course Phy 201 V{gʿnONassignment #017
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15:37:19 `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 --> Since we need to use the impulse momentum theorem we can say Fave`dt=m`dv.Since we need to find Fave we divide both sides by `dt, fave=m`dv/`dt.Fave=10kg(2m/s/.03s).Fave=667newtons.Since this is the force from the second mass on the first mass it is opposite the direction of the first mass it will be Fave=-667newtons confidence assessment: 2
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15:43:06 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 --> ok self critique assessment: 3
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15:56:52 `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 --> Since we know forces are equal and opposite of each other we can say that the force of mass one on mass two will be a +667newtons. Since we know `dt of.03s and Fwe acn find the `dp. 667newtons*.03s=20kg m/s.Now if we take `dp and divide by the mass of object two we can find the `dv. 20kg m/s/2kg=10m/s confidence assessment: 2
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15:58:54 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 --> ok,Iforgot to subtract the velocity from the second mass intial velocity self critique assessment: 2
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16:08:40 `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 --> if we use the formula KE=.5mv^2 we can find the KE. The 10kg mass is KE=.5*10kg*5m/s^2,KE=125joules. The 2kg mass will be,KE=.5*2kg*0^2)KE=0joules. So the total before the collision is 125 joules. Since the 10kg mass decreases to 3m/s then we have KE=.5*10kg*3m/s^2,KE=45joules.The 2kg mass has a velocity change of 10m/s so KE=.5*2kg*10m/s^2,KE=100joules. We take mass 1 + mass 2 we get. 45joules + 100joules=145joules which 20 greater then before the collision confidence assessment: 1
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20:09:50 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 --> ok self critique assessment: 3
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20:19:55 `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 --> If we use the equation p=mv we can find momentum. If we take mass one we get, p=10kg*5m/s.p=50kg m/s. Mass number two will have no momentum because there is not velocity so the total before collision is 50kg m/s.Mass number one after collision is, p=10kg*3m/s,p=30kg m/s. Mass number two is, p=2kg*10m/s,p=20kg m/s. So are total momentum after collision is 30kg m/s+20kg m/s=50kg m/s. The momentum after collision is the same as before collision confidence assessment: 3
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20:20:46 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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20:26:59 `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 --> Forces exerted by two interacting objects on one another are equal and opposite. It follows that the impulses applied by the object to one another are equal and opposite. Hence the net change in momentum is equal and opposite. any system in which the only forces exerted on the objects in the system are exerted by other objects. Impulses in such a system are applied in the form of equal and opposite forces, resulting in equal and opposite momentum changes. confidence assessment: 3
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20:27:43 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 --> ok self critique assessment: 3
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