Rate of Change of Momentum (College Board AP® Physics 1: Algebra-Based)

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Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Rate of change of momentum

  • Newton's second law of motion was originally postulated in terms of momentum

  • In terms of momentum, Newton's second law states

The rate of change of momentum is equal to the net external force exerted on an object (or system)

  • This can be written as:

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

  • Where:

    • F with rightwards arrow on top subscript n e t end subscript = net external force, measured in straight N

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

    • increment t = time interval, measured in straight s

  • The change in momentum is the difference between a system’s final momentum and its initial momentum, as written below:

change in momentum = final momentum − initial momentum

increment p with rightwards arrow on top space equals space p with rightwards arrow on top space minus space p with rightwards arrow on top subscript 0

  • The equation above can be used in situations where the mass of the body is not constant

  • When the mass is constant:

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 fraction numerator p with rightwards arrow on top space minus space p with rightwards arrow on top subscript 0 over denominator increment t end fraction space equals space fraction numerator m v with rightwards arrow on top space minus space m v with rightwards arrow on top subscript 0 over denominator increment t end fraction space equals space fraction numerator m open parentheses v with rightwards arrow on top space minus space v with rightwards arrow on top subscript 0 close parentheses over denominator increment t end fraction space equals space fraction numerator m increment v with rightwards arrow on top over denominator increment t end fraction

  • Since acceleration a with rightwards arrow on top is equal to the rate of change of velocity, the equation becomes:

F with rightwards arrow on top subscript n e t end subscript space equals space m a with rightwards arrow on top

  • Where:

    • m = mass of the body, measured in kg

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

    • 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

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

Worked Example

A professional tennis player hits a tennis ball of mass 60 g which travels at 34 m/s just before it makes contact with the racket. The ball is in contact with the racket for 5 ms before recoiling in the opposite direction at 40 m/s.

What is the average force exerted by the racket on the ball? Take the ball's initial velocity to be the positive direction.

A      –888 N

B      –72 N

C      72 N

D      888 N

The correct answer is A

Answer:

Step 1: Analyze the scenario

  • When the tennis ball comes into contact with the racket, the racket exerts a force on the ball over a time interval of 5 ms

  • This force changes the ball's momentum by

    • decelerating it from 34 m/s to 0 m/s in the positive direction

    • accelerating it from 0 m/s to 40 m/s in the opposite (negative) direction

Diagram showing a tennis ball before and after hitting a racket. Ball changes from +34 m/s to -40 m/s. Change in velocity calculated as -74 m/s.

Step 2: List the known quantities

  • Mass of the ball, m = 60 g = 0.06 kg

  • Time interval, increment t = 5 ms = 0.005 s

  • Initial velocity, v with rightwards arrow on top subscript 0 = +34 m/s (positive direction)

  • Final velocity, v with rightwards arrow on top = –40 m/s (negative direction)

Step 3: Determine an expression for the average force

  • Since force is equal to the rate of change of momentum:

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

Step 4: Calculate the average force

F with rightwards arrow on top space equals space fraction numerator 0.06 cross times open parentheses negative 40 space minus space 34 close parentheses over denominator 0.005 end fraction space equals space minus 888 space straight N

  • The answer is therefore A

Examiner Tips and Tricks

Be aware that multiple-choice options are designed to catch you out, so you need to be especially careful when vector quantities are involved! If an object changes direction, this must be reflected by the change in the sign of the velocity (and momentum).

Sometimes it can help to do a sense check, for example, in the worked example above, the ball's change in speed is (40 - 34) = 6 m/s, whereas the ball's change in velocity is (-40 - 34) = -74 m/s

  • The large magnitude reflects the fact that the ball's speed reduced to zero in the positive direction before it increased in the negative direction

  • The negative sign indicates the ball's final velocity is in the opposite direction to its initial velocity

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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.