In the practice problems using a force v time graph to find change in momentum, when a negative impulse was applied, a negative final velocity results, and the 'correct' answer states velocity decreases. But velocity increasing in the negative direction is not the same as a decrease velocity. In fact velocity can be increasing in the negative direction, and a negative impulse can result in an increase in velocity. If an object starts from rest and accelerates in the negative direction, would you say its velocity decreases??

You gotta be careful with the terms "velocity" and "speed". Speed is the absolute value of the velocity. In the questions, the velocity is decreasing, it is getting negative, but the speed is increasing (in the negative direction).

`During a safety test, a car hits a wall and stops in 0.55, The net force on the car is 6500 N during the collision. What is the magnitude of the change in momentum of the car?` Shouldn't the answer be divided by two? If the net force during the collision is 6500 N, and the car stops after 0.55 s, doesn't that mean the force goes from 6,500 N to 0 in 0.55 s, and the area under the force time graph (the impulse) is a triangle?

No, the force does not change. The force is constant during those 0.55 seconds. The force is equal to the change in momentum over the change in time. You know the force, you know the time, so you can solve for the change in momentum: Net force = (∆p)/(∆t) 6500 N = (∆p)/(0.55 s) ∆p = 3575 kg m/s. Hope this helps!

the article says"When the force is not constant, we use the average force to find the impulse."but in previous article was written that"For a constant force over time, the net external force is the same as the average external force over the time period."!

"When the force is not constant" "For a constant force" reread that please

Main content

Class 11 Physics (India)

Course: Class 11 Physics (India) > Unit 9

Lesson 14: Introduction to linear momentum and impulse

Impulse review

Google Classroom

Overview of key terms and equations related to impulse, including how impulse can be calculated from a force vs. time graph.

Key terms

Term (symbol)	Meaning
Impulse ( $F Δ t$ ‍)	Product of the average force exerted on an object and the time interval during which the force is exerted. Impulse is equal to the change in momentum ( $Δ p$ ‍) and is sometimes represented with the symbol $J$ ‍. Vector quantity with SI units of $N \cdot s$ ‍ or $\frac{kg \cdot m}{s}$ ‍.

Equations

Equation	Symbols	Meaning in words
$Δ p = F_{net} Δ t$ ‍	$Δ p$ ‍ is the change in momentum, $F_{net}$ ‍ is the net force, and $Δ t$ ‍ is the time period of the net force	Impulse is proportional to the constant net force acting on an object and the time period that the net force acts.

[How do we find impulse when force is not constant?]

How force changes momentum

If we take the impulse equation and solve for force, another relationship of the equation presents itself:

\begin{aligned} F Δ t & = Δ p \\ F & = \frac{Δ p}{Δ t} \end{aligned}

When a net force is exerted on an object, it changes that object's momentum over the time of the force exertion. In other words, force is the rate at which momentum changes. For example:

If an object experiences a large momentum change ( $Δ p$ ‍) over a short time duration ( $Δ t$ ‍), then there must have been a large net force ( $F$ ‍) applied to it.
Conversely, if an object experiences a small momentum change ( $Δ p$ ‍) over a long time duration ( $Δ t$ ‍), then there must have been a small net force ( $F$ ‍) applied to it.

[What about direction?]

How to find impulse from a force vs. time graph

Impulse is the area under the curve of the force vs. time graph. Areas above the time axis are positive

Δ p

and areas below the axis are negative

Δ p

. If the force is not constant, we can divide the graph into sections and add up the impulse in each section.

For example, to find the total impulse on the object in the force vs. time graph in Figure 1 over

t_{1} + t_{2}

, the areas of

A_{1}

and

A_{2}

can be added together.

A_{1}

is a rectangle of height

F_{0}

and width

t_{1}

A_{2}

is a triangle of height

F_{0}

and base

t_{2}

. The total impulse on the object over

t_{1} + t_{2}

\begin{aligned} Δ p & = A_{1} + A_{2} \\ = F_{0} t_{1} + \frac{1}{2} F_{0} t_{2} \end{aligned}

For worked examples of finding impulse or change in momentum from a force vs. time graph, watch our video about calculating impulse from force vs. time graphs.

Common mistakes and misconceptions

People forget what the sign of impulse means. Impulse is a vector, so a negative impulse means the net force is in the negative direction. Likewise, a positive impulse means the net force is in the positive direction.
People mistake impulse with work. Both impulse and work depend on the external net force, but they are different quantities. The properties of impulse and work are compared in the table below.

	Impulse	Work
Product of net force and...	time	displacement
Changes an object’s	momentum	energy
Quantity type	vector	scalar

Learn more

For deeper explanations of impulse, see our video about momentum and impulse.

To check your understanding and work toward mastering these concepts, check out our exercise on calculating change in momentum and speed from force vs. time graphs.

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Lisa Wolf
Posted 4 years ago. Direct link to Lisa Wolf's post “In the practice problems ...”
In the practice problems using a force v time graph to find change in momentum, when a negative impulse was applied, a negative final velocity results, and the 'correct' answer states velocity decreases. But velocity increasing in the negative direction is not the same as a decrease velocity. In fact velocity can be increasing in the negative direction, and a negative impulse can result in an increase in velocity. If an object starts from rest and accelerates in the negative direction, would you say its velocity decreases??
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(4 votes)
Answer
- Vinicius Coneglian
  Posted 4 years ago. Direct link to Vinicius Coneglian's post “You gotta be careful with...”
  You gotta be careful with the terms "velocity" and "speed". Speed is the absolute value of the velocity. In the questions, the velocity is decreasing, it is getting negative, but the speed is increasing (in the negative direction).
  Button navigates to signup page
  (4 votes)
shaul753
Posted 4 years ago. Direct link to shaul753's post “`During a safety test, a ...”
During a safety test, a car hits a wall and stops in 0.55, The net force on the car is 6500 N during the collision. What is the magnitude of the change in momentum of the car?

Shouldn't the answer be divided by two?
If the net force during the collision is 6500 N, and the car stops after 0.55 s, doesn't that mean the force goes from 6,500 N to 0 in 0.55 s, and the area under the force time graph (the impulse) is a triangle?
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(0 votes)
Answer
- obiwan kenobi
  Posted 4 years ago. Direct link to obiwan kenobi's post “No, the force does not ch...”
  No, the force does not change. The force is constant during those 0.55 seconds. The force is equal to the change in momentum over the change in time. You know the force, you know the time, so you can solve for the change in momentum:
  Net force = (∆p)/(∆t)
  
  6500 N = (∆p)/(0.55 s)
  
  ∆p = 3575 kg m/s. Hope this helps!
  Comment on obiwan kenobi's post “No, the force does not ch...”
  (4 votes)
Aditri
Posted 10 months ago. Direct link to Aditri's post “How do u get final veloci...”
How do u get final velocity from acceleration,time,and impulse?
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(1 vote)
Answer
James
Posted 3 years ago. Direct link to James's post “In the practice problem: ...”
In the practice problem:

A tennis player hits a ball and gives it a change in momentum of 3.3kgm/s over 5.0ms

What is the magnitude of the net force on the tennis ball?

Round answer to two significant digits.

I did Net force = (∆p)/(∆t)

to get 0.66 but the answer was wrong. Apparently there is a 10^-3s I should have multiplied the ∆t by.

Where does that come from?
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(1 vote)
Answer
- James
  Posted 3 years ago. Direct link to James's post “Oh, never mind, I see... ...”
  Oh, never mind, I see... it's the milliseconds part. 10^-3 = 0.001.
  Button navigates to signup page
  (1 vote)
mohamad
Posted 4 years ago. Direct link to mohamad's post “the article says"When the...”
the article says"When the force is not constant, we use the average force to find the impulse."but in previous article was written that"For a constant force over time, the net external force is the same as the average external force over the time period."!
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(0 votes)
Answer
- Omar K
  Posted 4 years ago. Direct link to Omar K's post “"When the force is not co...”
  "When the force is not constant"
  "For a constant force"
  reread that please
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  (3 votes)