A spring compressed with a thread is placed between the carts. If the thread is burned

A spring compressed with a thread is placed between the carts. If the thread is burned, then as a result of interaction with the spring, the carts will start moving. How will the speeds acquired by the bogies differ if the left bogie weighs 7.5 kg and the right bogie weighs 1.5 kg?

Given:

m1 = 7.5 kilograms – the mass of the first cart;

m2 = 1.5 kilograms – the mass of the second cart.

It is required to determine v2 / v1 – how the speed of the bogies will differ.

Since the condition of the problem is not specified, we neglect the forces of resistance to the movement of the trolleys. Then, according to the law of conservation of momentum (momentum), the impulses of the carts after interaction with the spring will be equal:

p1 = p2;

m1 * v1 = m2 * v2;

v2 = m1 * v1 / m2;

v2 / v1 = m1 / m2 = 7.5 / 1.5 = 75/15 = 5, or v2 = 5 * v1.

Answer: the speed of the second bogie (with a lower mass) will be 5 times greater than the speed of the first bogie.