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# A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ...

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1 months ago
 A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track when it collides with a stationary 6.50x10^3 kg caboose. If the two cars lock together upon impact, how fast are they moving after the collision. Read 196 times 6 Replies
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duddyduddy
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Posts: 7714
1 months ago
 - Master of Science in Biology- Bachelor of Science (Biology)- Bachelor of Education

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1 months ago
 I understand but is it possible if you could write the initial equation to what these numbers were plugged in?
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Staff Member
1 months ago
 Yes, sure.The initial equation is momentum initial = momentum final. Remember that momentum is mass times velocity.mass*velocity + mass*velocity = mass*velocityThen, I applied the givens the formula above, like this:4500(5.0) + 6500(0) = 11000(x)
 - Master of Science in Biology- Bachelor of Science (Biology)- Bachelor of Education
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1 months ago
 Thanks a lot sir
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Staff Member
1 months ago
 No worriesI'll mark this topic solved.
 - Master of Science in Biology- Bachelor of Science (Biology)- Bachelor of Education
wrote...
A month ago
 THANKS