## A 1340 kg Triumph TR-8 traveling north at 43 km/hr crashes head-on into a 1850 kg Checker cab moving west at 87 km/hr.

Question

**A 1340 kg Triumph TR-8 traveling north at 43 km/hr crashes head-on into a 1850 kg Checker cab moving west at 87 km/hr. The two cars become tangled together. What is their speed a moment after the collision?**

**If east is 0 degrees, north is 90 degrees, etc., at what angle are the entangled cars traveling just after their collision?**

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## Answer ( 1 )

Let

m1 = 1340 Kg; v1 = 43 km/h; m2 = 1850 kg; v2 = 87 km/h

mpmentum of m1 = P1; momentum of m2 = P2

I’ll break up the momenta into their vertical and horizontal components. I’ll use the sign convention that up and left are positive, down and right are negative.

P1 has no horizontal component so…

P1 = (m1v12) ĵ

P2 has no vertical component so…

P2 = (m2v22) î

so

P = (m2v22) î + (m1v12) ĵ

From Pythagoras

I P I = √ [(m2v12)2 + (m1v12)2]

I P I = 14220162 Kg m/s

This is equal to the mass * velocity2 of the cars after collision.

v^2 = 14220162 / (1850 + 1340)

v = √[14220162 / 3190]

v = 66.7 m/s

The direction of the momentum is given by

arc tan [(m1v12) / (m2v22)] = arc tan (0.177) = 10 degrees north of west.

If east is 0 degrees then the angle is 180 – 10 = 170 degrees