Respuesta :
Answer:
one half as large , initial velocity is two times larger
Explanation:
Momentum is conserved.
p₁ + p₂ = p₁' + p₂'
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
m₁ = m₂ =m , v₂= 0
v₁' =v₂'
mv₁ = 2mv₁'
v₁ = 2v₁'
The velocity of the combined cars after the collision is half the velocity of car1 before the collision.
Conservation of Momentum
According to the law of conservation of momentum when two bodies collide with each other then their momentum is conserved if no external force acts on the system. therefore, the combined momentum of the two objects before collision will be equal to the combined momentum of the two objects after the collision.
the momentum of the two objects before the collision
= momentum of the two objects after the collision
Given to us
- A moving freight car runs into an identical car at rest on the track.
- The cars couple together.
For Car 1,
mass of car 1 = [tex]m_1[/tex]
Velocity of car 1 = [tex]v_1[/tex]
For car 2,
as given the cars are identical. therefore,
the mass of car2 = [tex]m_2 = m_1[/tex]
Also, the second car is at rest,
the velocity of car 2 [tex]= v_2[/tex] = 0
After collision
the mass after the collision will be combined, as it is given that the cars couple together. therefore,
[tex]M = m_1+m_2[/tex]
and the velocity of the cars after the collision is V.
Conservation of Momentum
momentum of car1 + momentum of car2
= Momentum of Both cars together
[tex]m_1v_1+m_2v_2=MV\\ [/tex]
as the second car is at rest, [tex]v_2[/tex] = 0
[tex]m_1v_1+m_2(0)=MV\\ m_1v_1=MV\\[/tex]
substituting the value of M,
[tex]m_1v_1=(m_1+m_2)V\\[/tex]
we know [tex]m_2 = m_1[/tex],
therefore,
[tex]m_1v_1=(2m_1)V\\\\ m_1v_1=2\times m_1 \times V\\\\ V=\dfrac{m_1v_1}{2\times m_1}\\\\\\ V= \dfrac{v_1}{2}[/tex]
Hence, the velocity of the combined cars after the collision is half the velocity of car1 before the collision.
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