Ask Question
27 July, 23:47

A 1000-kg car approaches an intersection traveling north at 20.0 m/s. A 1200-kg car approaches the same intersection traveling east at 22.0 m/s. The two cars collide at the intersection and lock together. Ignoring any external forces that act on the cars during the collision, what is the velocity of the cars immediately after the collision?

+3
Answers (1)
  1. 28 July, 00:23
    0
    This problem applies momentum balance over the cars collided

    m1v1y + m2v2y = (m1 + m2) v'y

    0 + (1500 kg) (25.0 m/s) = (2500 kg) v'y

    v'y = 15.0 m/s

    Find the vector sum of the velocities:

    v^2 = v'x^2 + v'y^2 = (8.00 m/s) ^2 + (15.0 m/s) ^2

    v = 17.0 m/s

    find the angle: θ = tan^-1 (15.0 m/s / 8.00 m/s) = 61.9°

    now, the momentum: p = mv = (2500 kg) (17.0 m/s)

    p = 4.25 x 10^4 kg x m/s at 61.9° N of E
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 1000-kg car approaches an intersection traveling north at 20.0 m/s. A 1200-kg car approaches the same intersection traveling east at 22.0 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers