Three identical boxcars are coupled together and are moving at a constant speed of 28.0 m/sm/s 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. Required:
a. What is the speed of the four cars?
b. What percentage of the kinetic energy of the boxcars is dissipated in the collision?
c. What happened to this energy?

A) v = 21 m /s

B) 25%

C) ) on collision, this energy in the question appears in the form of the following namely; sound energy, heat energy etc

Explanation:

A) Let m be the mass of any of the cars

Thus:

mass of the three cars = 3m

Formula for kinetic energy = ½mv²

Thus, Kinetic energy of 3 identical and coupled cars = ½ x 3m x 28² = 1176 m

KE = 1176 m

Now mass of 4 coupled cars together = 4m

From conservation of linear momentum, we can find the speed of the four cars. Thus;

m1v1 = m2v2

We are told that the 3 coupled moved together with a speed of 28 m/s

Thus;

4m × v = 3m × 28

v = 3m x 28 / 4m

v = 21 m /s

B) from earlier, we saw the formula for kinetic energy. Thus, kinetic energy with of mass of 4 coupled cars together. Thus;

K = ½ x 4m x 21²

K = 882m

Loss of kinetic energy

ΔK = 1176 m - 882 m

ΔK = 294 m

Therefore, percentage of loss is;

%loss = (294 / 1176 ) x 100

%loss = 25 %

C) on collision, this energy in the question appears in the form of the following namely; sound energy, heat energy etc

the closer the planet is to the sun, the larger the gravity

your answer is, orbit around the nucleus