Linear Momentum

Momentum in Physics

The linear momentum \( \vec{p} \) of an object of mass \( m \) and velocity \( \vec{v} \) is given by:

\vec{p} = m \vec{v}

where \( \vec{v} \) is velocity in m/s, \( m \) is mass in kg, and \( \vec{p} \) is momentum in kg·m/s. Since mass is a scalar and velocity is a vector, momentum is a vector quantity sharing the same direction as velocity.

Example 1

Find the momentum (in kg·m/s) of a 1-ton car moving at:

a) \( v_1 = 18 \text{ km/hr} \) north

b) \( v_2 = 72 \text{ km/hr} \) south

c) \( v_3 = 90 \text{ km/hr} \) west

Solution

a) \( p_1 = m v_1 = 1000 \, \text{kg} \times \frac{18000 \, \text{m}}{3600 \, \text{s}} = 5000 \, \; \mathrm{kg\cdot m/s} \) north
b) \( p_2 = m v_2 = 1000 \, \text{kg} \times \frac{72000 \, \text{m}}{3600 \, \text{s}} = 20000 \, \; \mathrm{kg\cdot m/s} \) south
c) \( p_3 = m v_3 = 1000 \, \text{kg} \times \frac{90000 \, \text{m}}{3600 \, \text{s}} = 25000 \, \; \mathrm{kg\cdot m/s} \) west

Momentum of a Multi-Object System

For a system of objects with masses \( m_1, m_2, m_3, \dots \) and velocities \( \vec{v}_1, \vec{v}_2, \vec{v}_3, \dots \), the total momentum is the vector sum:

\vec{p}_{\text{total}} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 + \cdots

Example 2

Object O₁ (mass \( m_1 = 0.6 \, \text{kg} \)) moves north at \( |\vec{v}_1| = 10 \, \text{m/s} \).
Object O₂ (mass \( m_2 = 1 \, \text{kg} \)) moves east at \( |\vec{v}_2| = 8 \, \text{m/s} \).
Find the magnitude and direction of the total momentum.

Solution

Using coordinate axes (x: east, y: north):
\( \vec{p}_1 = m_1 \vec{v}_1 = (0, 0.6 \times 10) = (0, 6) \, \; \mathrm{kg\cdot m/s} \)
\( \vec{p}_2 = m_2 \vec{v}_2 = (1 \times 8, 0) = (8, 0) \, \; \mathrm{kg\cdot m/s} \)
Total: \( \vec{p} = \vec{p}_1 + \vec{p}_2 = (8, 6) \, \; \mathrm{kg\cdot m/s} \)

Magnitude: \( |\vec{p}| = \sqrt{8^2 + 6^2} = 10 \, \; \mathrm{kg\cdot m/s} \)
Direction: \( \theta = \arctan\left(\frac{6}{8}\right) \approx 36.9^\circ \) from east toward north.

Momentum and Kinetic Energy

The relationship between kinetic energy \( K \) and momentum magnitude \( |\vec{p}| \) is:

K = \frac{1}{2} m |\vec{v}|^2 = \frac{|\vec{p}|^2}{2m} \quad \text{and} \quad |\vec{p}| = \sqrt{2mK}

Example 3

Two objects A and B with masses \( m_1 \) and \( m_2 \) (\( m_1 < m_2 \)) have equal momenta.
Which has greater kinetic energy?

Solution

Given \( |\vec{p}_1| = |\vec{p}_2| \):
\( \sqrt{2 m_1 K_1} = \sqrt{2 m_2 K_2} \)
Squaring: \( 2 m_1 K_1 = 2 m_2 K_2 \)
Thus \( \frac{K_1}{K_2} = \frac{m_2}{m_1} > 1 \) since \( m_2 > m_1 \).
Therefore, the lighter object (mass \( m_1 \)) has higher kinetic energy.

References

1. Higher Level Physics - IB Diploma - Chris Hamper, Pearson
2. Physics - Raymond A. Serway and Jerry S. Faughn, Holt, Rinehart and Winston