#$&* course Phy 231 3/2 4pm 017. collisions
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Given Solution: By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = - 667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The force of the 10kg mass on the 2kg mass is equal and opposite to the force exerted by the 2kg mass on the 10kg. Therefore, the 2kg mass is subject to a +667N force. Therefore, the impulse of the force on the 2kg mass is 667N*.03s, or 20kg*m/s. This is the change in momentum. Since initial velocity is zero, change in momentum is equal to final momentum. 20kg*m/s=2kg*v v=10m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Initially, KE of the 10kg mass = .5(10kg)(5m/s)^2, and KE of the 2kg mass = 0. Total KE altogether is 125kg*m/s^2, or 125J, before the collision. After the collision, KE of the 10kg mass = .5(10kg)(3m/s)^2, and KE of the 2kg mass = .5(2kg)(10m/s)^2. Total KE altogether is 45J+100J, or 145J, after the collision. So the total KE after the collision is actually greater than before the collision, which I don't really understand. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So how likely is it that there would be such a situation, where KE actually increases? ------------------------------------------------ Self-critique rating:
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the change in momentum is equal to the average force times the duration of the force, and the force and its duration are of equal magnitude as applied to each object, that means that the momentum of each object changes by an equal but opposite amount; however much one object loses in momentum, the other one gains. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s. By how much did its momentum change? By how much did the momentum of the 3 kg mass change? How do the forces experienced by the two masses during the collision compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 2 kg mass begins with a momentum of 2kg*6m/s, or 12kg*m/s. After collision, it has a momentum of 2kg*4m/s, or 8kg*m/s. Its momentum changes by -4kg*m/s. By definition, this means the momentum of the 3 kg mass changes by +4kg*m/s. The two masses experience equal and opposite forces. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!