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A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. The first reaches the second two and all couple together. The mass of each is 3.05 ✕ 104 kg.(a)What is the speed (in m/s) of the three coupled cars after the first couples with the other two? (Round your answer to at least two decimal places.)Incorrect: Your answer is incorrect.What is the momentum of the two coupled cars? What is the momentum of the first car in terms of its mass and initial speed? Note all cars are initially traveling in the same direction. Apply conservation of momentum to find the final speed. m/s(b)Find the (absolute value of the) amount of kinetic energy (in J) converted to other forms during the collision.J

Respuesta :

Answer:

momentum of the coupled cars V =  1.77 m/s

kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J

Explanation:

given

m₁ =3.05 × 10⁴kg

u₁ =2.90m/s

m₂=6.10× 10⁴kg

u₂=1.20m/s

using law of conservation of momentum

m₁u₁ + m₂u₂ = (m₁ + m₂) V

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V

V =  1.617×10⁵/9.15×10⁴

V = 1.77m/s

K.E =1/2mV²

ΔK.E = K.E(final) - K.E(initial)

ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔK.E = -2.892×10⁴J

The final speed is 1.77 m/s

The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]

2.892×10⁴J of energy is converted.

Inelastic collision:

Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.

Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg

The initial speed of the railroad boxcar is u₁ = 2.90m/s

Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg

And the initial speed be u₂ = 1.20m/s

The initial momentum of the first car is:

m₁u₁ = 3.05 × 10⁴ × 2.90 =  8.84 × 10⁴ kgm/s

The initial momentum of the coupled car is:

m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s

Let the final speed after all the boxcars are coupled be v

From the law of conservation of momentum, we get:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv

v =  1.617×10⁵/9.15×10⁴

v = 1.77m/s

The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:

ΔKE = K.E(final) - K.E(initial)

ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔKE = -2.892×10⁴J

Learn more about inelastic collision:

https://brainly.com/question/13861542?referrer=searchResults

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