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A railroad car with a mass of 1.81 ✕ 104 kg moving at 2.88 m/s joins with two railroad cars already joined together, each with the same mass as the single car and initially moving in the same direction at 1.02 m/s. (a) What is the speed of the three joined cars after the collision? (b) What is the decrease in kinetic energy during the collision?

1 Answer

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Final answer:

(a) The final velocity of the three joined cars is approximately 1.64 m/s.

(b) The decrease in kinetic energy during the collision is approximately 1.73 x 10^7 J.

Step-by-step explanation:

To solve this problem, we can use the law of conservation of linear momentum to find the final velocity of the three joined cars. The formula for the conservation of linear momentum is:

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2 + m_3) v_f

where:

- m_1 is the mass of the first car,

- v_{1i} is the initial velocity of the first car,

- m_2 is the mass of the second car,

- v_{2i} is the initial velocity of the second car,

- m_3 is the mass of the third car (the one joining),

- v_f is the final velocity of the three joined cars.

After finding the final velocity, we can use it to calculate the decrease in kinetic energy using the formula:

ΔKE = KE_initial - KE_final

where:

KE = 1 / 2mv^2

Now, let's calculate it:

(a) Final Velocity v_f:

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2 + m_3) v_f

(1.81 x 10^4 kg x 2.88 m/s) + (2 x 1.81 x 10^4kg x 1.02 m/s) = (1.81 x 10^4 kg x 3) x v_f

Now, solve for v_f.

(b) Decrease in Kinetic Energy (ΔKE):

Δ KE = KE_initial - KE_final

ΔKE = 1 / 2m_1v_{1i}^2 + 1 / 2m_2v_{2i}^2 - 1 / 2(m_1 + m_2 + m_3)v_f^2

Now, plug in the values and calculate ΔKE.

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