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Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: s.h_math on Nov 25, 2021 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.
Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: duddy on Nov 25, 2021 Content hidden
Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: s.h_math on Nov 25, 2021 I understand but is it possible if you could write the initial equation to what these numbers were plugged in?
Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: duddy on Nov 25, 2021 Yes, sure.
The initial equation is momentum initial = momentum final. Remember that momentum is mass times velocity. mass*velocity + mass*velocity = mass*velocity Then, I applied the givens the formula above, like this: 4500(5.0) + 6500(0) = 11000(x) Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: s.h_math on Nov 26, 2021 Thanks a lot sir
Title: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: duddy on Nov 26, 2021 No worries
I'll mark this topic solved. Title: Re: A 4,50x10^3 kg railway car is moving east at a velocity of 5.0 m/s on a level frictionless track ... Post by: Mangal on Nov 26, 2021 THANKS
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