Assignment 17

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18:56:06 `q001. Note that this assignment contains 5 questions. . A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.

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RESPONSE --> 10kg * 5m/s = 50 10kg * 3m/s = 30 The change in Impulse was 20 since impulse and momentum might be equal in this case. 20 = Fnet * 'dt 20 = Fnet * .03 Fnet = 666.67N

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18:59:40 By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.

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RESPONSE --> Okay, I see why it should be negative.

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19:08:34 `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.

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RESPONSE --> 30 = Fnet * .03 Fnet = 1000N 667N *.03 = 2kg * v v = (667N * .03)/2 v = 10m/s approx

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19:13:27 Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second. This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object. A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s. Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.

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RESPONSE --> Newtons third law, Is this the force lost to moving the 2kg ball or more accurately, spent to move the ball? velocity checks okay.

The force arises as a result of the compression of the two balls. The force is zero at the instant of contact, then as the balls compress it rises to its maximum, then the balls spring back from the compression and the force diminishes back to 0.

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19:14:49 `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?

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RESPONSE --> less than, it lost some velocity.

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19:23:57 The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic) is involved.

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RESPONSE --> okay, so total KE was the sum of the KE of both objects.

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19:30:20 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?

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RESPONSE --> First Object Before 50 kg * m/s First Object After 30 kg * m/s Second Object Before 0 kg * m/s Second Object After 20 kg * m/s The total momentum is the same.

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19:30:35 The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision. The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second. The total momentum after collision is therefore equal to the total momentum before collision.

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RESPONSE --> check

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19:35:14 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?

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RESPONSE --> It ensures that Momentum is never lost but transfered, hence why we used 20 kg m/s for the second ball. This is my understanding of why it will always be equal, but there is probably a better one.

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19:38:42 Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.

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RESPONSE --> okay, Is my explenation valid?

You need to make it clear that since the forces are equal and opposite and act for the same time interval, then the resulting F `dt effects will also be equal and opposite.

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You did well on these questions. See my notes and let me know if you have questions.