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mixer [17]
3 years ago
6

Before Collision Consider a system to be one train car moving toward another train car at rest When the train cars collide, the

two cars stick together What is the total momentum of the system after the collision? O 800 kg . m/s m, = 600 kg V,= 4 m/s m = 400 kg v2 = 0 m/s 1,600 kg. m/s 0 2,400 kg • m/s 0 4,000 kg . m/s After Collision ​
Physics
2 answers:
lutik1710 [3]3 years ago
8 0

Answer:

llollk

Explanation:

erma4kov [3.2K]3 years ago
7 0

Answer:

2,400kg * m/s

Explanation:

You are missing some information in the question but the rest could be found some where else.

The question gives the masses and starting velocity of each car.

Car 1: m = 600kg and sv = 4m/s

Car 2: m 400kg and sv = 0m/s

Find the momentum of both cars.

Car 1: 600 * 4 = 2400

Car 2: 400 * 0 = 0

Add both.

2400 + 0 = 2400

Best of Luck!

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Answer:

a) t1 = v0/a0

b) t2 = v0/a0

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Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

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C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

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= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
3 years ago
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hope this helps :)

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3 years ago
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