the Given,
The mass of the 1st train car, M=15000 kg
The mass of the 2nd train car, m=14000 kg
The initial velocity of the 1st train car, u₁=4.0 m/s
The initial velocity of the 2nd train car, u₂=0.7 m/s
The time it takes for the trains to come to a stop, t=3 s
From the law of conservation of momentum, the total momentum of a system always remains constant.
Thus,
[tex]Mu_1+mu_2=(M+m)v_0_{}[/tex]Where v₀ is the velocity of the coupled trains after the collision.
On substituting the known values,
[tex]\begin{gathered} 15000\times4.0+14000\times0.7=(15000+14000)v_0 \\ v_0=\frac{15000\times4.0+14000\times0.7}{(15000+14000)} \\ =2.41\text{ m/s} \end{gathered}[/tex]Thus the final velocity of the trains is 2.41 m/s
Given that the trains come to a stop after 3 seconds of the collision.
That is the final velocity of the trains after the 3 seconds of collision is v=0 m/s
From the equation of motion,
[tex]d=\frac{1}{2}(v_0+v)t[/tex]Where d is the distance for which the trains slide before coming to a stop.
On substituting the known values,
[tex]\begin{gathered} d=\frac{1}{2}(2.41+0)\times3 \\ =3.62\text{ m} \end{gathered}[/tex]Thus the trains slide for 3.62 m after collision before coming to a stop.
