Question

12. A train car (mass 15,000 kg) moving right with a velocity of 4.0 m/s collides with a train car (mass 14,000 kg) moving right with a velocity of 0.7m/s. Theythe final velocity?If it takes 3 seconds to come to a stop, how far do the train cars slide?

Answer 1

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,

`Mu_1+mu_2=(M+m)v_0_{}`

Where v₀ is the velocity of the coupled trains after the collision.

On substituting the known values,

`\begin{gathered} 15000*4.0+14000*0.7=(15000+14000)v_0 \\ v_0=(15000*4.0+14000*0.7)/((15000+14000)) \\ =2.41\text{ m/s} \end{gathered}`

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,

`d=(1)/(2)(v_0+v)t`

Where d is the distance for which the trains slide before coming to a stop.

On substituting the known values,

`\begin{gathered} d=(1)/(2)(2.41+0)*3 \\ =3.62\text{ m} \end{gathered}`

Thus the trains slide for 3.62 m after collision before coming to a stop.