Impulse–Momentum Theorem (College Board AP® Physics 1: Algebra-Based)

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Katie M

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Reviewed by: Caroline Carroll

Impulse–momentum theorem

The momentum of an object remains constant unless an external net force acts upon it

  • An impulse is exerted when an external force is applied for a time

  • Therefore, it must also change the linear momentum of the system

    • This is known as the impulse-momentum theorem

  • The impulse-momentum theorem states that the impulse exerted on a system is equal to the change in momentum

J with rightwards arrow on top space equals space F with rightwards arrow on top subscript a v g end subscript increment t space equals space increment p with rightwards arrow on top

  • Where:

    • J with rightwards arrow on top = impulse, measured in straight N times straight s

    • F with rightwards arrow on top subscript a v g end subscript = average force exerted, measured in straight N

    • increment t = time interval over which the force acts, measured in straight s

    • increment p with rightwards arrow on top = change in momentum, measured in kg times straight m divided by straight s

  • In calculations, a more useful form of this equation is

J with rightwards arrow on top space equals space F with rightwards arrow on top subscript a v g end subscript increment t space equals space m open parentheses v with rightwards arrow on top space minus space v with rightwards arrow on top subscript 0 close parentheses

  • Where:

    • m = mass of the body, measured in kg

    • v with rightwards arrow on top = final velocity, measured in straight m divided by straight s

    • v with rightwards arrow on top subscript 0 = initial velocity, measured in straight m divided by straight s

  • Newton’s second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass

F with rightwards arrow on top subscript n e t end subscript space equals space fraction numerator increment p with rightwards arrow on top over denominator increment t end fraction space equals space m fraction numerator increment v with rightwards arrow on top over denominator increment t end fraction space equals space m a with rightwards arrow on top

  • Where:

    • increment v with rightwards arrow on top = change in velocity, measured in straight m divided by straight s

    • a with rightwards arrow on top = acceleration, measured in straight m divided by straight s squared

  • The impulse-momentum theorem tells us

    • for a given change in momentum, a small force acting over a long time has the same effect as a large force acting over a short time

    • for a constant force, applying the force over a longer time will lead to a greater change in momentum

    • for a specified time, a greater force will lead to a greater change in momentum

Worked Example

A rubber ball and a clay ball of the same mass are thrown horizontally at a wall. The rubber ball bounces off horizontally with the same speed it hits the wall. The clay ball strikes the wall and sticks to it.

If the impulse delivered to the rubber ball is J with rightwards arrow on top subscript r and the impulse delivered to the clay ball is J with rightwards arrow on top subscript c, what is the value of J with rightwards arrow on top subscript r over J with rightwards arrow on top subscript c?

A      negative 2

B      negative 1 half

C      plus 1 half

D      plus 2

The correct answer is D

Answer:

Step 1: Analyze the scenario

  • Initially, both balls travel:

    • in the same direction, which we can define as the positive direction

    • with the same velocity, which we can define as v

  • The rubber ball bounces off the wall, meaning:

    • it travels in the opposite direction, so its final velocity is negative v

    • its momentum becomes negative

  • The clay ball sticks to the wall, meaning:

    • it comes to rest, so its final velocity is zero

    • its momentum becomes zero

Diagram showing a clay ball and a rubber ball before and after collision with a wall. The clay ball stops after collision, while the rubber ball bounces back at speed v.

Step 2: Determine the impulse delivered to each object

  • The impulse delivered to an object is its change in momentum

J with rightwards arrow on top space equals space increment p with rightwards arrow on top space equals space m increment v with rightwards arrow on top

  • The impulse delivered to the rubber ball is

J with rightwards arrow on top subscript r space equals space increment p with rightwards arrow on top subscript r space equals space minus m v space minus space m v space equals space minus 2 m v

  • The impulse delivered to the clay ball is

J with rightwards arrow on top subscript c space equals space increment p with rightwards arrow on top subscript c space equals space 0 space minus space m v space equals space minus m v

Step 3: Determine the ratio of the impulses

  • The ratio of the impulse delivered to the rubber ball to that of the clay ball is:

J with rightwards arrow on top subscript r over J with rightwards arrow on top subscript c space equals space fraction numerator increment p with rightwards arrow on top subscript r over denominator increment p with rightwards arrow on top subscript c end fraction space equals space fraction numerator negative 2 m v over denominator negative m v end fraction space equals space 2

Worked Example

Car X is designed to include a crumple zone so that the front of the car collapses during impact. A similar car, Y, is designed without a crumple zone.

Both cars undergo a safety test in which they are driven into a solid concrete wall at the same speed. Analysis of the crash shows that the front of Car X collapses by 1.40 m, while the front of Car Y collapses by 15.0 cm.

(A) Is the change in momentum of Car X greater than, less than, or the same as the change in momentum of Car Y? Justify your answer.

(B) Is the average force on Car X greater than, less than, or the same as the average force on Car Y? Justify your answer.

(C) Is the time taken for Car X to come to rest greater than, less than, or the same as the time taken for Car Y to come to rest? Justify your answer.

(D) Use the information provided to justify whether crumple zones make cars safer in the event of a collision.

Answer:

Part (A)

Step 1: Analyze the scenario

  • The change in momentum is equal to the product of the mass and the change in velocity

increment p with rightwards arrow on top space equals space m increment v with rightwards arrow on top space equals space m open parentheses v with rightwards arrow on top space minus space v with rightwards arrow on top subscript 0 close parentheses

  • The final velocity of both cars is zero, so the change in momentum depends on the initial momentum of the car only

increment p with rightwards arrow on top space equals space m open parentheses 0 space minus space v subscript 0 close parentheses space equals space minus m v subscript 0

Step 2: Deduce and justify the relationship

  • The change in momentum of Car X is the same as the change in momentum of Car Y

  • Justification:

    • Both cars have the same mass and initial velocity

    • Therefore, increment p with rightwards arrow on top subscript straight X space equals space increment p with rightwards arrow on top subscript straight Y

Part (B)

Step 1: Analyze the scenario

  • According to the Work-Energy Theorem, the change in kinetic energy equals the work required to stop the car

increment K space equals space W space equals space 1 half m open parentheses v squared space minus space v subscript 0 squared close parentheses

  • The work done is equal to the product of the average force on the car and the distance moved while the force is applied

W space equals space F subscript a v g end subscript d space equals space 1 half m open parentheses 0 space minus space v subscript 0 squared close parentheses space equals space minus 1 half m v subscript 0 squared

  • The average force on the car is therefore equal to

F subscript a v g end subscript space equals space minus fraction numerator m v subscript 0 squared over denominator 2 d space end fraction

  • During the collision, the front of the car collapses, so the distance d is equal to the length of material that deforms

Step 2: Deduce and justify the relationship

  • The average force on Car X is less than the average force on Car Y

  • Justification:

    • The front of Car X collapses by d subscript straight X space equals space 1.40 space straight m whereas Car Y collapses by d subscript straight Y space equals space 0.15 space straight m

    • Therefore, Car X stops over a greater distance than Car Y, by a factor of d subscript straight X over d subscript Y space equals space fraction numerator 1.40 over denominator 0.15 end fraction space equals space 9.3

    • Both cars have the same mass and initial speed

    • Therefore, the force on Car X is smaller than the force on Car Y by a factor of F subscript straight X over F subscript straight Y space equals space d subscript straight Y over d subscript straight X space equals space fraction numerator 0.15 over denominator 1.40 end fraction space equals space 0.11

Part (C)

Step 1: Analyze the scenario

  • According to the Impulse-Momentum Theorem, the change in momentum is equal to the product of the net external force and the amount of time that the force is exerted

increment p with rightwards arrow on top space equals space F with rightwards arrow on top subscript a v g end subscript increment t

  • For a given change in momentum, the greater the net external force on a car, the shorter the collision time

Step 2: Deduce and justify the relationship

  • The time taken for Car X to come to rest is greater than the time taken for Car Y to come to rest

  • Justification:

    • In part (A), we established the change in momentum is the same for both cars

    • In part (B), we established Car X experiences a smaller average force than Car Y

    • Therefore, Car X stops over a longer time than Car Y, by a factor of fraction numerator increment t subscript straight X over denominator increment t subscript straight Y end fraction space equals space F subscript straight Y over F subscript straight X space equals space fraction numerator 1.40 over denominator 0.15 end fraction space equals space 9.3

Part (D)

Step 1: Analyze the scenario

  • As the car comes to a stop over the time period increment t, the driver will continue moving forward at the same speed v subscript 0 as the car before the crash, unless they are wearing a seatbelt

  • Assuming this is the case, the average force exerted by the seat belt on the driver is equal to the rate of change of momentum of the driver

F with rightwards arrow on top subscript a v g end subscript space equals space fraction numerator increment p with rightwards arrow on top over denominator increment t end fraction space equals space minus fraction numerator m v subscript 0 over denominator increment t end fraction

  • Where m is the mass of the driver

Step 2: Write a conclusion and justification for your claim

  • Crumple zones do make cars safer in the event of a collision

  • Justification:

    • The car with the crumple zone (Car X) was shown to increase both the distance and the time over which the collision occurred by a factor of 9.3 when compared to the car without a crumple zone (Car Y)

    • As a result, the force on the car, and therefore the driver, in Car X was about 9.3 times lower than the force on the driver in Car Y

    • This suggests the likelihood of injury is much lower in cars which have crumple zones

Examiner Tips and Tricks

Most practical applications of the impulse-momentum theorem involve a change in velocity with respect to time. In AP Physics 1, you will not be required to analyze systems where mass changes with respect to time quantitatively (i.e. involving calculations), but you could be asked to analyze such a system qualitatively. This could involve discussing proportionality relationships between the changing mass and another quantity, such as a change in momentum, force, or time.

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.