## Momentum in Physics

The linear momentum p of an object of mass m and velocity v is given by

p = m v

where v is the velocity in m/s, m the mass in Kg and p the momentum in Kg m/s.

Since m (mass) is a scalar and v (velocity) is a vector, the momentum p is a vector quantity and p and v have the same direction.

### Example 1

What is the momentum, in Kg m/s of a car of mass 1 ton moving at the following velocities

a) v1 = 18 km/hr to the north

b) v2 = 72 km/hr to the south

c) v3 = 90 km/hr to the west

__Solution to Example 1__

a) p1 = m v1 = 1 ton × 18 km/hr = 1000 Kg 18 000 m / (3600 s) = 5000 Kg m/s to the north

b) p2 = m v2 = 1 ton × 72 km/hr = 1000 Kg 72 000 m / (3600 s) = 20,000 Kg m/s to the south

c) p3 = m v3 = 1 ton × 90 km/hr = 1000 Kg 90 000 m / (3600 s) = 25,000 Kg m/s to the west

## Momentum of a System with More Than one Object

For a system with several objects of masses m1, m2, m3 etc. and corresponding velocities v1, v2, v3 etc., the total momentum p of the system is given by

p = m1 v1 + m2 v2 + m3 v3 +...

with m v1, m v2, m v3, ... being vector quantities so that p is a sum of vectors.

### Example 2

Two objects O1 and O2 with O1 moving to the north and O2 moving to the east, having masses m1 = 0.6 Kg and m2 = 1 Kg and velocities of magnitude |V1| = 10 m/s and |v2| = 8 m/s respectively. Find the magnitude and direction of the total momentum of the two objects.

__Solution to Example 2__

Let us use a system of rectangular axis with the x-axis along the east and the y-axis along the north to write the momentum in components form for each object

object 1 moving along the (north) y-axis: p1 = m1 v1 = m1 (0 , |v1|) = (0 , m1 |v1|)

object 2 moving along the (east) x-axis:: p2 = m2 v2 = m2 (0 , |v2|) = (m2 |v2| , 0)

the total momentum p of the two objects is the sum of vectors p1 and p2

p = p1 + p2 = (px , py) = (0 , m1 |v1|) + (m2 |v2| , 0) = (m2 |v2| , m1 |v1|) = (1 × 8 , 0.6 × 10) = (8 , 6)

magnitude: |p| = √ (8^{2} + 6^{2}) = 10 Kg m/s

Direction: Let θ be the angle made by p and the x-axis in the counter clockwise direction. Using the x and y components px and py of p, we have

tan θ = py / px = 6 / 8

θ = arctan(6 / 8) ≈ 36.9 °

Momentum p makes and angle of approximately 36.9 ° from east to north.

## Momentum and Kinetic Energy of an Object

We now quantify the relationship between the momentum p and the kinetic energy K of an object.

K = (1/2) m |v|^{2}

p = m v gives |v| = |p| / m

Substitute |v| by |p| / m in K to obtain

K = (1/2) m ( |p| / m) ^{2} = (1/2) |p| ^{2} / m

The above may also be solved for |p| to obtain

|p| = √(2 m K)

### Example 3

Two object A and B of masses m1 and m2 have equal momentums. Which of the two objects have the highest kinetic energy if m1 < m2?

__Solution to Example 3__

p1 = m1 v1 = √(2 m1 K1)

p2 = m2 v2 = √(2 m2 K2)

p1 = p2 , equal momentum

hence

√(2 m1 K1) = √(2 m2 K2)

Square bothe sides

2 m1 K1 = 2 m2 K2

divide both sides by 2 k1 m2 and simplify

m1 / m2 = K2 / K1

m1 < m2 means m1 / m2 < 1

K1 / K2 < 1

K2 < K1

The lighter object with mass m1 has higher kinetic energy.

## More References

1 - Higher Level Physics - IB Diploma - Chris Hamper - Pearson

2 - Physics - Raymond A. Serway and Jerry S. Faughn

Holt, Reinehart and Winston - Harcourt Education Company