Question

A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, 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 2.65 ✕ 104 kg. (a) What is the speed (in m/s) of the three coupled cars after the first couples with the other two?

Answer 1

Answer:

1.97 m/s

Explanation:

Recall that from the principle of conservation of energy, p = mv

$p_{1} + p_{23} = p_{123}$ ----- eqn 1

Total momentum before collision equals total momentum after collision

sub mv for p in eqn 1

$m_{1}v_{1} + m_{23}v_{23} = m_{123}v_{123}$ ----- eqn 2

Parameters

$m_{1} = m_{2} = m_{3} = 2.65 X 10^{4} kg$

$m_{23} = 5.3 X 10^{4} kg$

$m_{123} = 7.95 X 10^{4} kg$

$v_{1} = 3.50 m/s$

$v_{23} = 1.20 m/s$

$v_{123} = ?$

Where $v_{123}$ is the speed of the three coupled cars

substitute for all m and v in eqn 2

$(2.65 X 10^{4} X 3.5) + (5.3 X 10^{4} X 1.2) = (7.95 X 10^{4} X v_{123})$

$(9.275 X 10^{4} + 6.36 X 10^{4}) = (7.95 X 10^{4} X v_{123})$

$15.635 X 10^{4} = 7.95 X 10^{4} X v_{123}$

$v_{123}$ = $\frac{15.635 X 10^{4}}{7.95 X 10^{4}}$

$v_{123}$ = 1.9667 m/s

Therefore, the speed of the three cars after coupling is 1.97m/s

Answer 2

The speed of the three coupled cars after the first couples with the other two is 1.97 m/s.

Conservation of linear momentum

The speed of the three coupled cars is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(2.65 x 10⁴ x 3.5) + 1.2(2.65 + 2.65) x 10⁴ = v[(2.65 + 2.65 + 2.65)x 10⁴]

92,750 + 63,600 = v(79,500)

156,350 = 79,500v

v = 1.97 m/s

Thus, the speed of the three coupled cars after the first couples with the other two is 1.97 m/s.

Learn more about conservation of linear momentum here: https://brainly.com/question/7538238