## Tuesday, November 10, 2015

### Conservation of Linear Momentum Applications Problems and Solutions

Conservation of linear momentum

Linear momentum is a physical quantity which is mathematically defined as the product of mass and velocity of the body. It actually helps to understand that, a given body is able to transfer how much of the kinetic energy to the another body during the interaction. So linear momentum is a physical quantity which is helpful in understanding the ability of the body to transfer kinetic energy to the other bodies.

According to law conservation of linear momentum, if no external forces acting on a system, the linear momentum of a system always remains constant. This can be explained and derived basing on the Newton’s second law of motion. We know that according to Newton’s second law, force is defined as rate of change of momentum. If that external force is equal to 0, automatically rate of change of momentum will be equal to 0. That means change in the momentum is equal to 0 and it implies that the momentum of the system remains constant. This is called law conservation of linear momentum and it is valid in all situations. In fact linear momentum is a physical quantity that conserves in all the physical situations respect to of the any of the other conditions and the only  condition that it need is   there shall not be any external force acting on the system.

Recoil of a gun

If you bullet is fired from the gun, we know that the gun recoils in the opposite direction. This can be explained basing on the law conservation of linear momentum. In this case it is very clear that there is no external force acting on the gun and bullet system. We shall not consider the small force that you apply on the trigger as a significant external force which is making the bullet moving with a very higher velocity. It is only a triggering force and it is not causing actually that much of the velocity. The mechanism of the gun is giving its velocity. So we can assume that there is no external force acting on the system and hence linear momentum of the system has to remain constant.

Initially the entire system of bullet and gun is in the state of rest and hence its initial momentum is equal to 0. When the bullet motion the forward direction to keep the momentum of the system equal to 0 the gun motion the backward direction and it concerns the linear momentum as shown below.

Problem

A dog of mass 5 kg is standing on a flat boat so that he’s 10 m away from the shore. He walks 4 m towards the shore on the boat and then stops. If the boat mass is 20 kg and assuming there is no friction, what shall be the distance of the dog from the shore after it has travelled that distance?

Solution

Here dog and the boat can be treated as a system and very clearly there is no external force acting on the system. As there is no external force acting on the system linear momentum of the system is constant. Initially both dog and boat are in the state of rest therefore initial momentum of the system is equal to 0. Hence final momentum of the dog in forward direction is equal to the final momentum of the dog and boat in the reverse direction. There are equal in magnitude and there are only opposite indirection.

Basing on this we can prove that when the dog moves a distance of four meters in the forward direction, the system of the boat and dog moves in the backward direction by a distance of 0.8 m. As a result is not at a distance of 6 m from the shore, but as the boat has taken him back by a distance of 0.8 m therefore he’s at a distance of 6.8 m from the shore.

Explosion of a bomb into two pieces

If you bomb of certain mass who is in the rest state explodes into two pieces, this two pieces travels in opposite direction is and there were last is will be in satisfaction with the law conservation of linear momentum. We can find out the velocity of each part with respect to another part and we can also talk about the energy of the each piece respect to the total energy of the system as shown below.

Explosion of you bomb into three pieces

As the bomb is initially in the state of rest and later explodes into three pieces and there is no external forces acting on the system, the momentum of the system is conserved. If suppose the two pieces travels in perpendicular directions, then the third piece will travel in a resultant opposite direction and it can be derived as shown in.

Problem

A shell is fired from the gun with then known speed with an angle with respect to the horizontal. It explodes into three pieces of equal masses and the maximum height of the trajectory. One piece falls down vertically whereas the other piece retraces its path. Then what is the speed of the third piece?

Solution

It is very clear from the given situation that there is no external force acting on the system and hence linear momentum is conserved.

By the time the body reaches its maximum height it will have certain velocity and hence it will have certain initial momentum also. According to projectile motion we know that at the maximum height of the projectile, there will be only a horizontal component of velocity. As it has reached the maximum height along the vertical direction, velocity in that direction becomes zero. Hence it has momentum only along the X direction.

After the explosion one piece falls in the vertically downward direction, so it is not having any momentum along the x-axis. The second piece retraces its path therefore it will have the same velocity of the original body but in the opposite direction. By equating the momentum is of the system before and after the explosion we can find the velocity of the herpes as shown below.

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