# Momentum

Card 1:What is momentum?

## Momentum

Momentum is defined as the product of mass and velocity.

Formula: Unit: kgms-1

Type of quantity: Vector

Example 1
A student releases a ball with mass of 2 kg from a height of 5 m from the ground. What would be the momentum of the ball just before it hits the ground?

In order to find the momentum, we need to know the mass and the velocity of the ball right before it hits the ground.
It's given that the mass, m = 2kg.

The velocity is not given directly. However, we can determine the velocity, v, by using the linear equation of uniform acceleration.

This is a free falling motion,
The initial velocity, u = 0
The acceleration, a = gravirational acceleration, g = 10ms-2
The dispacement, s = high = 50m.
The final velocity = ?

From the equation
v2 = u2 + 2as
v2 = (0)2 + 2(10)(5)
v = 10ms-1

The momentum,
p = mv =(2)(10) = 20 kgms-1

Card 2: Principle of Conservation of Momentum

## 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

Card 3: Formula of Principle of Conservation of Momentum

## Formula Example 2: Both objects are in 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 3: Both objects 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 siredtion)

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

Card 4: What is elastic collision?

## Elastic Collision

Elastic collision is the collision where the kinetic energy is conserved after the collision.

Total Kinetic Energy before Collision
= Total Kinetic Energy after Collision

-In an elastic collision, the 2 objects seperated right after the collision, and
-the momentum is conserved after the collision.

Card 5: What is inelastic collision?

## Inelastic Collision

Inelastic collision is the collision where the kinetic energy is not conserved after the collision.

-In a perfectly elastic collision, the 2 objects attach together after the collision, and
-the momentum is also conserved after the collision.

Example 4: Perfectly Inelastic Collision
A lorry of mass 8000kg is moving with a velocity of 30 ms-1. The lorry is then accidentally collides with a car of mass 1500kg moving in the same direction with a velocity of 20 ms-1. After the collision, both the vehicles attach together and move with a speed of velocity v. Find the value of v.

(IMPORTANT: When 2 object attach together, they move with same speed.)

m1 = 8000kg
m2 = 1500kg
u1 = 30 ms-1
u2 = 20 ms-1
v1 = v
v2 = v

According to the principle of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2
(8,000)(30) + (1,500)(20) = (8,000)v+ (1,500)v
270,000 = 9500v
v = 28.42 ms-1

Card 6: Application of conservation of momentum 1 - Rocket

## Rocket

1. Mixture of hydrogen and oxygen fuels burn in the combustion chamber.
2. Hot gases are expelled through the exhausts at very high speed .
3. The high-speed hot gas produce a high momentum backwards.
4. By conservation of momentum, an equal and opposite momentum is produced and acted on the rocket, pushing the rocket upwards.
Card 7: Application of conservation of momentum 1 - Jet Engine

## Jet Engine

1. Air is taken in from the front and is compressed by the compressor.
2. Fuel is injected and burnt with the compressed air in the combustion chamber.
3. The hot gas is forced through the engine to turn the turbine blade, which turns the compressor.
4. High-speed hot gases are ejected from the back with high momentum.
5. This produces an equal and opposite momentum to push the jet plane forward.
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