Every where you notice the bodies moving with its own momentum. The ice skater, the gymnasts, the bullet getting fired. These all possess their own momentum. So how can you justify that their momentum is conserved or not? Lets see about it in this section. |

It states that the total momentum before collision is equal to the total momentum after the collision. It is a vector quantity and is always directed in the direction of velocity. The unit is kgm/s or Ns. It depends on mass and velocity of the body.

Take two bodies

**A**and**B**in an isolated system of masses**m**_{1}and**m**_{2 }with the initial velocities**u**_{1}and u_{2}collides each other attains final velocities v_{1}and v_{2}. ThusTotal momentum before collision = Total momentum after collision

Momentum before collision = m

_{1}u

_{1}+ m

_{2}u

_{2}

Momentum after collision = m

_{1}v

_{1}+ m

_{2}v

_{2}

So the Law of Conservation tells that m

_{1}u

_{1}+ m

_{2}u

_{2}= m

_{1}v

_{1}+ m

_{2}v

_{2}

_{1}and m

_{2}having initial velocities u

_{1}and u

_{2}. If these bodies collide for time t and their velocities change and attain v

_{1}and v

_{2}after collision.

From Newton's second law of motion we have

Rate of change of momentum of A, $\vec{F_A}$ = $\frac{\Delta \vec{P_1}}{t}$

Rate of change of momentum of B, $\vec{F_B}$ = $\frac{\Delta \vec{P_2}}{t}$

The Newtons third law is $\vec{F_B}$ = - $\vec{F_A}$

The force exerted by A on B is equal and opposite to the force exerted by B on A

$\frac{\Delta \vec{P_1}}{t}$ = - $\frac{\Delta \vec{P_2}}{t}$

The Momentum $\Delta{\vec{p_1}}$ = m

So m

(m

Thus total momentum before collision = total momentum after collision.

The vector sum of all the momenta for a closed system with no external force acting on it is a constant.

p

_{1}v_{1}- m_{1}u_{1}and $\Delta{\vec{p_2}}$ = m_{2}v_{2}- m_{2}u_{2}So m

_{1}v_{1}- m_{1}u_{1}= -(m_{2}v_{2}- m_{2}u_{2})(m

_{1}u_{1}+ m_{2}u_{2}) = (m_{1}v_{1}+ m_{2}v_{2})Thus total momentum before collision = total momentum after collision.

The vector sum of all the momenta for a closed system with no external force acting on it is a constant.

p

_{1}+ p_{2}+ p_{3}+ p_{4}+ . . . . .+ p_{i}= K, where K is a constant.The law of conservation of angular momentum states that in a closed
system the angular momentum always remains the same till no external torque is acts on the system. If a body of mass m is rotating with velocity v at a distance r from origin carrying momentum p. The angular momentum is given as

**L = mvr**

As momentum p = mv

L = rp

As momentum p = mv

L = rp

Example : During skating if the skater folds hands then he is able to have a greater rotational energy at a greater speed as compared to when he opens his hands wide. This tells about conservation of angular momentum.

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The linear momentum conservation is same as newtons first law. It tells that linear momentum will be conserved if no external force acts on it.

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For a particle the linear momentum is a vector quantity is defined as

$\vec{p}$ = m $\vec{v}$

The Newtons second law says that

$\vec{F}$ = m $vec{a}$ = m $\frac{d \vec{v}}{dt}$ = $\frac{d \vec{p}}{dt}$

$\vec{F}$ = $\vec{0}$

That implies that momentum p is a constant tells that momentum is conserved.

$\vec{p}$ = m $\vec{v}$

The Newtons second law says that

$\vec{F}$ = m $vec{a}$ = m $\frac{d \vec{v}}{dt}$ = $\frac{d \vec{p}}{dt}$

$\vec{F}$ = $\vec{0}$

That implies that momentum p is a constant tells that momentum is conserved.

The Newtons cradle is known for demonstrating the law of conservation of momentum shown in fig. Here you could see the n number of balls swinging in is equal to the n number of swinging out. This demonstrates the law of conservation of momentum through collision.

The trolley method can also be used to demonstrate the collisions in the law.

Here are some examples of conservation of momentum which illustrates you alot more about the how the law works:

**You could see the young**

**snooker aiming the yellow ball**

If the white ball collides the yellow one.The white ball loses the momentum whereas the yellow ball gains equal amount of momentum after collision. It tells that the total momentum before collision is same as that after collision.

**If you are at shoot with rifle or shotgun you are often told to hold the gun tightly against your shoul**

**der. Why is it so?**

when the bullet is fired it moves in the forward direction and the gun moves backward. It is due to the gaining of some momentum due to its velocity. To nullify this momentum gun moves takes a backward push such that it has same momentum magnitude to that of momentum of bullet with opposite direction.

**Here are some sample problems on conservation of momentum you can go through it:**

### Solved Examples

**Question 1:**A bomb of mass 8 kg initially at rest explodes into two fragments of masses 3 kg and 5 kg. If 3 kg mass moves with a velocity of 4 ms

^{-1}, what will be the velocity of 5 kg mass?

**Solution:**

Given: mass m

Since the bomb is initially at rest u = 0, v

The law of conservation of momentum is

m

0 = 3 $\times$ 4 + 5 v

0 = 12 + 5v

v

5 kg mass moves in the direction opposite to 3 kg mass. Hence v

_{1}= 3 kg and mass m_{2}= 5 kgSince the bomb is initially at rest u = 0, v

_{1}= 4 ms^{-1}and v_{2}=?The law of conservation of momentum is

m

_{1}u_{1 }+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}0 = 3 $\times$ 4 + 5 v

_{2}0 = 12 + 5v

_{2}v

_{2}= - $\frac{12}{5}$ = - 2.4 ms^{-1}5 kg mass moves in the direction opposite to 3 kg mass. Hence v

_{2}is negative.**Question 2:**An archer fires a 1 kg of arrow at 40 m/s. If both archer and bow weighs 70 kg. What will be the velocity of archer after firing the arrow?

**Solution:**

Given: Mass m

_{1}= 70 kg, mass m_{2}= 1 kg, Initial velocity of archer u_{1}= 0, initial velocity of arrow u_{2}= 0, Final velocity of archer v_{1}= 40 m/s, Final velocity of arrow = ?Momentum before collision p

m

0 = 70 kg $\times$ 40 m/s + 1 kg $\times$ v

Final velocity v

_{i }= Momentum after collision p_{f}m

_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}0 = 70 kg $\times$ 40 m/s + 1 kg $\times$ v

_{2}Final velocity v

_{2}= - 2800 m/s.