Understanding Force

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

 

Force Lessons

Newton's First Law of Motion
Newton's Second Law of Motion
Newton's Third Law of Motion
Weight and Mass
Force
Impulse
Momentum
Calculating Force
Friction

  

 

 

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