The Law of Conservation of Momentum and Examples of Problems

The Law of Conservation of Momentum and Examples of Problems – In the mid 1600s there were three scientists who investigated a Descartes assumption about the law of the conservation of motion. The three scientists are Wren, John Willis, and Huygens along with The Royal Society. The three succeeded in discovering the rules of physics called the Law of Conservation of Momentum. According to them, the total momentum of a system is always conserved. In addition, the understanding of momentum based on the three scientists is a vector quantity derived from the product of the object’s velocity with the object’s mass itself. There is also a reference to the rule of conservation of momentum which is often used as a high school exam question.

We can also observe this law in objects that collide with each other. For example, two marbles collide with each other. The marbles will bounce a few centimeters after the collision. This is because the two marbles have a certain speed. Then after the collision the two will have a constant velocity if the collision is perfectly elastic. This can happen because they do not lose energy. Now on this occasion I will explain about the rules of conservation of momentum along with references to the rules of conservation of momentum. For more details you can see below.

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The Law of Conservation of Momentum and Examples of Problems

The Law of Conservation of Momentum reads: “The total momentum of a system before the collision is equal to the total momentum after the collision, unless there is an external force.”. The external force that has a strong influence on the law of conservation of momentum is the frictional force. For example, two marbles colliding with the same speed will not experience the law of conservation of momentum if they do not hit the ground. This is because the frictional force is strong against these rules. The frictional force of the marbles on the ground is quite large, but the frictional force of the marbles on the ceramic/slick floor is small. The more aggressive the friction field, the greater the friction force.

The Law of Conservation of Momentum and Example Problems The Law of Conservation of Momentum and Example Problems

Below is a formula for the conservation of momentum that is used to make reference to the law of conservation of momentum. Here’s the formula:

The Law of Conservation of Momentum and Example Problems The Law of Conservation of Momentum and Example Problems

Information :
m1 = mass of the first object (kg)
m2 = mass of the second object (kg)
v1 = velocity of the first object before the collision (m/s)
v’1 = velocity of the first object after the collision (m/s)
v2 = velocity of the second object before the collision (m/s)
v’2 = velocity of the second object after the collision (m/s)

Example of the Law of Conservation of Momentum

1. A child is skateboarding with a speed of 6 m/s and the mass of the skate board is 4 kg. If the child has a mass of 20 kg. What is the speed of the skate board after the child jumps forward at 2 m/s?

Discussion
Given: m1 = 25 kg; v1 = v2 = 6 m/s; m2 = 4 kg; v’1 = 2 m/s
Asked: v’2 = ?
Answer:
m1 v1+ m2 v2 = m1 v’1 + m2 v’2
(25 x 6) + (4 x 6) = (25 x 2) + (4 x v’2)
150 + 24 = 50 + 4v’2
174 = 50 + 4v’2
4v’2 = 174 – 50
4v’2 = 124
v’2 = 124/4
v’2 = 31 m/s

2. A bullet having a mass of 10 grams is fired at the firing point. If the bullet is moving with a speed of 100 m/s and the mass of the gun is 1.5 kg. What is the recoil speed of the gun after opening fire?

Discussion
Given: mp = 10 g = 0.01 kg; vp = vs = 0 m/s; vp’ = 100 m/s’ ms = 1.5 kg
Asked : vs’ = ?
Answer:
mp vp+ ms vs = mp vp’ + ms vs’
(0.01 x 0) + (1.5 x 0) = (0.01 x 100) + (1.5 x vs’)
0 + 0 = 1 +1.5 vs’
-1.5 vs’ = 1
vs’ = -0.67 m/s
The sign (-) above means the gun is moving backwards.

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Thus a clarification of the conservation of momentum rule along with a reference to the law of conservation of momentum. Hopefully this article can add to your insight. Thank you.