A steel ball flying horizontally at a speed of 12 m / s hits a stone cube on ice.

A steel ball flying horizontally at a speed of 12 m / s hits a stone cube on ice. What speed did the cube acquire if the ball bounced back at a speed of 8 m / s? The weights of the cube ball are 0.2 and 4 kg, respectively.

To determine the speed of a stone cube after hitting a steel ball, we apply the law of conservation of momentum (we will take the direction of the initial velocity vector of the ball as a positive direction): Ush) / mk.

The values of the variables: msh is the mass of the steel ball (mw = 0.2 kg); Vsh – the speed of the ball before collision (Vsh = 12 m / s); Ush – rebound speed (Ush = 8 m / s); mk is the mass of a stone cube (mk = 4 kg).

Calculation: Uk = (mw * Vsh + mw * Ush) / mk = (0.2 * 12 + 0.2 * 8) / 4 = 1 m / s.

Answer: The stone cube has acquired a speed of 1 m / s.