A car with a mass of 4 tons moves horizontally at a speed of 0.3 m. S and a platform with a mass of 3 tons at a speed of 0.5 m, c moves behind a car. Find the speed of a car and a platform after automatic coupling.
m1 = 4 tons = 4000 kilograms – wagon weight;
v1 = 0.3 m / s (meters per second) – the speed of the carriage;
m = 3 tons = 3000 kilograms – platform weight;
v2 = 0.5 m / s – platform speed.
It is required to determine v (m / s) – the speed of movement of the car and platform after coupling.
According to the condition of the problem, the car and the platform before interaction (coupling) moved in the same direction. Then, in order to determine the speed of bodies after interaction, it is necessary to use the following formula (the law of conservation of momentum):
p1 + p2 = p;
m1 * v1 + m2 * v2 = (m1 + m2) * v, from here we find that:
v = (m1 * v1 + m2 * v2) / (m1 + m2);
v = (4000 * 0.3 + 3000 * 0.5) / (4000 + 3000) = (1200 + 1500) / 7000 = 2700/7000 =
= 27/70 = 0.4 m / s (the result has been rounded to one decimal place).
Answer: after coupling, the car and the platform will move at a speed of 0.4 m / s.