**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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