# Laws of Conservation of Momentum

Momentum is a commonly used word in sports. Often, sports commentators say “This team has winning momentum”, whenever such things are said. What they usually mean is that the team is on the move, it’s going to take some effort to stop their winning streaks. Whenever anything possesses momentum, that means that the thing is in motion and due to its inertia. It wants to stay in motion. Such a thing is assumed to possess momentum. Let’s look at this concept and its law of conservation in detail.

### Momentum

Momentum can be defined as a mass in motion. It is denoted by “p”. The amount of momentum possessed by an object depends upon two factors – its mass and its velocity. An object which does not have any mass will have zero momentum no matter how fast it moves. Similarly, a stationary object will always have zero momentum, whatever its mass may be. Let’s say the mass of the object is “m” and its velocity is “v”. Then the momentum is given by,

**p = mv**

The unit of momentum will be mass units times the velocity units. In the standard metric, the momentum is given by Kg-m/s.

**Momentum as a vector quantity**

A vector quantity is a quantity that has both magnitude and direction. Since momentum also depends on the velocity, it is a vector quantity. For example – a ball thrown towards the north will have a velocity towards the north. In that case, momentum will also point in the direction where the ball is moving. Its direction is the same as the direction of velocity.

### Law of Conservation of Momentum

A conserved quantity is a quantity that does not change. There are many conserved quantities in physics, and they are often useful for making predictions for the system in very complicated situations. The Law of conservation of momentum says that momentum is conserved for a system, but there is a catch. This law only applies to systems that are isolated. This means, there should not be any external forces acting on the system.

The law of conservation of momentum states that, when no external forces are acting on a system, then the momentum is conserved. Specifically, the total momentum of the system before and after any event remains the same.

Consider a system of two point masses m_{1} and m_{2}._{ }Initially, these bodies were moving with the velocity v_{1i} and v_{2i}. Now they collide with each other and their final velocities become v_{1f} and v_{2f}. So, according to the law of conservation of momentum,

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}

This assumes that there are only internal forces that are acting between the objects.

For a general system with n-particles. The law is given by the equation,

m_{1}v_{1i} + m_{2}v_{2i} + ….. + m_{2}v_{ni} = m_{1}v_{1f} + m_{2}v_{2f }+ ….m_{2}v_{nf}

**Why is momentum conserved? **

This is a consequence of Newton’s third law. In a collision between two objects A and B. Object A experiences a force F_{AB} which is due to B, similarly, object B experiences force F_{BA }which is due to A. These forces must be equal according to Newton’s third law. Since the collision was for a very short time .

Now, this is a very short time. So, this is considered an impulse. An impulse is equivalent to a change in momentum.

### Sample Problems

**Question 1: Calculate the momentum of a ball thrown at a speed of 100m/s and weighing 500g. **

**Solution: **

Given: M = 500g and V = 100 m/s

Momentum is given by,

p = MV

Plugging in the values in the formula

p = MV

⇒p = (500)(100)

⇒p = 50000 gm/s

⇒p = 5 × 10

^{4}gm/s

**Question 2: Calculate the momentum of a ball thrown at a speed of 10m/s and weighing 40g. **

**Solution: **

Given: M = 40g and V = 10 m/s

Momentum is given by,

p = MV

Plugging in the values in the formula

p = MV

⇒p = (40)(10)

⇒p = 400 gm/s

⇒p = 4 × 10

^{2}gm/s

**Question 3: Suppose two balls with mass 5Kg and 2Kg are moving in the same direction at 6 m/s and 2 m/s respectively. They collide, and after the collision, the 5Kg ball is moving at a speed of 5 m/s. What is the speed of the 2Kg ball?**

**Solution: **

Given: m

_{1}= 5000g and m_{2}= 2000gInitial Velocities: v

_{1i }= 6 m/s and v_{2i}= 2 m/sFinal Velocities: v

_{1f}= 5 m/s and v_{2f }According to the conservation of momentum law.

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}Plugging values into this equation,

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}⇒ (5000)(6) + (2000)(2) = (5000)(5) + 2000(v

_{2f})⇒ 1000 + 4000 = 2000(v

_{2f})⇒ 5000 = 2000(v

_{2f})⇒ 2.5m/s = v

_{2f}

**Question 4: Suppose two balls with mass 10Kg and 1Kg are moving in the same direction at 9 m/s and 3 m/s respectively. They collide, and after the collision, the 10Kg ball is moving at a speed of 7 m/s. What is the speed of the 1Kg ball?**

**Solution: **

Given: m

_{1}= 10000g and m_{2}= 1000gInitial Velocities: v

_{1i }= 9 m/s and v_{2i}= 3 m/sFinal Velocities: v

_{1f}= 7 m/s and v_{2f }According to the conservation of momentum law.

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}Plugging values into this equation,

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}⇒ (10000)(9) + (1000)(3) = (10000)(7) + 1000(v

_{2f})⇒ 2000 + 3000 = 1000(v

_{2f})⇒ 5000 = 1000(v

_{2f})⇒ 5 m/s = v

_{2f}

**Question 5: Suppose two balls with mass 10Kg and 5Kg are moving in the opposite direction at 9 m/s and 3 m/s respectively. They collide, and after the collision, the 10Kg ball is moving at a speed of 3 m/s. What is the speed of the 3Kg ball?**

**Solution: **

Given: m

_{1}= 10000g and m_{2}= 5000gInitial Velocities: v

_{1i }= 9 m/s and v_{2i}= -3 m/sFinal Velocities: v

_{1f}= 3 m/s and v_{2f }According to the conservation of momentum law.

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}Plugging values into this equation,

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}⇒ (10000)(9) – (1000)(3) = (10000)(3) + 1000(v

_{2f})⇒ 6000 – 3000 = 1000(v

_{2f})⇒ 3000 = 1000(v

_{2f})⇒ 3 m/s = v

_{2f}

**Question 6: Consider a cannon that weighs 500Kg. It fires cannon at the speed of 200m/s. The weight of the cannon is 2Kg. Find the speed of recoil for the cannon. **

**Solution: **

Given: m

_{1}= 500Kg and m_{2}= 2KgInitial Velocities: v

_{1i }= 0 m/s and v_{2i}= 0 m/sFinal Velocities: v

_{1f}and v_{2f}= 200m/s_{ }This is a classic problem which uses conservation of momentum law.

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}Plugging values into this equation,

m

_{1}v_{1i}+ m_{2}v_{2i}= m_{1}v_{1f}+ m_{2}v_{2f}⇒ (500)(0) + (2)(0) = (500)(v

_{1f}) + 2(200)⇒ – 400 = 500(v

_{2f})⇒ = v

_{2f}⇒- 0.8m/s = v

_{2f}The cannon gun will recoil at a speed of 0.8m/s after firing the cannon.