Linear Momentum
Newton’s Laws of Motion
Newton’s First Law (Law of Inertia)
A body at rest will remain at rest, and a body in motion will continue moving with constant velocity unless acted upon by an external force. The tendency of a body to resist changes in its motion is called inertia.
Newton’s Second Law
The rate of change of momentum of an object is directly proportional to the applied force and occurs in the direction of the force.
Newton’s Third Law
For every action, there is an equal and opposite reaction.
Mathematically: Fa = -Fb, where:
- Fa is the action force
- Fb is the reaction force
Linear Momentum and Impulse
Linear Momentum
The momentum (p) of an object is the product of its mass (m) and velocity (v).
Formula: $$ p = mv $$
Momentum is a vector quantity measured in kg·m/s.
Example 1:
Find the momentum of a 2500 kg truck traveling at 30 m/s eastward.
Solution:
p = 2500 × 30 = 75,000 kg·m/s
Example 2:
A boy (20 kg) is riding a bike (6 kg) at 4 m/s northward. Find the total momentum.
Solution:
Total mass = 20 + 6 = 26 kg
p = 26 × 4 = 104 kg·m/s
Impulse
Impulse (I) is the product of force and the time over which the force acts.
Formula:
$$ I = Ft $$
It is a vector quantity measured in Ns.
Example 3:
A 0.65 kg football is thrown at 10 m/s and caught by a stationary receiver in 0.050 s. Find the force exerted on the receiver’s hand.
Solution:
$$ F = \frac{m \times v}{t} $$
F = (0.65 × 10) / 0.05 = 130 N
Conservation of Linear Momentum and Collision
The principle of conservation of linear momentum states that if no external forces act on a system of two colliding objects, the total momentum before collision equals the total momentum after collision.
Elastic and Inelastic Collisions
Elastic Collision
An elastic collision is one in which both momentum and kinetic energy are conserved. The colliding objects separate after impact.
For objects A and B:
- $$ m_A u_A + m_B u_B = m_A v_A + m_B v_B $$
- $$ \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 $$
Inelastic Collision
In an inelastic collision, momentum is conserved but kinetic energy is not. The colliding objects stick together after impact.
For objects A and B:
$$ m_A u_A + m_B u_B = V(m_A + m_B) $$
If they move in opposite directions before collision:
$$ m_A u_A - m_B u_B = V(m_A + m_B) $$
Example 4:
A 2.0 kg rubber ball moving at 8.0 m/s to the right collides elastically with another identical ball at rest. The first ball stops after the collision. Find the velocity of the second ball after the collision and verify by calculating kinetic energy before and after impact.
Solution:
Given: mA = 2.0 kg, uA = 8.0 m/s, mB = 2.0 kg, uB = 0 m/s, vA = 0 m/s, vB = ?
Using momentum conservation:
(2 × 8.0) + 0 = 0 + (2 × vB)
16.0 = 2vB
vB = 8.0 m/s
Kinetic Energy Before and After Collision
Before Collision:
K.Eb = ½ mAuA2 + ½ mBuB2
K.Eb = [½ × 2 × 8²] + [½ × 2 × 0²] = 64 J
After Collision:
K.Ea = ½ mAvA2 + ½ mBvB2
K.Ea = [½ × 2 × 0²] + [½ × 2 × 8²] = 64 J
Since kinetic energy remains the same, the collision is elastic.