A ball with a mass of 100 g moves at a speed of 3 m / s and collides poorly

A ball with a mass of 100 g moves at a speed of 3 m / s and collides poorly with a ball with a mass of 0.5 kg. find changes in the kinetic energy of the system.

Given:

m1 = 100 grams = 0.1 kilograms – the mass of the first ball;

v1 = 3 m / s – speed of the first ball before collision;

m2 = 0.5 kilograms – the mass of the second ball.

It is required to determine dE (Joule) – the change in the kinetic energy of the system.

Before the collision, the kinetic energy of the system will be equal to the kinetic energy of the first ball, since the second ball is at rest:

E1 = m1 * v1 ^ 2/2 = 0.1 * 3 ^ 2/2 = 0.1 * 9/2 = 0.9 / 2 = 0.45 Joule.

Let us find the speed of the system after an inelastic collision:

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

v2 = m1 * v1 / (m1 + m2) = 0.1 * 3 / (0.1 + 0.5) = 0.3 / 0.6 = 0.5 m / s.

Then the kinetic energy of the system after the collision will be equal to:

E2 = (m1 + m2) * v2 ^ 2/2 = (0.1 + 0.5) * 0.5 ^ 2/2 = 0.6 * 0.25 / 2 =

= 0.3 / 2 = 0.15 Joules.

Then the change in kinetic energy will be equal to:

dE = E1 – E2 = 0.45 – 0.15 = 0.3 Joule.

Answer: the change in kinetic energy is 0.3 Joule.