Question

The three blocks (from left to right in the figure) have masses of 3.85 kg, 0.819 kg and 8.69 kg. The leftmost block has a velocity of 3.23 m/s toward the right, the center block is stationary, and the rightmost block as a velocity of 2.72 m/s toward the left. Assuming movement toward the right is positive, what is the velocity of the center-of-mass of this system as pictured (prior to collision)?

Answer 1

The velocity of the center-of-mass for the described system is calculated by summing the product of each mass and its velocity, and then dividing by the total mass, which will result in the center-of-mass velocity.

Step-by-step explanation:

The velocity of the center-of-mass of a system can be calculated by dividing the total momentum of the system by the total mass. To find the velocity of the center-of-mass of the system described in the question, we take the sum of the product of each mass and its corresponding velocity and divide it by the sum of the masses.

Using the formula:

Vcm = (m1v1 + m2v2 + m3v3) / (m1 + m2 + m3)

Where m1 = 3.85 kg, v1 = 3.23 m/s, m2 = 0.819 kg, v2 = 0 m/s (since it's stationary), m3 = 8.69 kg, and v3 = -2.72 m/s (since it's moving to the left).

The calculation will be:

Vcm = ((3.85 kg * 3.23 m/s) + (0.819 kg * 0 m/s) + (8.69 kg * -2.72 m/s)) / (3.85 kg + 0.819 kg + 8.69 kg)

The resulting velocity is the center-of-mass velocity.