A1.20 × 104 kg railroad car moving at 7.70 m/s to the north collides with and sticks to another railroad car of the same mass that is moving in the same direction at 1.84 m/s. what is the velocity of the joined cars after the collision?

4.77m/s

Explanation:

According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.

Let m1 and m2 be the mass of the first and second railroad cars

u1 and u2 be the velocities of the railroad cars

v be the common velocity

Using the formula

m1u1 + m2u2 = (m1 +m2)

m1 = 1.20×10⁴kg

m2 = 1.20×10⁴kg (body of same mass)

u1 = 7.70m/s

u2 = 1.84m/s

v = ?

(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v

9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v

11.45×10⁴ = 2.4×10⁴v

v = 11.45×10⁴/2.4×10⁴

v = 4.77m/s

The velocity of the cars after collision will be 4.77m/s

im guessing the west of station b