Linear Momentum

CSEC Physics Syllabus - Effective for examinations from May - June 2015

Section A - Mechanics

Dynamics: Motion in a Straight Line

Newton's Laws

Specific Objective 4.6

define linear momentum;

Specific Objective 4.7

describe situations that demonstrate the law of conservation of linear momentum;

Specific Objective 4.8

apply the law of conservation of linear momentum.

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Example 1

Calculate the momentum of an object if:

a)    its mass is 4.0 kg and its velocity is 8.0 m/s

b)    its mass is 500 g and its velocity is 3.0 km/s

c)    a force of 20 N is applied to it for 6.0 s and it moves from rest

d)    its mass is 2.0 kg and it falls from rest for 10 s (assume g = 10 m/s² or 10 N/kg).

    Solution

Momentum = Mass × Velocity

ρ = m  × v

a)    mass = 4.0 kg                    velocity = 8.0 m/s

ρ = m  × v

ρ = 4.0 kg × 8.0 m/s

⸫ ρ = 32 kgm/s 

b)    mass = 500 g                    velocity = 3.0 km/s

We must convert the object's mass and velocity:

500 g to kg 

1 000 g = 1 kg

1 g = 0.001 kg

⸫ 500 g = 0.5 kg

Recall kilo (k) = 1 000

3.0 km/s = 3.0 × 1 000 m/s

3.0 km/s = 3 000 m/s

Therefore,

ρ = m  × v

ρ = 0.5 kg × 3 000 m/s

⸫ ρ = 1 500 kgm/s 

c)    Using the relation of Newton's second law and Momentum:

Force × time = Change in Momentum

F × t = (m × v) - (m × u)

where, 

Force = F

time = t

mass = m

v = Final Velocity 

u = Initial Velocity

F = 20 N                      t = 6.0 s                     u = 0

F × t = (m × v) - (m × u)

20 N × 6.0 s = (m × v) - (m × 0)

20 N × 6.0 s = (m × v) 

120 Ns = m × v

Recall Momentum, ρ =  m × v

⸫ ρ = 120 Ns

d)    Using the relation of Newton's second law and Momentum:

Force × time = Change in Momentum

F × t = (m × v) - (m × u)

m = 2.0 kg                      t = 10 s                     u = 0                    g = 10 N/kg

The Force applied to the object is its weight, 

Weight = mass × gravity

W = 2.0 kg × 10 N/kg

⸫ W = 20 N

F × t = (m × v) - (m × u)

20 N × 10 s = (m × v) - (m × 0)

20 N × 10 s = (m × v) 

200 Ns = m × v

Recall Momentum, ρ =  m × v

⸫ ρ = 200 Ns

Example 2

If a constant force acts for 5.0 s on an object of mass 20 kg and increases its momentum by 30 Ns, calculate:

a)    the magnitude of the force

b)    the acceleration of the object. 

    Solution

a)    Using the relation of Newton's second law and Momentum:

Force × time = Change in Momentum

F × 5.0 s = 30 Ns

F = 30 Ns ÷ 5.0 s

⸫ F = 6 N

b)    Using Newton's second law:

Force = Mass × Acceleration

F = m × a

6.0 N = 20 kg × a

a = 6.0 N ÷ 20 kg

⸫ a = 0.3 N/kg

Example 3

A railway truck traveling along a level track at 9 m/s collides with and becomes coupled to a stationary truck. Find the velocity of the coupled trucks immediately after the collision if the stationary truck has a mass which is:

a)    equal to the mass of the moving truck

b)    twice the mass of the moving truck.

    Solution

a)    Using the Law of Conservation of Linear Momentum: 

Total Momentum before Collision = Total Momentum after Collision

Momentum of railway truck in motion + Momentum of stationary truck = Momentum of coupled trucks

(m₁ × u₁) + (m₂ × u₂) = (m₁ + m₂) × v

Recall the mass of the stationary truck is equal to the mass of the moving truck,

(m₁ × 9 m/s) + (m₁ × 0) = (m₁ + m₁) × v

(m₁ × 9 m/s) = 2m₁ × v

v = (m₁ × 9 m/s) ÷ 2m₁

 ⸫ v = 4.5 m/s

b)    Using the Law of Conservation of Linear Momentum: 

Total Momentum before Collision = Total Momentum after Collision

Momentum of railway truck in motion + Momentum of stationary truck = Momentum of coupled trucks

(m₁ × u₁) + (m₂ × u₂) = (m₁ + m₂) × v

Recall the mass of the stationary truck is twice the mass of the moving truck,

(m₁ × 9 m/s) + (2m₁ × 0) = (m₁ + 2m₁) × v

(m₁ × 9 m/s) = 3m₁ × v

v = (m₁ × 9 m/s) ÷ 3m₁

 ⸫ v = 3.0 m/s

Example 4

Two planes of the same mass collide head-on and become tangled so that they move on together. If the engines of both were stopped at the moment of impact and the speeds of the plane at impact were 120 m/s and 200 m/s, find the joint velocity immediately after the collision. 

    Solution

Using the Law of Conservation of Linear Momentum: 

Total Momentum before Collision = Total Momentum after Collision

Momentum of left moving plane + Momentum of right moving plane = Momentum of tangled planes

(m₁ × u₁) + (m₂ × u₂) = (m₁ + m₂) × v

(m₁ × 200 m/s) + (m₂ × 120 m/s) = (m₁ + m₂) × v

Recall the planes have the same mass,

(m₁ × 200 m/s) + (m₁ × 120 m/s) = (m₁ + m₁) × v

(m₁ × 200 m/s) + (m₁ × 120 m/s) = 2m₁ × v

We can factor m₁ from the left side of the equation,

m₁ × (200 m/s + 120 m/s) = 2m₁ × v

m₁ × 320 m/s = 2m₁ × v

v = (m₁ × 320 m/s) ÷ 2m₁

 ⸫ v = 160 m/s