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 force required to move the cart is divided by three or 33.33 n.

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.