Help? :( A 9,300 kg. a railroad car traveling at a velocity of 15m/s strikes a second boxcar at rest. If the two cars stick together and move off with a velocity of 6m/s, what is the mass of the second car?

The mass of the second railroad car which was initially at rest is 13950kg.

Given the data in the question;

Mass of railroad car2;

Since railroad car2 was initially at rest,

After the strike, the wo cars stick together and moved

To determine the mass of the second railroad car, we use conservation of linear momentum:

We substitute our given values into the equation

Therefore, the mass of the second railroad car which was initially at rest is 13950kg.

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m2 = 13950 [kg]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which is defined as the product of mass by velocity, and can be determined by means of the following expression:

P = m*v

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now that momentum is conserved, we need to analyze the conditions before the collision and after the collision. In this way we must propose the following equation:

ΣPbefore = ΣPafter

(m1*v1) + (m2*v2) = (m1+m2)*v3

where:

m1 = mass of the railroad car = 9300 [kg]

v1 = velocity of the railroad car before the collision = 15 [m/s]

m2 = mass of the box car [kg]

v2 = velocity of the boxcar before the collision = 0 (the car is at rest)

(m1 + m2) = combined mass of the cars, they stick together after the collision [kg]

v3 = combined velocity of the cars after the collision = 6 [m/s]

(9300*15) + (m2*0) = (m1+m2)*6

139500 = (6*9300) + 6*m2

83700 = 6*m2

m2 = 13950 [kg]

0.11 m/s and 0.28 m/s should be correct

reserve is the answer