Answer:
1.97 m/s
Explanation:
Recall that from the principle of conservation of energy, p = mv
[tex]p_{1} + p_{23} = p_{123}[/tex] ----- eqn 1
Total momentum before collision equals total momentum after collision
sub mv for p in eqn 1
[tex]m_{1}v_{1} + m_{23}v_{23} = m_{123}v_{123}[/tex] ----- eqn 2
Parameters
[tex]m_{1} = m_{2} = m_{3} = 2.65 X 10^{4} kg[/tex]
[tex]m_{23} = 5.3 X 10^{4} kg[/tex]
[tex]m_{123} = 7.95 X 10^{4} kg[/tex]
[tex]v_{1} = 3.50 m/s[/tex]
[tex]v_{23} = 1.20 m/s[/tex]
[tex]v_{123} = ?[/tex]
Where [tex]v_{123}[/tex] is the speed of the three coupled cars
substitute for all m and v in eqn 2
[tex](2.65 X 10^{4} X 3.5) + (5.3 X 10^{4} X 1.2) = (7.95 X 10^{4} X v_{123})[/tex]
[tex](9.275 X 10^{4} + 6.36 X 10^{4}) = (7.95 X 10^{4} X v_{123})[/tex]
[tex]15.635 X 10^{4} = 7.95 X 10^{4} X v_{123}[/tex]
[tex]v_{123}[/tex] = [tex]\frac{15.635 X 10^{4}}{7.95 X 10^{4}}[/tex]
[tex]v_{123}[/tex] = 1.9667 m/s
Therefore, the speed of the three cars after coupling is 1.97m/s
The speed of the three coupled cars after the first couples with the other two is 1.97 m/s.
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
