### Principle of Conservation of Momentum

The principle of conservation of momentum states that in a system make out of objects that react (collide or explode), the total momentum is constant if no external force is acted upon the system.

Sum of Momentum Before Reaction

= Sum of Momentum After Reaction

### Formula

Example – Both Objects are in the Same Direction before Collision

A Car A of mass 600 kg moving at 40 ms^{-1} collides with a car B of mass 800 kg moving at 20 ms^{-1} in the same direction. If car B moves forwards at 30 ms^{-1} by the impact, what is the velocity, v, of the car A immediately after the crash?**Answer**:

m_{1} = 600kg

m_{2} = 800kg

u_{1} = 40 ms^{-1}

u_{2} = 20 ms^{-1}

v_{1} = ?

v_{2} = 30 ms^{-1}

According to the principle of conservation of momentum,

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

(600)(40) + (800)(20) = (600)v_{1} + (800)(30)

40000 = 600v_{1} + 24000

600v_{1} = 16000

v_{1} = 26.67 ms^{-1}

Example – Both Object are in opposite direction Before Collision

A 0.50kg ball traveling at 6.0 ms^{-1} collides head-on with a 1.0 kg ball moving in the opposite direction at a speed of 12.0 ms^{-1}. The 0.50kg ball moves backward at 14.0 ms^{-1} after the collision. Find the velocity of the second ball after collision.

**Answer**:

m_{1} = 0.5 kg

m_{2} = 1.0 kg

u_{1} = 6.0 ms^{-1}

u_{2} = -12.0 ms^{-1}

v_{1} = -14.0 ms^{-1}

v_{2} = ?

(IMPORTANT: velocity is negative when the object move in opposite direction)

According to the principle of conservation of momentum,

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v_{2}

-9 = – 7 + 1v_{2}

v_{2} = -2 ms^{-1}

### Explosion

Before explosion both object stick together and at rest. | After collision, both object move at opposite direction. |

Total Momentum before collision Is zero | Total Momentum after collision : m |

From the law of conservation of momentum: Total Momentum Before collision = Total Momentum after collision 0 = m m (-ve sign means opposite direction) |

**Example**:

A man fires a rifle which has mass of 2.5 kg. If the mass of the bullet is 10 g and it reaches a velocity of 250 m/s after shooting, what is the recoil velocity of the pistol?**Answer**:

This is a typical question of explosion.

m_{1} = 2.5 kg

m_{2} = 0.01 kg

u_{1} = 0 ms^{-1}

u_{2} = 0 ms^{-1}

v_{1} = ?

v_{2} = 250 ms^{-1}

By using the equation of conservation of momentum principle

0 = m_{1}v_{1} + m_{2}v_{2}

0 = (2.5)v_{1} + (0.01)(250)

(2.5)v_{1} = -2.5v_{1} = -1 ms^{-1}

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