The speed of A and B immediately after collision is 5.28m/s
Explanation:
Mass of A is 6275kg
Speed of A is 6.5m/s
Mass of B is 5155kg
Speed of B is 3.8m/s
Track is frictionless.
A and B stick together.
speed of attached A and B = ?
mₐsₐ + mᵇsᵇ = (mₐ + mb) s
[tex]6275 X 6.5 + 5155 X 3.8 = ( 6275 + 5155) X s\\\\s = \frac{40787.5 + 19589}{11430}\\ \\s = \frac{60376.5}{11430}\\ \\s = 5.28m/s[/tex]
Therefore, The speed of A and B immediately after collision is 5.28m/s
The conservation of momentum is used measure the final velocity. The velocity of the two attached railroad cars after collisision is 5.28 m/s.
The velocity of the two attached railroad cars can be calculated by the conservation of momentum formula,
[tex]m_a v_a + m_b v_b = (m_a +m_b )v_f[/tex]
Where,
[tex]m_a[/tex]- mass of Railroad car A = 6275 kg
[tex]v_a[/tex]- velocity of the Railroad car A = 6.5 m/s
[tex]m_b[/tex] - mass of the railroad car B = 5155 kg
[tex]v_b[/tex] - velocity of the railroad car B = 3.8 m/s
Put the values in the formula and calculate for [tex]v _f[/tex]
[tex]6275 \times 6.5 + 5155 \times 3.8 = (19589) v_f\\\\v_f = \dfrac {40787.5 + 5155 \times 3.8} { (11430}\\\\v_f = 5.28 \rm \ m/s[/tex]
Therefore, the velocity of the two attached railroad cars after collisision is 5.28 m/s.
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