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Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the final velocities
Question
Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of objects 1 and 2? Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express the velocities in terms of v.
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2021-08-26T02:11:25+00:00
2021-08-26T02:11:25+00:00 1 Answers
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Answers ( )
Answer:
(v₁, v₂) = [(v/3), (4v/3)]
Or
(v₁, v₂) = (v, 0)
Explanation:
In elastic collisions, the momentum and kinetic energy is usually conserved.
The momentum before collision = momentum after collision
And
Kinetic energy before collision = Kinetic energy after collision
Momentum of object 1 before collision = (2m)v = 2mv
Momentum of object 2 before collision = (m)(0) = 0
Momentum of object 1 after collision = (2m)(v₁) = 2mv₁
Momentum of object 2 after collision = (m)(v₂) = mv₂
So, we have
2mv = 2mv₁ + mv₂
2v = 2v₁ + v₂
v₂ = 2v – 2v₁ (eqn 1)
Kinetic energy of object 1 before collision = (1/2)(2m)(v²) = mv²
Kinetic energy of object 2 before collision = (1/2)(m)(0²) = 0
Kinetic energy of object 1 after collision = (1/2)(2m)(v₁²) = mv₁²
Kinetic energy of object 2 after collision = (1/2)(m)(v₁²) = (mv₂²/2)
So, we have,
mv² = mv₁² + (mv₂²/2)
v² = v₁² + (v₂²/2)
2v² = 2v₁² + v₂² (eqn 2)
Substitute (v₂ = 2v – 2v₁) from (eqn 1) into (eqn 2)
2v² = 2v₁² + (2v – 2v₁)²
2v² = 2v₁² + 4v² – 8vv₁ + 4v₁²
6v₁² – 8vv₁ + 2v² = 0
6v₁² – 6vv₁ – 2vv₁ + 2v² = 0
6v₁(v₁ – v) – 2v(v₁ – v) = 0
(6v₁ – 2v)(v₁ – v) = 0
6v₁ = 2v or v₁ = v
v₁ = (v/3) or v₁ = v
If v₁ = (v/3)
From (eqn 1)
v₂ = 2v – 2v₁
v₂ = 2v – 2(v/3)
v₂ = 2v – (2v/3)
v₂ = (4v/3)
If v₁ = v,
From eqn 1,
v₂ = 2v – 2v₁
v₂ = 2v – 2v = 0
(v₁, v₂) = [(v/3), (4v/3)]
Or
(v₁, v₂) = (v, 0)