Three coupled railroad cars roll along and couple with a fourth car, which is initially at rest. These four cars roll along and couple with a fifth car initially at rest. This process continues until the speed of the final collection of railroad cars is one-fifth the speed of the initial three railroad cars. All the cars are identical. Ignoring friction, how many cars are in the final collection?
let us assume mass of each guard s M and we want is the initial velocity of three cars. So we have we do equals to one by five of we want so plan momentum, consideration In acceleration we can write even initial X equals two p two x So from here we can right three multiplied by leaving equals two nm multiplied by we please niche a number of cars and end our final number of cars So massive three car will be three and massive and car will be and so weaken Put the value of V two in the situation. So we get three women equals to end. The two is one by five of the so from here and is equal to 15 so we can say that there are total 15 cars