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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. |
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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.
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.
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 |
This section contains several examples involving the impulse.
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
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,
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