A 2000.-kg limousine moving east at 10.0 m/s collides with a 1000.-kg honda moving west at 26.0 m/s. the collision is completely inelastic an takes place on an icy (frictonless road) ?

To solve this problem, we can use the principle of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.

First, calculate the total momentum before the collision: Total momentum before = (mass of limousine * velocity of limousine) + (mass of Honda * velocity of Honda)

Total momentum before = (2000 kg * 10.0 m/s) + (1000 kg * (-26.0 m/s))

Next, calculate the total mass after the collision: Total mass after = mass of limousine + mass of Honda

Total mass after = 2000 kg + 1000 kg

Now, use the conservation of momentum to find the velocity of the combined system after the collision: Total momentum after = Total mass after * final velocity

Set the initial momentum equal to the final momentum and solve for the final velocity.

Once you have the final velocity, you can determine the direction based on the signs of the velocities before the collision. If both objects move in the same direction, the final velocity will be positive; if they move in opposite directions, the final velocity will be negative.

To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Total momentum before collision: (p_{\text{initial}} = m_1v_1 + m_2v_2)

Where: (m_1 = 2000 , \text{kg}) (mass of the limousine) (v_1 = 10.0 , \text{m/s}) (velocity of the limousine) (m_2 = 1000 , \text{kg}) (mass of the Honda) (v_2 = -26.0 , \text{m/s}) (velocity of the Honda, negative because it's moving west)

Total momentum after collision: (p_{\text{final}} = (m_1 + m_2) \cdot v_f)

Where: (v_f) is the final velocity of the combined mass after the collision.

Using the conservation of momentum equation: (p_{\text{initial}} = p_{\text{final}})

(m_1v_1 + m_2v_2 = (m_1 + m_2) \cdot v_f)

Substituting the given values: (2000 , \text{kg} \times 10.0 , \text{m/s} + 1000 , \text{kg} \times (-26.0 , \text{m/s}) = (2000 , \text{kg} + 1000 , \text{kg}) \times v_f)

(20000 , \text{kg} \cdot \text{m/s} - 26000 , \text{kg} \cdot \text{m/s} = 3000 , \text{kg} \cdot \text{m/s} \times v_f)

(-6000 , \text{kg} \cdot \text{m/s} = 3000 , \text{kg} \cdot \text{m/s} \times v_f)

(v_f = \frac{-6000 , \text{kg} \cdot \text{m/s}}{3000 , \text{kg}})

(v_f = -2.0 , \text{m/s})

Therefore, the final velocity of the combined mass after the collision is (-2.0 , \text{m/s}), which means it's moving westward.