How does the impulse affect the ball's speed?

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kissafilipino
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1. I understand I need to find the area underneath the force time graph, however I do not understand how to use the impulse = m(delta)v

2. the question says:

A 0.145 kg ball, moving in the positive direction at 13 m/s, is acted on by the impulse shown in the graph in Figure 9-12 What is the ball's speed at 4.0 s?


Here is the image: http://www.webassign.net/glencoe99/9-12.gif

3. all the info is relevant, however it confuses me because I don't understand the underthegraph thing that is suppose to help me solve
 
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  • Force versus time graph: F (N) on the vertical axis from -2 to 2, t (s) on the horizontal axis from 0 to 4. The force is zero until t=1, rises linearly to 2 N at t=2, falls back to zero at t=3, and stays at zero to t=4.
    9-12.gif
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kissafilipino said:
1. I understand I need to find the area underneath the force time graph, however I do not understand how to use the impulse = m(delta)v

2. the question says:

A 0.145 kg ball, moving in the positive direction at 13 m/s, is acted on by the impulse shown in the graph in Figure 9-12 What is the ball's speed at 4.0 s?


3. all the info is relevant, however it confuses me because I don't understand the underthegraph thing that is suppose to help me solve
What does the 'the (area) underthegraph thing' represent? It is a function of force and time. Note also that before 1 second and after 3 seconds, there is no longer any force applied.
 
I realize that this might be a bit late, but here's some help if you want it.

In the graph, the area of the triangle created by the line is actually the impulse used during that period of time, as F*t is measured in Ns or Kg m/s.

So you start out with a certain amount of momentum, then a bit more is added, and now you have the total momentum of the ball.

Initial Momentum = 0.145*13 = 1.885
Area/Impulse = 2
Mass of Ball = 0.145

1.885 + 2 = Answer
0.145