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s.h_math s.h_math
wrote...
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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 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.
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duddyduddy
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1 months ago
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Hi s.h_math,

Railway car mass: 4500 kg
Railway car speed: 5.0 m/s

Stationary railway car mass: 6500 kg
Stationary railway car speed: 0 m/s

Combined mass of two cars: 4500 + 6500 = 11000 kg
Combined speed of two cars: unknown (let's call it x)

Using the law of conservation of linear momentum problem:

4500(5.0) + 6500(0) = 11000(x)

4500(5.0) + 0 = 11000(x)

Divide both sides by 11000:

22500 / 11000 = x

x = 2.0 m/s [east]

How is that?
1
- Master of Science in Biology
- Bachelor of Science (Biology)
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wrote...
1 months ago
I understand but is it possible if you could write the initial equation to what these numbers were plugged in?
wrote...
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*velocity

Then, 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
wrote...
1 months ago
Thanks a lot sir
wrote...
Staff Member
1 months ago
No worries

I'll mark this topic solved.
- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
wrote...
A month ago
THANKS
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