Object A has mass m_4 = 10 kg and initial momentum F_a, j = kg. m/s. just before
ID: 1998295 • Letter: O
Question
Object A has mass m_4 = 10 kg and initial momentum F_a, j = kg. m/s. just before it strikes object B, which has mass m_s = 12 kg. Just before the collision object 8 has initial momentum F_as = kg. m/s Consider a system consisting of both objects A and B. What is the total initial momentum of this system, just before the collision? F_sus j = (19, -2, 0) kg. m/s The forces that A and B exert on each other are very large but last for a very short time. If we choose a time interval from just before to just after the collision, what is the approximate value of the impulse applied to the two-object system due to forces exerted on the system by objects outside the system? Therefore, what does the momentum principle predict that the total final momentum of the system will be, just after the collision? Just after the collision, object a is observed to have momentum F_aj = kg. m/s. What is the momentum of object B past after the collision? At this point we've learned all that we can from applying the momentum principle. Next we'll see what additional information we can obtain by using the energy principle. Before the collision, what was the magnitude of the momentum of object A? Before the collision, what was the kinetic energy of object A? Remember that you can calculate kinetic energy not only from K = (1/2)m |F|^3 but more directly from K = (1/2)|F|^2/m. Before the collision, what was the magnitude of the momentum of object B?Explanation / Answer
According to the given problem,
A)First, we are talking about Momentum, which is conserved (meaning the sum will always stay the same when comparing before and after collisions), and is additive / subtractive. The seem to be calling this the "Momentum Principle".
It's written in older style vector notation = < x, y, z >. Since the z component of each is zero, this reduces down to two-dimensional vectors -- it all happens in two dimensions, not three.
To get the total momentum, just add the two vectors (add their x, y and z components -- actually just x and y, since the z is zero, and can thus be ignored): < 15+4, -7+5> = <19, -2> kg•m/s
B) On the second part, according to the Momentum Principle, P = Fnett. So your answer is just your change in momentum. Since the Law of Conservation of Momentum states that a momentum of the given system will not change,
Your answer Fnet (or P) = <0, 0, 0> N·s
C) Momentum is conserved. So this is the same as the answer to question A (momentum before collision = momentum after)
<19, -2> kg•m/s
D) Again, momentum is conserved. So if object A has momentum of <13, 4 > the components of object B must add with it to get the total of < 19, -2 > . I could rearrange this algebraically starting from <13, 4 > + < x, y > = < 19, -2 >, but I won't. I can tell just by looking at it that 13 + x = 19 must give me 6, and 4 + y = -2 must give me -6. So the answer is < 6, -6 > kg•m/s
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