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  1. Scalars & Vectors | Week 1
    5 Topics
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics

Lesson 9, Topic 5
In Progress

Application of Newton’s Laws & Conservation of Momentum Laws

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1. Recoil of a Gun:

When a gun is fired, the bullet moves forward with a velocity of V1, and mass M1, if the gun has mass M2, it also moves with a velocity, V2. From conservation of momentum.

The momentum of the gun and bullet before and after firing is given as

Recoil of a Gun Easy

Before firing

The gun and the bullet are both at rest, so the momentum (mv) of the gun and the bullet are both zero. So, the total momentum of the system before the bullet is fired is zero.

After firing

After the gun is fired, the bullet gets a velocity to the right, as shown in the diagram above. This means that the bullet now has momentum, M1V1, to the right. Since there was zero total momentum in the system before the gun was fired, there must be zero total momentum now. This means that the gun must have an equal and opposite momentum to the left. The momentum of the gun is given by M2V2. Since the momentum must be conserved we have,

M1V1 + M2V2 = 0

M1V1 = – M2V2

Note: The negative sign shows that the momentum of the gun is in the opposite direction of the bullet. therefore the gun jerks backwards of recoils.

M1 = Mass of the bullet
M2 = Mass of gun
V1 = Velocity of bullet
V2 = Velocity of gun.


A machine gun with a mass of 5 kg fires a 50 g bullet at speed of 100ms-1. Calculate the recoil speed of the machine gun.


Mass of bullet (M1) = \( \frac{50}{1000} \scriptsize = 0.05\:kg\)

Speed of bullet (V1) = 100ms-1

Mass of gun (M2) = 5kg

Recoil speed of machine gun (V2) = ?

Momentum before firing the machine gun is 0

Momentum after firing;

M1V1 = – M2V2

0.05 × 100 = – (5 × V2)

5 = -5V2

V2 = \(– \frac{5}{5} \)

V2 = -1 ms-1

The recoil velocity = 1 ms-1 (directed opposite to that of the bullet)

2. Jet and Rocket Propulsion:

The propulsion of all rockets including jet engines, rockets used for launching satellites, and deflating balloons, is explained by the same physical principle – Newton’s third law of motion.
Gases are contained in the combustion chambers of rocket engines. When the gas is burnt in the chamber, it expands and exerts pressure within the chamber. The hot gas is expelled downwards through a nozzle at a very high speed from the rocket. The ejected gas has mass and velocity and therefore has momentum. The momentum of the ejected gas produces equal and opposite momentum on the rocket which propels the rocket forward.

3. Why Walking Is Possible:

Human beings walk by pushing their foot against the ground. According to the principle of conservation law of linear momentum and Newton’s third law of motion, as a person pushes the ground with his or her foot (action force ), the ground in turn exerts an equal and opposite force (reaction force) on the person. Therefore a person walking is pushed forward by the reaction force of the ground on him or her, and not by the person’s own push on the ground.


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