course Phy 201 ???????L?????assignment #017
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13:46:52 `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 --> p=mv=momentum Initial momentum=10kg*5m/s=50kgm/s Final momentum=10kg*3m/s=30m/s Change=-20m/s `dp=Fave*`dt Fave=`dp/`dt -20kgm/s/.03s=-667kgm/s^2 For the force exerted by the first object on the second. It is opposite for the force exerted by the second object on the first. 667kgm/s^2=667N confidence assessment: 2
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13:48:37 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 --> Am confused about when it is exerted on the first object or on the second. Fave*`dt=m`dv Fave=m`dv/`dt=10kg*(-2m/s)/.03s=-667N self critique assessment: 2
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13:53:25 `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 --> Would be opposite to the -667kgm/s=667kgm/s mv=p 2kg*v=667kgm/s divide by 2kg 333.5m/s=v confidence assessment: 2
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13:56:53 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 --> F`dt=667N*.03s=20kgm/s Momentum change of 20kgm/s on a 2kg object= change in velocity of 20kgm/s/2kg=10m/s v0=0 0 + 10m/s=10m/s Should have used the change in momentum for p instead of the force. self critique assessment: 2
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14:03:55 `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 --> KE before=.5*10kg*(5m/s)^2+ .5*2kg*0=125J KE after=.5*10kg*(3m/s)^2 +.5(2kg)(10m/s)^2=145J confidence assessment: 1
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14:05:18 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, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved.
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RESPONSE --> The total kinetic energy after the collision is greater than the total before the collision, suggesting that some other source of energy must have converted PE to kinetic self critique assessment: 2
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14:07:26 `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 --> Before=10kg*5m/s + 2kg*0=50kgm/s After=10kg*3m/s +2kg*10m/s=50kgm/s They are the same confidence assessment: 2
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14:07:41 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 --> ok self critique assessment: 3
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14:11:00 `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 --> The time interval is the same for both forces The force exerted by the first object on the second is equal and opposite to that exerted by the second object on the first. confidence assessment: 1
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14:12:50 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 --> Because the forces exerted by the 2 objects on one another is equal and opposite and over the same time, we have equal and opposite forces acting for equal time intervals. Since the changes in momentum are equal and opposite, total momentum change in zero. The momentum before and the momentum after collision is equal. self critique assessment: 2
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