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ValentinkaMS [17]
3 years ago
12

Two identical objects in outer space have a head-on collision and stick together. If, before the collision, one had been moving

at 2 m/s and the other at 1 m/s, their combined speed after the collision would be?
Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

1.5 m/s

Explanation:

Momentum is conserved and conservation of momentum is

p₁ + p₂ = p'₁ + p'₂

or

m₁v₁ + m₂v₂ = m₁v'₁  + m₂v'₂

In our problem, after collision v'₁ will be equal to v'₂.

Since objects are identical m₁ = m₂

m(v₁+ v₂) = 2m x v'₁

(2m/s + 1m/s) = 2v'₁

v'₁ = v'₂ = 1.5 m/s

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You have to take note of the individual directions of the plane. Since one is heading east, and the other is heading west, the planes are heading at opposite directions. So, it means that their distance between each other would be equal to 1,200 miles which accounts for the sum of their individual distances. The equation is as follows:

Total Distance = Distance of slower plane + Distance of faster plane
1,200 miles = st + (30+s)(t)
where
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3 years ago
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lilavasa [31]

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6 0
3 years ago
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Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

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Explanation:

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6 0
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