momentum

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Dynamics is a subset of mechanics, which is the study of motion. Whereas kinetics studies that motion itself, dynamics is concerned about the CAUSES of motion. In particular, it involves the concepts of force, momentum and energy. This video gives an overview of what dynamics is, and is useful as an intro to its study, or as a review

Momentum is intricately connected to the concept of force. In fact in Newton's Principia he described his second law of motion in terms of momentum.
In classical physics the concept of momentum is relatively straightforward. It's simply an object mass times its velocity.

momentum = mass x velocity
p = m x v
unit : kg.m/s

So for example a heavy moving object will have a large momentum due to its mass. But an object can also have a large momentum if it's velocity is particularly high. So for example a bullet can have a relatively large momentum by way of its velocity.

A Change in Momentum
The analysis of motion is more concerned why an object's motion changes,. In other words, it has a change in velocity, it accelerates
As a result, there is a change in momentum
A change in momentum is called its impulse.

Watch the video as I explain using the 'eggsample' of an egg falling
(get it), as well as cars safety, the concepts of momentum and impulse.

Check your understanding

Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.

Practice Problems

1. Calculate the magnitude of the momentum (p = mv) in the following cases:

(a) A proton (mass = 1.67 × 10^-27 kg) moving at 5 × 10⁶ m/s
(b) A 15 g bullet moving at 500 m/s
(c) A 75 kg sprinter running at 12 m/s
(d) The Earth (mass = 5.98 × 10²⁴ kg) moving at an orbital speed of 2.98 × 10⁴ m/s

(a) Proton: p = (1.67 × 10^-27 kg) × (5 × 10⁶ m/s) = 8.35 × 10^-21 kg·m/s (or N·s)

(b) Bullet: Convert 15 g = 0.015 kg → p = 0.015 × 500 = 7.5 kg·m/s

(c) Sprinter: p = 75 kg × 12 m/s = 900 kg·m/s

(d) Earth: p = (5.98 × 10²⁴ kg) × (2.98 × 10⁴ m/s) = 1.78 × 10²⁹ kg·m/s

2. An 18,000 kg van is moving at a speed of 15 m/s. If Superman is to stop the van in 0.5 s, what average force must he exert on the van?

Given: m = 18,000 kg, v_i = 15 m/s, v_f = 0 m/s, Δt = 0.5 s

Step 1 (Calculate Impulse / Change in Momentum): Δp = m(v_f - v_i) = 18,000 × (0 - 15) = -270,000 kg·m/s

Step 2 (Apply Impulse-Momentum Theorem): F_avg × Δt = Δp → F_avg = -270,000 / 0.5

Answer: -540,000 N (540,000 N opposite to motion)

3. A 0.15-kg baseball is thrown at a speed of 20 m/s. It is hit straight back at the pitcher with a final speed of 22 m/s.

(a) What impulse is delivered to the ball?
(b) Find the average force exerted by the bat on the ball if the two are in contact for 2 × 10^-3 s.

Given: m = 0.15 kg, v_i = -20 m/s (towards batter), v_f = +22 m/s (towards pitcher)

(a) Impulse: J = Δp = m(v_f - v_i) = 0.15 × (22 - (-20)) = 0.15 × 42 = 6.3 N·s (towards pitcher)

(b) Average Force: F_avg = J / Δt = 6.3 / (2 × 10^-3 s) = 3,150 N (towards pitcher)

Here another worksheet on impulse, includes answer - worked out solutions

CONSERVATION OF MOMENTUM

According to Newton's Third Law of Motion, whenever an object exerts a force on a second object, the second object exerts an equal and opposite force back on the first. Consequently, when two objects interact or collide, both experience equal and opposite changes in momentum.

When no external net forces act on a system, the total momentum remains constant:

Total Momentum Before Collision = Total Momentum After Collision

This fundamental principle is known as the Law of Conservation of Momentum.

In the following video, using the crash between a bus and a bee I show how we derive and understand Conservation of Momentum

Interactive
Now its an opportunity to play.
Using the animation from pHET, start by exploring LCM rectilinearly, that is, in one dimension. click the intro TAB.
Leave the check off on kinetic energy and keep the elasticity to 100%
adjust the initial velocities ands then observe both the resulting collision but also examine the vector diagrams in the momenta diagram in the bottom right.
You can also do this in the Explore 1D TAB. You will note that total momentum of the system remains constant.

Since momentum is a vector quantity, the same is true for two dimension situations.
Once you are confident in one dimension, explore two dimensions. Again , the total momentum of the system remains constant, irrespective of the individual momenta of each individual object

Sample problems
We now have three sample problems. The first in 1D, the second two are 2D
Have a go at each before watching video

Here are some more problems to try

1. A 40-kg child standing on a frozen pond throws a 0.5-kg stone to the east with a speed of 5 m/s. Find the recoil velocity of the child.

Given: m_child = 40 kg, m_stone = 0.5 kg, v_stone = +5 m/s (East)

Conservation of Momentum: p_initial = p_final = 0

0 = (m_child × v_child) + (m_stone × v_stone)

0 = (40 × v_child) + (0.5 × 5)

40 × v_child = -2.5 → v_child = -2.5 / 40 = -0.0625 m/s

Answer: 0.0625 m/s West

2. A rifle with a weight of 30 N fires a 5-g bullet with a speed of 300 m/s. Find the recoil speed of the rifle, if it is held loosely by the shooter.

Given: Weight of rifle W = 30 N → m_rifle = W / g = 30 / 9.8 ≈ 3.061 kg

Bullet mass m_bullet = 5 g = 0.005 kg, v_bullet = 300 m/s

Conservation of Momentum: p_initial = p_final = 0

0 = (m_rifle × v_rifle) + (m_bullet × v_bullet)

0 = (3.061 × v_rifle) + (0.005 × 300)

3.061 × v_rifle = -1.5 → v_rifle = -1.5 / 3.061 ≈ -0.49 m/s

Answer: 0.49 m/s (in opposite direction)

3. A 65-kg boy and a 40-kg girl, both wearing skates, face each other at rest on a skating rink. The boy pushes the girl, sending her to the east with a speed of 4 m/s. What is the velocity (speed and direction) of the boy?

Given: m_boy = 65 kg, m_girl = 40 kg, v_girl = +4 m/s (East)

Conservation of Momentum: p_initial = p_final = 0

0 = (m_boy × v_boy) + (m_girl × v_girl)

0 = (65 × v_boy) + (40 × 4)

65 × v_boy = -160 → v_boy = -160 / 65 ≈ -2.46 m/s

Answer: 2.46 m/s West

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Momentum

MOMENTUM

CONSERVATION OF MOMENTUM

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