Physics, 12.02.2020 22:26 Baby010391

Railroad car A, with mass 6275 kg, is traveling east at 6.5 m/s along a straight track. It strikes railroad car B, with mass 5155 kg, which is also traveling east along the same track at 3.8 m/s. The two railroad cars stick together. Assume the track is frictionless.

How Fast do the two attached railroad cars move immediately after the collision?

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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,

m_a v_a + m_b v_b = (m_a +m_b )v_f

Where,

m_a - mass of Railroad car A = 6275 kg

v_a - velocity of the Railroad car A = 6.5 m/s

m_b - mass of the railroad car B = 5155 kg

v_b - velocity of the railroad car B = 3.8 m/s

Put the values in the formula and calculate for v _f

$6275 \times 6.5 + 5155 \times 3.8 = (19589) v_f\\\\v_f = \dfrac {40787.5 + 5155 \times 3.8} { (11430}\\\\v_f = 5.28 \rm \ m/s$

Therefore, the velocity of the two attached railroad cars after collisision is 5.28 m/s.

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Answer from: hella33revj

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

$6275 X 6.5 + 5155 X 3.8 = ( 6275 + 5155) X s\\\\s = \frac{40787.5 + 19589}{11430}\\ \\s = \frac{60376.5}{11430}\\ \\s = 5.28m/s$

Therefore, The speed of A and B immediately after collision is 5.28m/s

Answer from: Quest

i sometimes feel that way too. i feel that i should hurt someone because they hurt me, but i wouldnt do it. that would make things so much worse i have so much experience with this stuff

explanation:

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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Railroad car A, with mass 6275 kg, is traveling east at 6.5 m/s along a straight track. It strikes r...