course Phy 231 G??b??????V???assignment #017
......!!!!!!!!...................................
20:13:42 `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.
......!!!!!!!!...................................
RESPONSE --> We know that Fnet * 'dt = Impulse = 'dp; therefore, if we want to find the Fnet we solve the equation to look like this Fnet = 'dp/ 'dt. This means that with the given information we can easily find the average force involved by taking the momentum of the first object, which also represents the total momentum present due to being the only mass with velocity, and find that (10 kg * 5 m/s)/ .03 s = 1666.67 N. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:15:45 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.
......!!!!!!!!...................................
RESPONSE --> Ok, I used the initial velocity of the first mass instead of the final velocity vector. I misunderstood the application of which velocity was relevant to the force; however, I know realize that the force during impact due to the combined masses and changes in velocity was the relevant factor. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:24:14 `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.
......!!!!!!!!...................................
RESPONSE --> This force would now be determined by using the 3 m/s of the first object after collision giving us a F = ( 10 kg * 3 m/s)/ 0.03 s = 1000 N. Also we can find the final velocity of the 2 kg mass with the knowledge that the momentum of a system is conserved; therefore, m1 * v01 + m2 * v02 = m1 * vf1 + m2 * vf2, which when we fill in the known values we find that (10 kg * 5 m/s) + ( 2 kg * 0 m/s) = ( 10 kg * 3 m/s) + ( 2 kg * vf2). This gives us 50 kg*m/s = 30 kg*m/s + 2 kg * vf2, which gives us 20 kg*m/s = 2 kg * vf2, which find will give us the final velocity of the second object vf2 = 10 m/s. confidence assessment:
.................................................
......!!!!!!!!...................................
20:39:38 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.
......!!!!!!!!...................................
RESPONSE --> I can't believe that I forgot to apply Newtons third law, instead of what I did with the my previous calculation. Lesson learned. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:41:31 `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?
......!!!!!!!!...................................
RESPONSE --> We know for fact that the total moment is conserved; therefore, since value of momentum is determined by the mass and velocity, which are the same values used in the determination of KE that the total KE is indeed the same before and after the collision. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:43:26 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.
......!!!!!!!!...................................
RESPONSE --> Oh, I believe incorrectly that simply because the values were the same that the conservation would be maintained; however, I now see that since the variables are manipulated differently that the KE of the system does indeed change while the momentum is conserved. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:44:37 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
......!!!!!!!!...................................
RESPONSE --> We used the Impulse-Momentum theory to calculate the final velocity of the second mass of 2 kg by making the assumption that the before and after momentum values are indeed the same. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:44:47 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.
......!!!!!!!!...................................
RESPONSE --> Ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:47:35 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
......!!!!!!!!...................................
RESPONSE --> It implies that the changes in velocity, and in some cases mass, of one object involved in a collision will be dependent upon each other to maintain a fixed amount. Simply that when velocity of one mass decreases and assuming both masses are constant that the velocity of the other object must increase to compensate by an amount dependant upon the mass of the second object in relation to the first and vice versa. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:03:52 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.
......!!!!!!!!...................................
RESPONSE --> Ok self critique assessment: 3
.................................................