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4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

cars collide and entangle. What is the resulting velocity of the wreckage?

1 Answer

2 votes

Answer:-2.61 m/s

Step-by-step explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum
p_(o) must be equal to the final momentum
p_(f):


p_(o)=p_(f) (1)

Where:


p_(o)=mV_(o)+MU_(o) (2)


p_(f)=(m+M)V_(f) (3)


m=1200 kg is the mass of the first car


V_(o)=20 m/s is the velocity of the first car, to the North


M=1400 kg is the mass of the second car


U_(o)=-22 m/s is the mass of the second car, to the South


V_(f) is the final velocity of both cars after the collision


mV_(o)+MU_(o)=(m+M)V_(f) (4)

Isolating
V_(f):


V_(f)=(mV_(o)+MU_(o))/(m+M) (5)


V_(f)=((1200 kg)(20 m/s)+(1400 kg)(-22 m/s))/(1200 kg+1400 kg) (6)

Finally:


V_(f)=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

User KFleischer
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