Afreight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car, which is initially at rest. if the speed of the cars after they couple together is 36 km/h, what is the mass of the second car?

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right