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Our teacher gave us several options, but the only ones that don't violate conservation of momentum are:

Cart A and Cart B move together with the same velocity

Car A and Car B move with Car B having a greater velocity than Car A

Car A stops and Car B moves with the same velocity that Car A had before the collision

The third option is the only one that definitely does not violate conservation of momentum, the other two options could either violate it or not violate it. Can anyone explain what happens in this situation and why? We are told that rolling friction is negligible here and should be ignored.

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For simplicity I'm assigning mass $2kg$ to each and initial velocity of the moving cart to be $2ms^{-1}$.You can find particular cases for all your options where momentum will be conserved , suppose both carts stick and move with velocity $1ms^{-1}$ , and obviously the last option also conserves momentum, so you need more data to determine the end result of the collision , here comes the use of coefficient of restitution ,it is an experimentally determined quantity which is nothing but the ratio of relative velocities of separation and approach , for elastic collision , the ones in which kinetic energy is also conserved its value is 1.You can now obtain a second relation from this data which will help you determine velocities after collision and conclude the last option to be correct.

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Assuming your question refers to elastic collisions, from UCSB website: http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/24.12.html If we have two carts, their total momentum before and after collision is m1v1i + m2v2i = m1v1f + m2v2f, where the subscripts 1 and 2 refer to the two carts, and i and f refer to velocity before collision and after collision, respectively. Since kinetic energy is conserved in an elastic collision, we also have (1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + m2v2f2. We can rewrite these equations as m1(v1i - v1f) = m2(v2f - v2i), and m1(v1i2 - v1f2) = m2(v2f2 - v2i2). Assuming that the initial and final velocities for each cart are not equal, if we divide the second equation by the first, we have v1i + v1f = v2f + v2i, or v1i - v2i = v2f - v1f. We see that in this type of collision, the relative velocity of the two carts after collision equals the opposite of their relative velocity before collision.

From the last equation(s), we can get expressions for the final velocities, v1f = v2f + v2i - v1i, and v2f = v1i + v1f - v2i. Inserting these into the momentum equation above gives: v1f = [(m1 - m2)/(m1 + m2)]v1i + [2m2/(m1 + m2)]v2i, and v2f = [2m1/(m1 + m2)]v1i + [(m2 - m1)/(m1 + m2)]v2i.

These equations indicate that there are several interesting special cases. First, if both carts have equal mass, one term in each equation disappears, and the other equals 1, giving v1f = v2i, and v2f = v1i. Thus, the two carts exchange momentum, each going off with the velocity that the other had had before the collision.

Next, we examine what happens when one of the carts is initially at rest, say m2. Now, v2i = 0, and v1f = [(m1 - m2)/(m1 + m2)]v1i, and v2f = [2m1/(m1 + m2)]v1i. If the two carts have the same mass, we have v1f = 0, and v2f = v1i. Thus, after collision, the first cart stops while the second one takes off with the velocity that the first one had had before collision.

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  • $\begingroup$ what is an elastic collision? $\endgroup$
    – Jodast
    Jun 2, 2019 at 18:57
  • $\begingroup$ It’s an idealization to make the math simple. Kinetic energy is conserved, the carts do not stick together. $\endgroup$
    – 511mev
    Jun 2, 2019 at 21:04

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