Video: Finding the Momentum Change of an Object from a Force-Time Graph

The total force acting on an object as a function of time is given in the graph. What is the magnitude of the change in momentum of the object between 𝑡 = 0 s and 𝑡 = 0.8 s? [A] 24 kg⋅m/s. [B] 18 kg⋅m/s. [C] 9 kg⋅m/s. [D] 12 kg⋅m/s. [E] 6 kg⋅m/s.

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Video Transcript

The total force acting on an object as a function of time is given in the graph. What is the magnitude of the change in momentum of the object between 𝑡 equals zero seconds and 𝑡 equals 0.8 seconds? (a) 24 kilograms times meters per second. (b) 18 kilograms times meters per second. (c) Nine kilograms times meters per second. (d) 12 kilograms times meters per second. (e) Six kilograms times meters per second.

In our graph, the force is plotted on the 𝑦-axis and the time is plotted on the 𝑥-axis. Between zero and 0.6 seconds, our force increases at a constant rate from zero newtons to 10 newtons. Between 0.6 seconds and 0.8 seconds, our force stays at a constant value of 30 newtons.

The problem asks us to find the change in momentum of our object from zero seconds to 0.8 seconds. We use a pink line to highlight this time frame on the graph. We use Δ𝑃 to represent the change in momentum. The problem gives us information in graphical form of force and time and asks us about the change in momentum.

To solve the problem, we must figure out what the relationship is between change in momentum, force, and time. We can start by remembering the impulse–momentum theorem. The impulse–momentum theorem states that the impulse, 𝐽, is equal to the change in momentum, Δ𝑃. This still does not provide a connection between change in momentum, force, and time. We need to define what impulse is.

Impulse is the product of force, 𝐹, and change in time, Δ𝑡. We can see that change in momentum is equal to impulse, and force times change in time is equal to impulse. Therefore, we can say that they’re equal to each other. Change in momentum equals force times change in time. The only problem with this expression is that it assumes a constant force.

Looking back at our graph, we can see that, between zero and 0.6 seconds, our force is changing. This means we’d have to find force times change in time for every little segment along that line between zero and 0.6 seconds. If we were to do this mathematically, this would require us to do integration, which is a calculus term.

There is another way to solve the problem that involves doing it graphically. Recall that change in momentum is equal to the area under a force-versus-time graph. Looking back at our graph, this means we need to find the area of the shape highlighted in pink as it refers to the time frame between zero and 0.8 seconds.

Remember that when we’re talking about the area, it is the area between our curve and our 𝑥-axis. To make calculating this area easier, we can break down our big shape into two smaller shapes. Between zero and 0.6 seconds, the area between our line and our 𝑥-axis forms a triangle, as highlighted in blue. Between 0.6 and 0.8 seconds, the area between our line and the 𝑥-axis forms a rectangle, as highlighted in yellow.

Recall that the area of a triangle, represented here as 𝐴 subscript T, is equal to one-half base, 𝑏, times height, ℎ, of the triangle. And that the area of a rectangle, represented here as 𝐴 subscript R, is equal to base, 𝑏, times ℎ, height, of the rectangle. We can substitute in the values from our graph to determine the area of both the triangle and the rectangle.

To determine the area of the triangle, we replace 𝑏 with 0.6 seconds as the base of our triangle goes from zero to 0.6 seconds. And we replace ℎ with 10 newtons as the height of our triangle goes from zero to 10 newtons. When we multiply out one-half by 0.6 seconds by 10 newtons, we get the area of our triangle to be equal to three kilograms times meters per second.

Now we need to find the area of the rectangle. We replace 𝑏 with 0.2 seconds as the base of our rectangle goes from 0.6 to 0.8 seconds, leaving us a difference of 0.2 seconds. And we replace ℎ with 30 newtons because the height of our rectangle goes from zero newtons to 30 newtons. When we multiply out 0.2 seconds by 30 newtons, we get that the area of the rectangle is equal to six kilograms times meters per second.

Our change in momentum for our object is gonna be equal to the area of the triangle plus the area of the rectangle. Adding together three kilograms times meters per second plus six kilograms times meters per second, we get a change in momentum of nine kilograms times meters per second. Looking back at our answer choices, we see that answer choice (c), nine kilograms times meters per second, correctly matches with the magnitude of the change in momentum of the object between zero seconds and 0.8 seconds.

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