# Physics Problems with Solutions # Linear Momentum

## What is a momentum?

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| = √ (82 + 62) = 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