Physics reference frames

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A student runs an experiment with two carts on a low-friction track. As measured in the Earth reference frame, cart 1 (m = 0.57kg ) moves from left to right at 1.0 m/s as the student walks along next to it at the same velocity. Let the +x direction be to the right. What velocity v⃗ E2,i in the Earth reference frame must cart 2 (m = 0.19kg ) have before the collision if, in the student's reference frame, cart 2 comes to rest right after the collision and cart 1 reverses direction and travels from right to left at 0.33 m/s? What does the student measure for the momentum of the two-cart system?

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@volleyballgurl48,

In Earth reference, the total momentum looks like:

Before contact
m1v1 + m2v2
(.57kg)(1m/s) + (.19kg) X

After contact the student is moving right at 1m/s. Student sees cart 2 as stationary which means that it is also moving right at 1m/s. The student sees cart one moving backwards at .33m/s but since the student is moving forward at 1m/s, from an Earth reference, cart one is moving at -.33 + 1 = .67m/s. This means that the momentum from the Earth perspective is:
(.57kg)(.67m/s) + (.19kg)(1m/s)

Since momentum is conserved, set those equal to each other and solve for X. X = 0.01 m/s (moving to the right from an Earth reference).

You can compute the momentum from a student reference for either the before or after situation, but let's use the given data for the after situation since it doesn't depend on my calculations. From the student's view:

(.57kg)(-.33m/s) + (.19kg)( 0 ) = -0.188 kg m/s

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Physics reference frames

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