Impulse and Force-Time Graph

[Qube]

Homework Statement

Given the graph (see picture), determine:

a) magnitude of impulse (max error of +/- 25%)
b) direction of impulse
c) mass of the object that the impulse acted on.

Note that initial velocity is 3.8 m/s to the west and that final velocity is 4.8 m/s to the east.

Homework Equations

1) Define an coordinate system. West is negative. East is positive. So that means initial velocity is -3.8 m/s and final velocity is + 4.8 m/s

2) Impulse is mass * change in velocity or force * change in time. Thus, impulse is the area under a force-time graph.

The Attempt at a Solution

Unfortunately, the graph of force-time is non-linear. To determine impulse, then, I used two triangles and estimated the sum of the areas of these two triangles.

This gives me the impulse. The magnitude of the impulse is 0.875 N*s to the east since the change in velocity is positive (and thus to the east).

Because impulse is equal to mass * change in velocity, I can now easily solve for the mass of the object.

Attached please find a picture of my work. I would appreciate it if you guys could verify the accuracy of my method and results. Note that the blue rectangle in the picture of the graph below represents my attempt to apply a Reimann sum to find impulse ... I think doing the Reimann sum would have been considerably more difficult and likely less accurate to boot!

http://i.minus.com/jbjAWztX3spxVC.JPG

 

[Andrew Mason]

Homework Statement

Given the graph (see picture), determine:

a) magnitude of impulse (max error of +/- 25%)
b) direction of impulse
c) mass of the object that the impulse acted on.

Note that initial velocity is 3.8 m/s to the west and that final velocity is 4.8 m/s to the east.

Homework Equations

1) Define an coordinate system. West is negative. East is positive. So that means initial velocity is -3.8 m/s and final velocity is + 4.8 m/s

2) Impulse is mass * change in velocity or force * change in time. Thus, impulse is the area under a force-time graph.

The impulse is the area under the F t graph, but the answer is very easy if you relate it to mΔv instead.

AM

 

[Qube]

Right, but I don't have mass. Mass is an unknown in this example problem. I do have the change in velocity though.

 

[Chestermiller]

Looks good. Of course, you didn't have to split it into two separate triangles. The area of the big triangle is bh/2, where b is the full base and h is the altitude.

Chet

 

[HalfLostAndSearching]

Your method for determining the impulse and the mass of the object is correct. The impulse is indeed equal to the area under the force-time graph, and using two triangles to estimate the area is an appropriate approach. Your calculation of the mass is also correct, assuming that the units for impulse are N*s. However, it would be helpful to include units in your final answer for the mass, which would be in kilograms (kg).

Additionally, it would be beneficial to include a brief explanation of how you determined the direction of the impulse. You correctly noted that the change in velocity is positive, indicating that the impulse is in the same direction as the final velocity (to the east). This could be explicitly stated in your response.

Overall, your response is accurate and demonstrates a good understanding of the concepts of impulse and force-time graphs. Keep up the good work!