Center of mass: A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts?

Question

Quizplus · Accepted Answer

The conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision in the absence of any external forces.

Let's define the positive direction as the direction of the 310-g cart, and the negative direction as the direction of the 260-g cart. Using the conservation of momentum, we have:

(m1*v1) + (m2*v2) = (m1+m2) * Vcom

where m1 and v1 are the mass and velocity of the 310-g cart, m2 and v2 are the mass and velocity of the 260-g cart, and Vcom is the velocity of the center of mass of the two carts.

Plugging in the given values, we get:

(0.310 kg) * (1.25 m/s) + (0.260 kg) * (-1.33 m/s) = (0.310 kg + 0.260 kg) * Vcom

0.38875 kg*m/s - 0.3458 kg*m/s = 0.570 kg * Vcom

0.04295 kg*m/s = 0.570 kg * Vcom

Vcom = 0.04295 kg*m/s / 0.570 kg

Vcom = 0.0732 m/s

Therefore, the speed of the center of mass of the two carts is 0.0732 m/s, which is answer choice B.

Center of Mass: a 310-G Air Track Cart Is Traveling