 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 xaxis along the east and the yaxis along the north to write the momentum in components form for each object
object 1 moving along the (north) yaxis: p1 = m1 v1 = m1 (0 , v1) = (0 , m1 v1)
object 2 moving along the (east) xaxis:: 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 xaxis 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
