#$&* course PHY 201 017. collisions
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Given Solution: By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m1 = 10 kg, v01 = 5 m/s, vf1 = 2 m/s, m2 = 2 kg, v2 = 0 m/s, 'dt = .03 s Momentum = mass * velocity Impulse = net force * change in time Momentum = impuse m*v = fAve * 'dt fAve = +667 N v = fAve * 'dt / m = (667 kg*m/s^2 * 0.03 s) / 2 kg = 10 m/s confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: for 2 kg mass KEf = 1/2mvf^2 = 0.5(2 kg)(10 m/s)^2 = 100 J KE0 = 1/2mv0^2 = 0.5(2 kg)(0 m/s)^2 = 0 J For 10 kg mass KEf = 1/2mvf^2 = 0.5(10 kg)(3 m/s)^2 = 45 J KE0 = 1/2mv0^2 = 0.5(10 kg)(5 m/s)^2 = 125 J Initial KE totals 125 J + 0 J = 125 J Final KE total 45 J + 100 J = 145 J The KE after collision is more than the KE before collision confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum = mass * velocity Impulse = net force * change in time Before momentum 1 = m1 * v1 = 10 kg * 5 m/s = 50 kg*m/s (N) momentum 2 = m2 * 'v2 = 10 kg * 0 m/s = 0 kg*m/s (N) Total Before Momentum = 50 N After momentum 1 = 10 kg * 3 m/s = 30 N momentum 2 = 2 kg * 10 m/s = 20 N Total After Momentum = 50 N The momentum before is equal to the momentum after. confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The two forces acting on each other are equal in magnitude and acting on each other in opposite direction with the same 'dt (because they are acting on each other) Therefore, they both share the same impulse and momentum = impulse. confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s. By how much did its momentum change? By how much did the momentum of the 3 kg mass change? How do the forces experienced by the two masses during the collision compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m1 = 2 kg, v01 = 6 m/s, m2 = 3 kg, vf1 = 4 m/s, v02 = 0 m/s m1 = 2 kg * (4 m/s - 6 m/s) = -4 N m2 = 3 kg * 0 m/s = 0 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: m1 = 2 kg, v01 = 6 m/s, m2 = 3 kg, vf1 = 4 m/s, v02 = 0 m/s m1 = 2 kg * (4 m/s - 6 m/s) = -4 N m2 = 3 kg * 0 m/s = 0 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!