Conservation of Momentum

There are several real-life examples of conservation of momentum.

• Linear Momentum: When a rocket is launched into space, it starts with a net zero momentum. During its journey, the downward momentum of the burning fuel is equal to the upward momentum of the rocket. Therefore, the net momentum of the rocket-propulsion system remains zero.
• Angular Momentum: A skater slowly rotates when spreading her arms. However, when she drops her arms, she rotates faster. The product of angular velocity and radius gives the angular momentum. When the skater spreads her arms, the radius is longer, and the angular velocity is lower. When she pulls her arms, the radius is smaller, and the angular velocity is higher. In both cases, the angular momentum remains unchanged and is conserved.

Consider a rigid body comprising different points. If p_1i, p_2i, …, p_ni represents the initial momentum of the 1^st, 2^nd, …, and nth particle, and p_1f, p_2f, …, p_nf represents their final momentum, the law of conservation of momentum can be mathematically represented by

Recall that momentum is the product of mass and velocity. If m₁, m₂, …, m_n represents the masses, v_1i, v_2i, …, v_ni represent the initial velocities, and v_1f, v_2f, …, v_nf represent the final velocities, the above expression reduces to 

The above expression for momentum conservation applies to one dimension only. For three-dimensional representation, we use vectors.

To summarize,

The conservation principle easily applies to the collision of particles, which is discussed in another article. Collisions can be classified into different types based on kinetic energy transfer. Two types of collisions are:

• Elastic collision – both momentum and kinetic energy are conserved 
• Inelastic collision – only momentum is conserved, not kinetic energy 

In addition, there are perfectly inelastic collisions, where objects stick together after the collision, and only momentum is conserved.

Problem 1: Two cars, Car A and Car B, collide head-on. Car A has a mass of 1000 kg and is initially traveling at 20 m/s to the right, while Car B has a mass of 1200 kg and is initially traveling at 15 m/s to the left. Calculate the velocities of Car A and Car B after the collision.

Solution

Given, m_A = 1000 kg, v_Ai = 20 m/s, m_B = 1200 kg, and v_Bi = 15 m/s

The velocity of car A is

v_Af = (m_A – m_B)/(m_A + m_B) v_Ai + 2m_B/(m_A + m_B) v_Bi

=> v_Af = (1000 kg – 1200 kg)/(1000 kg + 1200 kg) x 20 m/s + 2 x 1000 kg/(1000 kg + 1200 kg) x 15 m/s

=> v_Af = 11.8 m/s

The velocity of car B is given by

v_Bf = 2m_A/(m_A + m_B) v_Ai – (m_A – m_B)/(m_A + m_B) v_Bi

=> v_Bf = 2 x 1000 kg/(1000 kg + 1200 kg) x 20 m/s – (1000 kg – 1200 kg)/(1000 kg + 1200 kg) x 15 m/s

=> v_Bf = 19.54 m/s

Both cars travel right after the collision.

Problem 2: A billiard ball with a mass of 0.2 kg is moving at a velocity of 3 m/s to the right. It collides with another stationary billiard ball of the same mass. Determine the velocities of both balls after the collision.

Solution

Given, m_A = 0.2 kg, v_Ai = 3 m/s, m_B = 0.2 kg, and v_Bi = 0 m/s

When two similar masses collide, they exchange velocities. Therefore, the first ball will come to rest, and the velocity of the second ball is 3 m/s.

Problem 3: A rocket of mass 5000 kg is initially at rest on the launchpad. The exhaust gases are expelled from the rocket with a velocity of 1000 m/s relative to the rocket. If the mass of the expelled gases is 100 kg, what is the rocket’s velocity after the gases are expelled?

Solution

Given m_rocket = 5000 kg, v_{rocket, initial} = 0, m_gas = 100 kg, v_{gas, initial} = 0 and v_{gas, final} = 1000 m/s

From the law of conservation of momentum

m_rocket x v_{rocket, initial} + m_gas x v_{gas, initial} = m_rocket x v_{rocket, final} + m_gas x v_{gas, final}

=> 5000 kg x 0 + 100 kg x 0 = 5000 kg x v_{rocket, final} – 100 kg x 1000 m/s

=> 0 = 5000 kg x v_{rocket, final} – 100000 kg ˑ m/s

=> v_{rocket, final} = 20 m/s

Problem 4: A hunter fires a rifle with a mass of 3 kg. The bullet, with a mass of 0.05 kg, is expelled from the rifle with a velocity of 400 m/s. Determine the recoil velocity of the rifle.

Solution

Given m_rifle = 3 kg, v_{rifle, initial} = 0, m_bullet = 0.05 kg, m_{bullet, initial} = 0 m/s, v_{bullet, final} = 400 m/s

From the law of conservation of momentum

m_rifle x v_{rifle, initial} + m_bullet x v_{bullet, initial} = m_rifle x v_{rifle, final} + m_bullet x v_{bullet, final}

=> 3 kg x 0 + 0.05 kg x 0 = 3 kg x v_{rifle, final} – 0.05 kg x 400 m/s

=> 0 = 3 kg x v_{rifle, final} – 20 kg m/s

=> v_rifle = 6.67 m/s