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 = 0.48 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 +xdirection be to the right.

Part A What velocity v 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? Express your answer with the appropriate units.

Part B What does the student measure for the momentum of the two-cart system? Express your answer with the appropriate units.

Part C What does a person standing in the Earth reference frame measure for the momentum of each cart before the collision? Enter your answers numerically separated by a comma.

Explanation / Answer

From the law of conservation of momentum, the total momentum before and after the collision remains constant.

Before collision, we have, m1v1+m2v2 = (0.48*1)+(0.16*x)

Now after collision, the student sees the cart 2 as stationary. It means the cart 2 has velocity same as that of the student. Therefore, after collision the velocity of cart 2 is 1m/s. The student sees cart 1 moving opposite at .33m/s but since the student is moving forward at 1m/s, from an Earth reference, cart 1 is moving at -.33 + 1 = .67m/s. This means that the momentum from the Earth perspective is:
(.48kg)(.67m/s) + (.16kg)(1m/s) = 0.4816 which should be equal to the momentum before collision.

Therefore, 0.16x = 0.4816 - 0.48 => x = 0.01m/s (moving along the positive x axis from left to right)

Part B: The momentum from student's reference for the system can be given by calcuating the total momentum after collision from the given data.

(0.48*(-0.33))+(0.16*0) = -0.1584 Kg m/s

Part C: Before collision:

Momentum of cart 1 in earth's reference frame is 0.48 * 1 = 0.48 Kg m/s

Momentum of cart 2 in earth's refernce frame is 0.16 * 0.01 = 0.0016 Kg m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.