#$&* course PHY 121 7/22 8:30 017. collisions
.............................................
Given Solution: By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have used the correct formula and managed to get the correct answer. At first, I didn't understand how I got the correct answer (which I didn't trust) with the 2 equations, but I can see that solving for acceleration first gave me the `dv and `dt. Plugging acceleration into the F = m * a then gave me the m & F parts, so I had the pieces, just not in one neat package. I understand how it should have worked, though. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fave * `dt = m `dv `dv = Fave * `dt / m `dv = 666.7 N * .03 s / 2 kg `dv = 20.001 Ns / 2 kg `dv = 10 m/s The initial velocity for the object was 0 m/s, so the final velocity must be 10 m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique rating: 2 ********************************************* 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: The total kinetic energy after the collision should equal the total kinetic energy before the collision. Before: KE = .5 m v^2 KE = .5 (10 kg) (5 m/s)^2 KE = .5 (10 kg ) (25 m^2/s^2) KE = 125 Joules KE = .5 (2kg) (0 m/s)^2 KE = 0 Joules 125 Joules + 0 Joules = 125 Joules before the collision. After: KE = .5mv^2 KE = .5 (10 kg) (3 m/s)^2 KE = .5 (10 kg) (9 m^2/s^2) KE = 45 Joules KE = .5 (2 kg) (10 m/s)^2 KE = .5 (2 kg) (100 m^2/s^2) KE = 100 Joules 45 Joules + 100 Joules = 145 Joules I thought they would match. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum before: m * v 10 kg * 5 m/s = 50 kg m/s 2 kg * 0 m/s = 0 kg m/s 50 kg m/s + 0 kg m/s = 50 kg m/s Momentum after: 10 kg * 3 m/s = 30 kg m/s 2 kg * 10 m/s = 20 kg m/s 30 kg m/s + 20 kg m/s = 50 kg m/s They are the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* 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 impulse-momentum theorum tells us that the change in the momentum of an object is equal to the product of the force on the object times the time over which it acts. Forces exerted are equal and opposite, so the changes in the momentum will be equal and opposite. Therefore, the total momentum of the 2 objects will be unchanged in the collision. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q006. A 2 kg mass moving at 6 m/s collides with a 3 kg mass, after which the 2 kg mass is moving at 4 m/s. By how much did its momentum change? By how much did the momentum of the 3 kg mass change? How do the forces experienced by the two masses during the collision compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum before: 2 kg * 6 m/s = 12 kg m/s Momentum after: 2 kg * 4 m/s = 8 kg m/s 12 kg m/s - 8 kg m/s = 4 kg m/s Momentum before: 3 kg * 0 m/s = 0 kg m/s Momentum after: 4 kg m/s 0 kg m/s + 4 kg m/s = 4 kg m/s The momentum of the 2 kg object decreased by 4 kg m/s and the momentum of the 3 kg object increased by 4 kg m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!