A car with a mass of 3 tons, moving along a horizontal section of the road, collides with an automatic
A car with a mass of 3 tons, moving along a horizontal section of the road, collides with an automatic coupler with a fixed platform weighing 17 tons. The speed of the joint movement of a car and a platform is 0.3 m / s. What is the speed of the car before colliding with the platform?
mw = 3 t = 3000 kg.
mp = 17 t = 17000 kg.
V “= 0.3 m / s.
Vp = 0 m / s.
Since during coupling the car interacts only with the platform, they can be considered a closed system of interacting bodies, for which the law of conservation of momentum is valid.
mw * Vw + mp * Vp = mw * Vw “+ mp * Vp”.
Before the hitch, the platform was at rest, so Vp = 0 m / s.
Since after coupling the car with the platform moves at the same speed, then Vw “= Vp” = V “.
mw * Vw = (mw + mp) * V “.
Vv = (mv + mp) * V “/ mv.
Vw = (3000 kg + 17000 kg) * 0.3 m / s / 3000 kg = 2 m / s.
Answer: before the coupling, the car was moving at a speed of Vw = 2 m / s.