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The Physics of Impact: Understanding Force in Different Scenarios

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Article Content
I’m writing this article to help explain momentum-impulse questions that deal karate chops, kicking a ball, or any other type of collision. It’s often misunderstood when and were we “follow through” or “snap back” after striking an object. This should clear it up.

Introduction

When it comes to applying force, context is everything.

Whether it’s a soccer player aiming for a goal, a karate master breaking a board, or the safety considerations in a car crash, the principles of physics govern the outcome.

The key variables are the amount of force applied, the duration of the impact, and the desired result.

While the impulse-momentum theorem remains a cornerstone in all scenarios, the application of this principle varies significantly depending on the goal.

After reading this, try to answer the question at the end and see if you get it right!

Soccer Ball: The Art of Follow-Through

In sports, the objective often involves maximizing the transfer of energy to the ball.

For a soccer player, this means imparting as much kinetic energy as possible, allowing the ball to travel a great distance or at a high speed.

The technique involves a follow-through after contact with the ball. This action ensures that the foot remains in contact with the ball for the longest possible time, extending the impact duration and thus increasing the impulse (the product of force and time). The longer the foot pushes against the ball, the more energy is transferred, resulting in a faster and further traveling ball.

The follow-through is essential here; it ensures that the force isn’t just a quick snap but a sustained push that gives the ball its velocity.

Follow Through – Backed by formulas

Let’s prove the follow thorugh method with some formulas.

Remember, when a soccer player kicks a ball, they’re applying a force over a period of time to give the ball as much speed as possible. The impulse imparted to the ball can be expressed by the formula:

[katex] \text{Impulse} = F \times \Delta t [/katex]

Where [katex] F [/katex] is the force applied, and [katex] \Delta t [/katex] is the time over which the force is applied. The impulse is also equal to the change in momentum ([katex]\Delta p [/katex]) of the ball:

[katex] F \times \Delta t = \Delta p = m \times \Delta v [/katex]

Here, [katex] m [/katex] is the mass of the soccer ball, and [katex] \Delta v [/katex] is the change in velocity of the ball. By following through, the player increases [katex] \Delta t [/katex], which increases the impulse and therefore the change in velocity, making the ball go faster and farther.

Karate Chop: The Sudden Break

Contrast this with the technique a karate master uses to break a wooden board.

The goal is not to send the board flying but to create a break.

The board’s structural integrity can only be overcome by a forceful, high-impact blow delivered in the shortest time frame. The snapping back of the hand is less about the physical retraction and more about the approach to the strike. It’s a biomechanical cue for the karate master to deliver a high-speed, explosive force.

The speed at the moment of impact, rather than the time over which the force is applied, is the critical factor. The snap-back motion signifies a maximization of the force at the moment of contact and ensures that the time of impact is minimal, which is essential for breaking the board.

Proof in Numbers

Let’s prove this with numbers.

Remember, for a karate master breaking a board, the focus is on applying a large amount of force in as short a time as possible. The force applied can be described as:

[katex] ​F = \frac{\Delta p}{\Delta t} [/katex]

In this scenario, [katex] \Delta t [/katex] is very small, which increases the force [katex] F [/katex] since momentum change [katex] \Delta p [/katex] (which equals the mass times the change in velocity of the hand) is relatively constant for a given technique. The smaller the time of impact, the larger the force applied to the board. The goal is to maximize [katex] F [/katex] by minimizing [katex] \Delta t [/katex], thus creating the necessary force to break the board.

Automotive Safety: Prolonging the Impact

In the context of automotive safety, the objective during a collision is to reduce the force experienced by the occupants. Here, extending the time over which a collision occurs is beneficial, as it reduces the force imparted. Car safety features, such as crumple zones and airbags, are designed to lengthen the duration of a person’s deceleration during a crash, thereby lowering the impact force. Unlike the soccer ball or the karate chop, the goal here is to absorb and distribute energy over time rather than to concentrate it.

Proving it

Once again, let’s turn to the formulas.

Remember, in automotive safety, the force experienced by passengers during a crash is minimized by increasing the time over which the impact occurs, utilizing the formula:

[katex] ​F = \frac{\Delta p}{\Delta t} [/katex]​

Crumple zones and airbags increase [katex] \Delta t [/katex], the time over which the change in momentum (or the deceleration of passengers) takes place. This reduces the average force [katex] F [/katex] experienced by the passengers. The longer the collision time, the less force is felt at any given moment, as the change in momentum is distributed over a longer period.

Comparative Analysis

The soccer ball scenario is governed by the need for momentum transfer over a longer duration, maximizing velocity. The karate chop, however, is all about the rate of force delivery – the higher the rate, the more effective the break. In car safety, the principle is inverted; extending the impact time decreases the force, protecting the passengers.

Test your knowledge

Difficulty - Advanced
Solve Type - Conceptual

A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.

  1. Strike the wood perpendicular to its surface with as much force as possible.

  2. Strike the wood and snap her hand back as quickly as possible.

  3. Strike the wood and not attempt to snap her hand back.

  4. Strike the wood at an angle to the normal with as much force as possible.

  5. Strike the wood and follow through with her hand.

View Full Question and Explanation

Conclusion

The physics of force and impact is nuanced and varies with the scenario. Understanding the relationship between force, time, and energy is crucial in optimizing outcomes, whether in sports, martial arts, or safety engineering. Each scenario demands a tailored approach to manage the forces involved effectively. By applying the right principles in the right context, one can achieve the desired result, be it scoring a goal, breaking a board, or saving lives in a crash.

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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