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A 313-g air track cart is traveling at 1.52 m/s, and a 306-g cart is traveling in the opposite direction at 1.02 m/s. What is the velocity (sign matters) of the center of mass of the two carts?

Answer :

The velocity of the center of mass of the two carts is approximately 0.265 m/s.

To find the velocity of the center of mass of the two carts, we can use the principle of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.

The momentum (p) of an object is given by the product of its mass (m) and velocity (v):

p = m * v

Let's denote the mass of the first cart as m₁, the velocity of the first cart as v₁, the mass of the second cart as m₂, and the velocity of the second cart as v₂.

The initial momentum of the system is the sum of the individual momenta:

p_initial = m₁ * v₁ + m₂ * v₂

Given that m₁ = 313 g = 0.313 kg, v₁ = 1.52 m/s, m₂ = 306 g = 0.306 kg, and v₂ = -1.02 m/s (opposite direction), we can substitute these values into the equation for p_initial:

p_initial = (0.313 kg) * (1.52 m/s) + (0.306 kg) * (-1.02 m/s)

Calculating this, we find:

p_initial ≈ 0.476 kg·m/s - 0.31212 kg·m/s ≈ 0.16388 kg·m/s

Since momentum is conserved, the total momentum after the collision is also 0.16388 kg·m/s.

To find the velocity of the center of mass (v_cm), we divide the total momentum by the total mass (m_total) of the system:

v_cm = p_total / m_total

The total mass (m_total) is the sum of the individual masses:

m_total = m₁ + m₂

Substituting the given values:

m_total = 0.313 kg + 0.306 kg

Calculating this, we find:

m_total ≈ 0.619 kg

Now we can calculate the velocity of the center of mass:

v_cm = (0.16388 kg·m/s) / (0.619 kg)

Calculating this, we find:

v_cm ≈ 0.265 m/s

To know more about velocity refer here:

https://brainly.com/question/30559316#

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Rewritten by : Brahmana