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The comments in Listing 1 explain the steps involved in finding the solution.

The output produced by Listing 1 is shown in Figure 1 with the magnitude and angle of the vector that describes the change of momentum at the end.

Figure 1 . Solution results.
Start Script The givens.weight = 10000 kg speed1 = 40 m/sangle 1 = 90 degrees speed2 = 20 m/sangle 2 = 0 degrees Computed mass.mass = 1020 kg Components of momentum vectors.P1x = 0 P1y = 40816P2x = 20408 P2y = 0Components of momentum change vectors. deltaPx = 20408deltaPy = -40816 Magnitude and angle of change vector.deltaPm = 45634 m kg/s deltaPa = -63 degreesEnd Script

I find it interesting that the magnitude of the change in momentum is greater than the magnitude of either the initial or final momentum.

Change in momentum due to change in speed only

What happens if the car in the previous example changes speed but doesn't change direction.

Solution:

Change the given conditions in the script in Listing 1 to those shown at the beginning of Figure 2 .

Figure 2 . Change in speed only.
Start Script The givens.weight = 10000 kg speed1 = 40 m/sangle 1 = 90 degrees speed2 = 20 m/sangle 2 = 90 degrees Computed mass.mass = 1020 kg Components of momentum vectors.P1x = 0 P1y = 40816P2x = 0 P2y = 20408Components of momentum change vectors. deltaPx = -0deltaPy = -20408 Magnitude and angle of change vector.deltaPm = 20408 m kg/s deltaPa = 270 degreesEnd Script

This change causes the car to slow down, but to continue in the same direction. As a result, the angle of the change in momentum is an angle thatis opposite to the direction that the car is moving. The magnitude of the change in momentum depends entirely on the initial and final speeds.

Change in momentum due to change in direction only

What happens if the car in the previous example changes direction but doesn't change speed?

Solution:

Change the given conditions in the script in Listing 1 to those shown at the beginning of Figure 3 . This scenario simulates the car making a 10-degree turn to the right without changing speed.

Figure 3 . Change in direction only.
Start Script The givens.weight = 10000 kg speed1 = 40 m/sangle 1 = 90 degrees speed2 = 40 m/sangle 2 = 80 degrees Computed mass.mass = 1020 kg Components of momentum vectors.P1x = 0 P1y = 40816P2x = 7088 P2y = 40196Components of momentum change vectors. deltaPx = 7088deltaPy = -620 Magnitude and angle of change vector.deltaPm = 7115 m kg/s deltaPa = -5 degreesEnd Script

Impulse examples

This section contains several examples involving the impulse.

Pushing a wagon part 1

1. What is the impulse experienced by pushing a 10-kg wagon that was initially at rest, with a constant force of 2 newtons for a period of 3 seconds?

Answer:

The impulse is given by the product of force and time. The mass of the wagon is superfluous for this question.

impulse = 2 N * 3 s = 6*N*s

Pushing a wagon part 2

2. What is the acceleration of the wagon in question 1 above?

Answer:

Now we do need to know the mass.

The most straightforward solution comes from the fact that we know the mass and that the force is uniform. Therefore,

Questions & Answers

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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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