Principle of Conservation of Momentum

Posted by Sek | 12:32 AM | | 1 comments »

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 1 - Both Object 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?

m1 = 600kg
m2 = 800kg
u1 = 40 ms-1
u2 = 20 ms-1
v1 = ?
v2 = 30 ms-1

According to the principle of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2
(600)(40) + (800)(20) = (600)v1 + (800)(30)
40000 = 600v1 + 24000
600v1 = 16000
v1 = 26.67 ms-1

Example 2 - 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.

m1 = 0.5 kg
m2 = 1.0 kg
u1 = 6.0 ms-1
u2 = -12.0 ms-1
v1 = -14.0 ms-1
v2 = ?

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

According to the principle of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2
(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v2
-9 = - 7 + 1v2
v2 = -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 :m1v1 + m2v2 From the law of conservation of momentum:Total Momentum Before collision = Total Momentum after collision0 = m1v1 + m2v2m1v1 = - m2v2(-ve sign means opposite direction)

Examples or Application of Conservation of Momentum in Explosion
1. Fire a pistol or rifle
2. Launching a rocket
3. Application in jet engine
4. Fan boat

Example 3
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?

This is a typical question of explosion.

m1 = 2.5 kg
m2 = 0.01 kg
u1 = 0 ms-1
u2 = 0 ms-1
v1 = ?
v2 = 250 ms-1

By using the equation of conservation of momentum principle
0 = m1v1 + m2v2
0 = (2.5)v1 + (0.01)(250)
(2.5)v1 = -2.5v1 = -1 ms-1