Three identical boxcars are coupled together and are moving at a constant speed of 28.0 on a level, frictionless track. They collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be neglected. Part AWhat is the speed of the four cars? Part BWhat percentage of the kinetic energy of the boxcars is dissipated in the collision? Part CWhat happened to this energy?

Question

Physics

Selina Gould

21 December, 04:00

Three identical boxcars are coupled together and are moving at a constant speed of 28.0 on a level, frictionless track. They collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be neglected. Part AWhat is the speed of the four cars? Part BWhat percentage of the kinetic energy of the boxcars is dissipated in the collision? Part CWhat happened to this energy?

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Answers (1)

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Caroline Rivers · Answer 1

Let m be the mass of each car

mass of three cars = 3m

Kinetic energy of 3 coupled cars = 1/2 x 3m x 28²

= 1176 m

mass of 4 coupled cars = 4m

Its velocity v = 3m x 28 / 4m (using conservation of momentum law)

= 21 m / s

Kinetic energy of 4m mass

= 1/2 x 4m x 21²

= 882 m

Loss of energy

= 1176 m - 882 m = 294 m

percentage of loss

= (294 / 1176) x 100

= 25 %

This energy appears in the form of sound, heat etc on collision.