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?

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 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?

kainui · Answer

Since momentum is conserved,$$m_1 v_1 + m_2 v_2 = m_1v_{\rm new} + m_2v_{\rm new}  = (m_1 + m_2)v_{\rm new}$$Plug the values in.$$m_1 = 1000 kg$$$$v_1 = 20\hat j$$$$m_2 = 1200 kg$$$$v_2 = 22\hat i$$

kainui · Answer

Do you need further help with this? Have you studied vectors?

anonymous · Answer

I have but I came up with an answer of 15.1 and an angle of 52.8 east of north

kainui · Answer

let me check.$$2200v = 20,000\hat j + 26,400\hat i$$$$v=\frac{100}{11}\hat i +12\hat j$$