The mass (m) and velocity (v) of an item are used to calculate linear momentum. It is more difficult to halt an item with more momentum. p = m v is the formula for linear momentum. Conservation of momentum refers to the fact that the overall quantity of momentum never changes. Let’s learn more about linear momentum and momentum conservation.

Linear Momentum of System of Particles

We know that the particle’s linear momentum is:

p = m v

where p represents the particle’s momentum.

For a single particle, Newton’s second law is:

F = dP ⁄ dt,

where F represents the particle’s force.

The total linear momentum of ‘n‘ particles is:

So, when we compare these equations, we get:

P = M V

As a result, we may state that, a system of particles’ total linear momentum is equal to the product of the system’s total mass and the velocity of its Centre of mass.

Differentiate the above equation:

dP ⁄ dt = M dV ⁄ dt

dV ⁄ dt is the acceleration of the centre of mass.

Assume A be the acceleration of the centre of mass.

Therefore, dP ⁄ dt = M A

M A is the external force, so, dP ⁄ dt = F_ext

This equation is just Newton’s second law applied to a system of particles. If the total external force operating on the system is zero, the system is said to be in equilibrium.

dP ⁄ dt = 0 if F_ext = 0.

P = constant, as a result of this.

When the entire force exerted on a particle’s system is equal to zero, the system’s total linear momentum is constant or conserved. This is the law of total linear momentum conservation for a system of particles.

If there is no external force, the individual particles’ momenta may fluctuate, but their total remains constant. The term “momentum” refers to a vector quantity.

Conservation of momentum of two particles system

Conservation of Total Linear Momentum of a System of Particles

Consider the case of radioactive decay. What is the definition of radioactive decay? It’s a process in which one unstable nucleus divides into two relatively stable nuclei, releasing massive amounts of energy in the process. If a parent nucleus is unstable and wishes to become stable, it will emit a particle and another daughter nucleus to achieve stability. Compared to the parent nucleus, this daughter nucleus is significantly more stable. This is the definition of radioactive decay. Assume that the parent nucleus is at rest, that the mass of the daughter nucleus is m, and that the mass of the daughter nucleus is M.

As a result, the parent nucleus’ mass will be m + M. Everything that happens in this situation is due to an internal force rather than an external force.If F_ext = 0, we may conclude that dP ⁄ dt = 0. 

So, P = constant.

If a body’s net external force is zero, the rate of change of momentum is likewise zero, implying that there is no change in momentum.

Conservation Of Linear Momentum Example

With velocities, v and V, two bodies of mass m and M are travelling in opposing directions. We must determine the system’s velocity if they collide and move together after the impact.

Momentum will be conserved since there is no external force acting on the system of two bodies.

Initial momentum = Final momentum

M V – m v = (M + m) v_final

We can simply calculate the system’s final velocity using this equation.

Conservation Of Linear Momentum Applications

• The launching of rockets is one of the applications of conservation of momentum. The exhaust gases are forced downwards when the rocket fuel burns and the rocket is propelled higher as a result.
• Motorboats operate on the same premise; they push the water backwards and are pushed forwards in response to maintain propulsion.
• Consider the third law of motion, which describes the motion of an air-filled balloon. Air escapes from the balloon as soon as it is released, and it has momentum. The balloon moves in the opposite direction of the air rushing out to conserve momentum.

Sample Problems

Problem1: Two balls of equal masses are thrown upwards along the same vertical line at an interval of 3 seconds with the same initial velocity of 44.1 m s⁻¹. Find the total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic.

Solution:

Given:

Initial velocity, u = 44.1 m s⁻¹

Time interval between flights of two balls = 3 s

When the particles collide the height of both particles will be same,

For first particle, time will be ‘t’ and for second particle time will be (t – 3).

h = u t − (1 ⁄ 2) g t² = u (t − 3) − (1 ⁄ 2) g (t − 3)²

 u t − (1 ⁄ 2) g t² = u t − 3 u − (1 ⁄ 2) g (t² − 6 t + 9)

0 = − 3 u + 3 g t − (9 ⁄ 2) g

t = (u ⁄ g) + (3 ⁄ 2)

t = (44.1 ⁄ 9.8) s + 1.5 s = 6 s

By solving the quadratic equation in t we get the time of first collision to be 6 s.

h = 44.1 × 6 − (1 ⁄ 2) × 9.8 × 6²

= 264.6 m − 176.4 m

= 88.2 m

Now by momentum conservation, velocity of particles after collision:

m (u − g t) + m (u − g (t − 3)) = (m + m) v′

m u − 6 m g + m u − 3 m g = 2 m v′ 

v′ = u − (9 ⁄ 2) g

= 0 m ⁄ s

Acceleration, a = − g

Now, assume time taken for collision with ground be t₁

h = (1 ⁄ 2) g t₁²

t₁ = √(2 h ⁄ g)

= √(2 × 88.2  ⁄ 9.8) s

= √18 s

Hence, total time of flights is (6 + √18) s​ and (3 + √18) s​.

Problem 2: Give few practical applications of the law of conservation of linear momentum?

Solution:

Few applications of conservation of linear momentum is listed below:

• By blowing air out of its mouth or tossing an object in the opposite direction of the direction in which he wishes to go, the person remaining on the frictionless surface can get away from it.
• When a guy leaps from a boat on the beach, the boat is pushed away from the coast somewhat.
• The recoil of a rifle.

Problem 3: A compressed spring binds two unequal weights together. When the cord is burnt by a matchstick, the spring is released, what quantity of the two masses flying apart will be equal?

Solution:

Two unequal masses are first joined by a compressed spring. The cord is then burned with a matchstick, and the spring is released, causing the two masses to separate and acquire velocities that are inversely proportional to their masses, resulting in equal momentum.

Problem 4: Consider two skaters who began at a rest and then pushed off against each other on ice with reduced friction. The woman weighs 45 kg, whereas the man weighs 60 kg. The woman is moving away at a speed of 4 m ⁄ s. What is man’s recoil velocity?

Solution:

Given:

Mass of woman, m = 45 kg

Mass of man, M = 60 kg

Initial velocity = 0

Final speed of woman, v = 3.5 m ⁄ s

According to law of conservation of momentum,

0 = (45 × 4 + 60 V) kg m ⁄ s

V = – 3 m ⁄ s

The negative sign shows that the man will move in opposite direction to the woman.

Hence, the man’s recoil velocity is 3 m ⁄ s.

Problem 5: What is the condition for a system’s momentum to be conserved?

Solution:

If a body’s net external force is zero, the rate of change of momentum is likewise zero, implying that there is no change in momentum.

If Fext = 0, we may conclude that dP ⁄ dt = 0. So, P = constant.