Lesson Explainer: Force as Rate of Momentum Change Physics • 9th Grade

In this explainer, we will learn how to relate the rate of change of momentum of an object to the force acting on it, using the formula Δ𝑝=𝐹Δ𝑑.

It is possible to show that the change in the momentum of an object is the product of the force on the object and the time that the force acts for.

A force acting on an object accelerates the object. The acceleration of an object is the time rate of change of the velocity of the object. We can express this as π‘Ž=Δ𝑣Δ𝑑, where π‘Ž is the acceleration of the object, Δ𝑣 is the change in the velocity of the object, and Δ𝑑 is the time interval during which the velocity of the object changes.

The acceleration of an object due to a force is described by Newton’s second law of motion, 𝐹=π‘šπ‘Ž, where 𝐹 is the force acting on the object and π‘š is the mass of the object.

We can substitute the value of π‘Ž in terms of 𝑣 and 𝑑 into Newton’s second law of motion to obtain the following expression: 𝐹=π‘šΞ”π‘£Ξ”π‘‘.

If this expression is multiplied by Δ𝑑, we see that 𝐹Δ𝑑=π‘šΞ”π‘£.

This expression is equal to the product of the force and the time interval during which it acts.

The momentum of an object is given by 𝑝=π‘šπ‘£, where 𝑝 is the momentum of the object and 𝑣 is the velocity of the object.

The change in the momentum of an object is the product of the mass of the object and the change in the velocity of the object. This relationship can be expressed as Δ𝑝=π‘šΞ”π‘£.

From this, we see that 𝐹Δ𝑑=Δ𝑝, showing that it is indeed the case that the product of a force and the time interval during which the force acts equals the change in the momentum of the object.

Understanding the relationship between force and momentum gives us a better understanding of what momentum is useful for describing.

The momentum of a moving object can represent the two following quantities:

  • the amount of force that must act in a given time to bring a moving object to rest,
  • the amount of time that a given force must act for to bring a moving object to rest.

It is just as valid to say that a given momentum can represent the two following quantities:

  • the amount of force that must act in a given time to bring an object initially at rest to a given velocity,
  • the amount of time that a given force must act for to bring an object initially at rest to a given velocity.

Let us look at an example where the time for which a force acts depends on the force and the change in momentum due to the force.

Example 1: Determining the Time Interval Required to Bring an Object to Rest

An object with an initial momentum of 20 kgβ‹…m/s is acted on by an average force of 2.5 N and comes to rest. For how long does the force act on the object?

Answer

This is an example where using the quantity of momentum makes it easier to determine the time interval that a given force must act in to bring a moving object to rest.

The relationship between a force and the time that it acts in to change the momentum of an object is given by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force that acts, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in momentum.

To obtain Δ𝑑, we can divide this formula by 𝐹, which gives us 𝐹Δ𝑑𝐹=Δ𝑑=Δ𝑝𝐹.

An object at rest must have a momentum of zero. The initial momentum of the object is 20 kgβ‹…m/s, so the change in momentum is given by Δ𝑝=π‘βˆ’π‘=0βˆ’20=βˆ’20β‹…/.finalinitialkgms

The force that acts has a magnitude of 2.5 N. The force must act in the direction of the change of the momentum of the object; so, including its sign, the force, 𝐹, is βˆ’2.5 N.

Substituting the values for momentum change and force into the formula Δ𝑑=Δ𝑝𝐹, we find that Δ𝑑=βˆ’20βˆ’2.5=8.seconds

Let us look at an example in which a force that acts depends on the time for which the force acts and the change in momentum due to the force.

Example 2: Determining the Force Required to Increase the Velocity of an Object over a Time Interval

A frog of mass 30 g jumping into the air from a resting position accelerates to a speed of 12 cm/s in 0.025 s. How many newtons of force did the frog’s legs exert in the jump?

Answer

This is an example where using the quantity of momentum makes it easier to determine the amount of force that must act in a given time to bring a moving object to rest.

The relationship between a force and the time that it acts in to change the momentum of an object is given by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force that acts, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in momentum.

To obtain 𝐹, we can divide this formula by Δ𝑑, which gives us 𝐹Δ𝑑Δ𝑑=𝐹=Δ𝑝Δ𝑑.

An object at rest must have a momentum of zero. The initial momentum of the frog is zero. The final momentum of the frog is given by 𝑝=π‘šπ‘£.

It is convenient to express momentum in units of kilogram-metres per second (kgβ‹…m/s), so the mass of the frog is converted from grams to kilograms and the final velocity of the frog is converted from centimetres per second to metres per second.

The mass of the frog in kilograms is 0.030 kg and its final velocity in metres per second is 0.12 m/s. This gives a value for the momentum of the frog of 𝑝=0.030Γ—0.12=0.0036β‹…/.kgms

The change in momentum of the frog is given by Δ𝑝=π‘βˆ’π‘=0.0036βˆ’0=0.0036β‹…/.finalinitialkgms

The time interval in which the frog’s legs exert force is 0.025 seconds long. Substituting the values for momentum change and time into the formula 𝐹=Δ𝑝Δ𝑑, we find that 𝐹=0.00360.025=0.144.N

If we took this force as equivalent to a weight, using the formula π‘Š=π‘šπ‘”, where 𝑔 is 9.8 m/s2, we find that a mass that would produce this weight force is approximately 15 grams.

Assuming that the frog jumps vertically, this means that the frog’s legs pushing down on a surface would act on that surface equivalently to the weight of an object of mass 15 grams at rest on the surface. The mass of the frog is stated to be 30 grams, so when the frog jumps, the force it exerts downward is equivalent to approximately half its body weight.

Let us now look at an example involving determining the momentum change due to the action of a force.

Example 3: Determining the Momentum Change due to a Force Acting over a Time Interval

An average force of 12.5 N is applied by moving air to a bicycle and its rider, of total mass 80 kg, while it is windy that day. The force is applied throughout a time interval of 0.8 s as the gust of wind is very brief. The wind blows from directly behind the bicycle. What is the total momentum change?

Answer

Let us take the direction of the wind as the positive direction. The wind blows from directly behind the bicycle, so the motion of the bicycle is also in the positive direction.

The relationship between a force and the time that it acts in to change the momentum of an object is given by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force that acts, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in momentum.

Substituting the values of the force and the time for which the force acts, we see that the momentum change of the bicycle is given by Δ𝑝=𝐹Δ𝑑=12.5Γ—0.8=10β‹…/.kgms

It is worth noting that the value of the mass of the bicycle is not required to determine the momentum change of the bicycle.

Let us now look at an example where a force acts in the opposite direction to the direction of the velocity of an object to change the momentum of the object.

Example 4: Determining the Force Required to Produce a Momentum Change over a Time Interval

A man running on a road had a momentum of 360 kgβ‹…m/s. The road ahead of the man was covered by sand. The man ran through the sand for 12 seconds and his momentum when he had crossed the sand-covered part of the road was 330 kgβ‹…m/s. What average force did the sand apply to the man while he ran across it?

Answer

Let us take the direction of the man as the positive direction.

The relationship between a force and the time that it acts in to change the momentum of an object is given by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force that acts, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in momentum.

To obtain 𝐹, we can divide this formula by Δ𝑑, which gives us 𝐹Δ𝑑Δ𝑑=𝐹=Δ𝑝Δ𝑑.

The change in momentum of the man is given by Δ𝑝=π‘βˆ’π‘=330βˆ’360=βˆ’30β‹…/.finalinitialkgms

Substituting the values for momentum change and time into the formula 𝐹=Δ𝑝Δ𝑑, we find that 𝐹=βˆ’3012=βˆ’2.5.N

The force acts in the opposite direction to the velocity of the man, so it is negative.

Let us now look at an example in which the momentum of an object changes due to the action of a force, but the momentum of the object is not directly given.

Example 5: Determining the Time Interval Required for an Object to Change in Velocity

A tennis racquet hits a tennis ball that has a mass of 60.5 g and applies a constant 75 N force to it. The tennis ball changes its velocity by 30 m/s during the collision. For how many seconds does the collision last?

Answer

The relationship between a force and the time that it acts in to change the momentum of an object is given by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force that acts, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in momentum.

The momentum of the tennis ball is not stated. The velocity of the tennis ball is also not given at any instant, and the tennis ball cannot be assumed to be at rest before or after the collision.

The change in the velocity of the tennis ball is given, however, so the change in its momentum can be determined by using the formula Δ𝑝=π‘šΞ”π‘£, where the mass of the tennis ball is given as 60.5 grams and the change in velocity is given as 30 m/s. It is convenient to express momentum in units of kilogram-metres per second (kgβ‹…m/s), so the mass of the tennis ball is converted from grams to kilograms and is 0.0605 kg. We have then that Δ𝑝=0.0605Γ—30=1.815β‹…/.kgms

To obtain the time that the tennis racquet is in contact with the tennis ball, Δ𝑑, we can divide the formula 𝐹Δ𝑑=Δ𝑝 by 𝐹, which gives us 𝐹Δ𝑑𝐹=Δ𝑑=Δ𝑝𝐹.

Substituting the values for momentum change and force into the formula Δ𝑑=Δ𝑝𝐹, we find that Δ𝑑=1.81575=0.0242.seconds

Let us summarize what has been learned in these examples.

Key Points

  • The change in the momentum of an object is related to the force that acts on the object and the time for which the force acts by the formula 𝐹Δ𝑑=Δ𝑝, where 𝐹 is the force on the object, Δ𝑑 is the time for which the force acts, and Δ𝑝 is the change in the momentum of the object.
  • The mass of an object that changes momentum does not need to be known to determine the relationship between the change in the momentum of an object, the force that acts on the object, and the time for which the force acts.
  • The relationship between the change in the momentum of an object, the force that acts on the object, and the time for which the force acts is useful in determining
    • the amount of force that must act in a given time to bring a moving object to rest,
    • the amount of time that a given force must act for to bring a moving object to rest,
    • the amount of force that must act in a given time to bring an object initially at rest to a given velocity,
    • the amount of time that a given force must act for to bring an object initially at rest to a given velocity.

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