Two balls weighing 6 kg and 4 kg move on a horizontal surface at speeds of 8 m / s and 3 m / s.

Two balls weighing 6 kg and 4 kg move on a horizontal surface at speeds of 8 m / s and 3 m / s. With what speed they will move after an inelastic impact (as a whole), if: a) the first ball overtakes the second; b) Are the balls moving towards each other?

Given:

m1 = 6 kilograms – the mass of the first ball;

m2 = 4 kilograms – the mass of the second ball;

v1 = 8 m / s – speed of the first ball;

v2 = 3 m / s – speed of the second ball.

It is required to determine the speed of the balls v (m / s) after inelastic collision in the following cases:

a) the first ball will catch up with the second;

b) balls are moving towards each other.

In both cases, the solution will be in accordance with the law of conservation of momentum.

In case a):

The velocities of the balls before impact are directed in one direction, then:

m1 * v1 + m2 * v2 = v * (m1 + m2);

v = (m1 * v1 + m2 * v2) / (m1 + m2) = (6 * 8 + 4 * 3) / (6 + 4) = 60/10 = 6 m / s.

In case b):

The velocities of the balls are opposite to each other, then:

m1 * v1 – m2 * v2 = v * (m1 + m2);

v = (m1 * v1 – m2 * v2) / (m1 + m2) = (6 * 8 – 4 * 3) / (6 + 4) = 36/10 = 3.6 m / s.

Answer: the speed of the balls will be in case a) – 6 m / s, in case b) – 3.6 m / s.