A platform weighing 10 tons moves at a speed of 1.5 m / s along a horizontal section of the railway track.

A platform weighing 10 tons moves at a speed of 1.5 m / s along a horizontal section of the railway track. It is caught up by a platform weighing 12 tons, moving at a speed of 3 m / s. How fast will the platforms move after hitching? How far will the platforms travel in 2 seconds after hitching? Friction should be neglected

Given: M1 = 10 kg – the mass of the first platform;
M2 is the mass of the second platform;
v = 1.5 m / s – speed of the first platform; u = 3 m / s – speed of the second platform.
Decision:
To solve the problem, we will use the law of conservation of momentum in the projection onto the axis, which is aligned with the vectors v and u.

Let’s make the equation:
M1v + M2u = (M1 + M2) U
Let us express from the equation U.
U = (M1v + M2u) / (M1 + M2).
By solving this equation, we find out the speed of the platforms after coupling.
We plug in the known data and calculate the speed.
U = (15 + 36) / 22 ≈ 2.3 m / s

Second part. Since, by condition, we can neglect friction, we have the right to assume that no forces act on any of the platforms along the horizontal axis. This means that the platforms move evenly. From this it follows that the path that they will cover in 2 seconds after coupling can be found by the formula:
S = U t
Substituting the known data there, we get:
S = 2.3 * 2 = 4.6m
Answer: 2.3 m / s and 4.6 m