A student runs an experiment with two carts on a low-friction track. As measured

Question

A student runs an experiment with two carts on a low-friction track. As measured in the Earth reference frame, cart 1 (m = 0.54 kg ) 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 vE2,i in the Earth reference frame must cart 2 (m = 0.18 kg ) have before the collision if, in the student's reference frame, cart 2 comes to rest right after the collision and cart 1 travels from right to left at 0.33 m/s ?

What does the student measure for the momentum of the two-cart system?

What does a person standing in the Earth reference frame measure for the momentum of each cart before the collision?

Explanation / Answer

Solution:

Let us go to the basics first.

In Earth reference, the total momentum looks like:

Before contact
m1v1 + m2v2
(0.54 kg)(1m/s) + (0.18kg) 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:
(0.54kg)(.67m/s) + (.18kg)(1m/s)

Since momentum is conserved, set above two equations 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. From the student's view:

(.54kg)(-.33m/s) + (.18kg)( 0 ) = -0.1782 kg m/s                 (Answer)

In the Earth reference frame measure for the momentum of each cart before the collision:

m1v1 =(0.54 kg)(1m/s) = 0.54 kg m/s (Answer)

m2v2 = (0.18kg) 0.01 m/s = 0.0018 kg m/s (Answer)

Thanks!!!

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