Momentum with Two Carts on a Low-Friction Track

In summary, the conversation discusses a student's experiment with two carts on a low-friction track and their velocities in the Earth and student's reference frames. The student is asked to calculate the velocity of cart 2 in the Earth frame before a collision, the momentum of the two-cart system as measured by the student, and the momentum of each cart as measured by a person in the Earth frame. The conversation also mentions the use of different velocities for cart 1 after the collision. The solution involves picking a frame of reference and ensuring all numbers are for that frame only.
  • #1
xxphysics
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Homework Statement


A student runs an experiment with two carts on a low-friction track. As measured in the Earth reference frame, cart 1 (m = 0.48 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.

A. What velocity v⃗ E2,i in the Earth reference frame must cart 2 (m = 0.16 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?
B. What does the student measure for the momentum of the two-cart system?
C. What does a person standing in the Earth reference frame measure for the momentum of each cart before the collision?

Homework Equations


(m1)(Ve1x,i) + (m2)(Ve2x,i) = (m1)(Ve1x,f) + (m2)(Ve2x,f)

The Attempt at a Solution


A. (0.48 kg)(1 m/s) + (0.16 kg)(Ve2x,i) = (0.48 kg)(-(1/3)m/s) + (0.16 kg)(0 m/s)
I got Ve2x,i = -2.0 m/s which doesn't make sense and it is in fact wrong. Where am I messing up? Also our teacher said to use (1/3) for the velocity of cart 1 after the collision instead of 0.33. I did try both, 0.33 gave me 3.99 which I entered as 4 and it was wrong.
 
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  • #2
Check your frame of references. Your relationship is a mixture of velocities in the Earth frame and in the students frame. You must pick a frame and ensure all numbers are for that frame only.
 
  • #3
rpthomps said:
Check your frame of references. Your relationship is a mixture of velocities in the Earth frame and in the students frame. You must pick a frame and ensure all numbers are for that frame only.
How do you manipulate the equation to do so? I'm sorry we've just started reference frames and I don't totally understand how to make sure the equation distinguishes from the two.
 
  • #4
Well, the person is walking at 1 m/s and they notice that a cart is stationary after a collision. How fast must the cart be moving in the Earth's frame in order for the person to see this?
 
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1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and velocity, and is a vector quantity with both magnitude and direction.

2. How does momentum change when two carts collide on a low-friction track?

According to the law of conservation of momentum, the total momentum of a closed system remains constant. This means that when two carts collide on a low-friction track, the total momentum of the system before and after the collision must be the same.

3. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. In equation form, it is represented as p = mv, where p is momentum, m is mass, and v is velocity. The SI unit for momentum is kg * m/s.

4. What is the relationship between force and momentum?

Force is directly proportional to the rate of change of momentum. In other words, the greater the force applied to an object, the greater its change in momentum will be. This relationship is represented by Newton's second law of motion, F = ma, where F is force, m is mass, and a is acceleration.

5. How does friction affect the momentum of a system?

Friction is a force that opposes the motion of an object. In a low-friction track, there is minimal friction between the two carts, which allows for the conservation of momentum to hold true. However, if there is significant friction present, it can decrease the overall momentum of the system by converting some of it into heat energy.

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