0.05 x 100

V = 0.05 x 100/5 = 1

 The recoil speed of the machine gun is 1 ms^-1.

Note: the mass of the bullet has to be converted from grams to kilograms. i.e., from 50 g to 0.05 kg.

Question 5.

If fuel was consumed at a steady rate of 5.0 x 10^-2 kg per second in a rocket engine and ejected as a gas with a speed of 4 x 10³ ms^-1, what is the thrust on the rocket?

Solution:

The thrust on the rocket is equal to the momentum of ejected gas.

Momentum of ejected gas = mv

                                      = 5.0 x 10^-2 x 4 x 10³

                                = 2.0 x 10²

                                = 200 Ns