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:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

$E=0.5mV^2$

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

$E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2$

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

$E2=0.5(3)(m)V^2$

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

$0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2$

$V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3}$

$V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s$

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

Answer 2

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